Mathematical induction and binomial theorem

Class 11 Mathematics, unit 7, book pages 175–194

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Mathematical induction is a way to prove that a statement is true for every positive integer: show it is true at the start (the basis step) and that each case passes to the next (the inductive step), like a line of falling dominoes. The binomial theorem then gives the expansion of $(a + b)^{n}$ for a positive integer $n$ , with the coefficients read from Pascal's triangle or calculated as $(r n)$ , and the binomial series extends the idea to other indices. The unit ends with uses of the theorem: unit digits, remainders, last digits, comparing large numbers and divisibility tests.

Learning outcomes (11)

Describe a mathematical argument, identify the base case, induction hypothesis and precise conclusion.
Apply the principle of mathematical induction to prove statements, identities, divisibility of numbers and summation formulae.
Evaluate and justify a conclusion, communicating a position clearly in an appropriate mathematical form in daily life.
State and apply the binomial theorem to expand expressions of the form $(a + b)^{n}$ , where $n$ is a positive integer.
Describe the binomial theorem as the expansion of binomial powers restricted to the set of natural numbers.
Calculate binomial coefficients using Pascal's triangle.
Expand using the binomial theorem, and use appropriate techniques to simplify the expression.
Find an approximate value using the binomial theorem. Applications of the binomial theorem.
Use the binomial theorem to find the remainder when a number to some large exponent is divided by a number.
Use the binomial theorem to find the last digit of a number, test the divisibility by a number and compare two large numbers.
Apply the concept of mathematical induction and the binomial theorem to real world problems such as (puzzles, domino effects, Pascal's triangle, economic forecasting, rankings, variables subletting).

7.1Mathematical inductionOften examined

To see the idea of mathematical induction, picture an infinite line of dominoes. Whenever one domino falls backward, it knocks the next one down as well. Now suppose the first domino is made to fall. What happens next? They all fall down (Fig. 7.1).

Figure 7.1: A line of dominoes. If the

k

th domino falls, it pushes the

(k + 1)

th, and the first domino is made to fall, so every domino falls (the

k

th domino and the push are the accent).

If the $k$ th domino falls backward, it will also push the $(k + 1)$ th domino backward.

To connect this picture with the principle of mathematical induction, consider the sentence "the $n$ th domino falls backward" and call it $P (n)$ . Two facts are known. First, for every $k \geq 1$ , if $P (k)$ is true (the $k$ th domino falls), then $P (k + 1)$ is also true (the $(k + 1)$ th domino falls). Second, $P (1)$ is true (the first domino falls).

Hence, by the principle of mathematical induction, $P (n)$ , that is, "the $n$ th domino falls backward", is true for every integer $n \geq 1$ .

How the book's domino picture matches induction
Domino idea	Mathematical statement
The first domino is made to fall	$P (1)$ is true (the starting case)
Any domino that falls knocks down the next	If $P (k)$ is true then $P (k + 1)$ is true, for every $k \geq 1$
Every domino falls	$P (n)$ is true for every integer $n \geq 1$

7.1.2Principle of mathematical inductionOften examined

A proof by mathematical induction shows that a statement $P (n)$ is true for every positive integer $n$ . It has two steps, and both must be done.

Basis step: Show that the statement $P (1)$ is true (the first domino falls).
Inductive step: Show that $P (k + 1)$ is true whenever $P (k)$ is true. This means: assume $P (k)$ is true for $n = k$ , and prove from it that $P (k + 1)$ is true.

Example 1

Use the method of mathematical induction to prove that

1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots + n (n + 1) = \frac{n ( n + 1 ) ( n + 2 )}{3}

for all positive integers

n

Write the proposition $P (n)$ .
$P (n) : 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots + n (n + 1) = \frac{n ( n + 1 ) ( n + 2 )}{3}$
Step 1 (basis step). Put $n = 1$ . The left side has only the first term, $1 \cdot 2$ . Compare it with the right side.
$LHS = 1 \cdot 2 = 2, RHS = \frac{1 ( 1 + 1 ) ( 1 + 2 )}{3} = \frac{6}{3} = 2$
Both sides are equal, so $P (1)$ is true.
Step 2 (inductive step). Assume $P (k)$ is true for $n = k$ .
$1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots + k (k + 1) = \frac{k ( k + 1 ) ( k + 2 )}{3}$
We must prove that $P (k + 1)$ is true. The next term of the series is $(k + 1) ((k + 1) + 1) = (k + 1) (k + 2)$ . Add it to both sides of the assumption.
$1 \cdot 2 + \dots + k (k + 1) + (k + 1) (k + 2) = \frac{k ( k + 1 ) ( k + 2 )}{3} + (k + 1) (k + 2)$
Take the common factor $(k + 1) (k + 2)$ out of the right side and add the fractions in the bracket.
$= (k + 1) (k + 2) [\frac{k}{3} + 1] = (k + 1) (k + 2) \cdot \frac{k + 3}{3}$
Write the result in the form of $P (n)$ with $n = k + 1$ .
$= \frac{( k + 1 ) ( ( k + 1 ) + 1 ) ( ( k + 1 ) + 2 )}{3}$
This is exactly $P (k + 1)$ . So $P (k + 1)$ is true whenever $P (k)$ is true. By the principle of mathematical induction, $P (n)$ is true for all positive integers $n$ .

1 \cdot 2 + 2 \cdot 3 + \dots + n (n + 1) = \frac{n ( n + 1 ) ( n + 2 )}{3} \forall n \geq 1

Example 2

Use the method of mathematical induction to prove that

(1 - \frac{1}{2}) (1 - \frac{1}{3}) (1 - \frac{1}{4}) \dots (1 - \frac{1}{n + 1}) = \frac{1}{n + 1}

for all positive integers

n

Write the proposition $P (n)$ .
$P (n) : (1 - \frac{1}{2}) (1 - \frac{1}{3}) \dots (1 - \frac{1}{n + 1}) = \frac{1}{n + 1}$
Step 1 (basis step). For $n = 1$ the product has one factor.
$1 - \frac{1}{2} = \frac{1}{1 + 1} \Rightarrow \frac{1}{2} = \frac{1}{2}$
So $P (1)$ is true.
Step 2 (inductive step). Assume $P (k)$ is true for $n = k$ .
$(1 - \frac{1}{2}) (1 - \frac{1}{3}) \dots (1 - \frac{1}{k + 1}) = \frac{1}{k + 1}$
The next factor is $1 - \frac{1}{( k + 1 ) + 1}$ . Multiply both sides of the assumption by it.
$(1 - \frac{1}{2}) \dots (1 - \frac{1}{k + 1}) (1 - \frac{1}{k + 2}) = \frac{1}{k + 1} (1 - \frac{1}{k + 2})$
Simplify the bracket on the right with a common denominator, then cancel $(k + 1)$ .
$= \frac{1}{k + 1} \cdot \frac{k + 2 - 1}{k + 2} = \frac{1}{k + 1} \cdot \frac{k + 1}{k + 2} = \frac{1}{k + 2}$
Since $k + 2 = (k + 1) + 1$ , this is the statement $P (k + 1)$ .
$\frac{1}{k + 2} = \frac{1}{( k + 1 ) + 1}$
So $P (k + 1)$ is true whenever $P (k)$ is true. By induction, $P (n)$ is true for all positive integers $n$ .

j = 2 \prod n + 1 (1 - \frac{1}{j}) = \frac{1}{n + 1} \forall n \geq 1

Example 3

Use the method of mathematical induction to show that

n^{2} - 3 n + 4

is an even number (divisible by

2

) for all positive integers

n

The proposition $P (n)$ is: $n^{2} - 3 n + 4$ is an even number.
Step 1 (basis step). For $n = 1$ :
$1^{2} - 3 (1) + 4 = 2$
$2$ is even, so $P (1)$ is true.
Step 2 (inductive step). Assume $P (k)$ is true: $k^{2} - 3 k + 4$ is even.
Expand $P (k + 1)$ , that is $(k + 1)^{2} - 3 (k + 1) + 4$ .
$(k + 1)^{2} - 3 (k + 1) + 4 = k^{2} + 2 k + 1 - 3 k - 3 + 4$
Regroup so that the assumed expression $k^{2} - 3 k + 4$ appears. The rest is $2 k + 1 - 3 = 2 k - 2$ .
$= (k^{2} - 3 k + 4) + (2 k + 1 - 3) = (k^{2} - 3 k + 4) + 2 (k - 1)$
The first bracket is even by the assumption, and $2 (k - 1)$ is even because it has the factor $2$ . The sum of two even numbers is even, so $(k + 1)^{2} - 3 (k + 1) + 4$ is even.
Hence $P (k + 1)$ is true whenever $P (k)$ is true. By induction, $n^{2} - 3 n + 4$ is even for all positive integers $n$ .

n^{2} - 3 n + 4 is even \forall n \geq 1

Example 4

Use the method of mathematical induction to show that

3^{n} > n^{2}

for all positive integers

n

The proposition $P (n)$ is $3^{n} > n^{2}$ .
Step 1 (basis step). For $n = 1$ :
$3^{1} > 1^{2} \Rightarrow 3 > 1$
So $P (1)$ is true.
Step 2 (inductive step). Assume $P (k)$ is true: $3^{k} > k^{2}$ .
Write $3^{k + 1}$ as $3 \cdot 3^{k} = 3^{k} + 3^{k} + 3^{k}$ and replace one $3^{k}$ by the smaller $k^{2}$ (assumption).
$3^{k + 1} = 3 \times 3^{k} = 3^{k} + 3^{k} + 3^{k} > 3^{k} + 3^{k} \Rightarrow 3^{k + 1} > k^{2} + 3^{k} (∵ 3^{k} > k^{2})$
Now use the fact $3^{k} > 2 k + 1$ for $k > 1$ (for $k = 2$ : $9 > 5$ , and it keeps holding because $3^{k}$ is multiplied by $3$ at each step while $2 k + 1$ grows by only $2$ ).
$3^{k + 1} > k^{2} + 2 k + 1$
Recognise the perfect square.
$3^{k + 1} > (k + 1)^{2}$
This is $P (k + 1)$ . So $P (k + 1)$ is true whenever $P (k)$ is true, and $P (n)$ holds for all positive integers $n$ .

3^{n} > n^{2} \forall n \geq 1

Example 5

Use the method of mathematical induction to show that

4 + 4 \cdot 6 + 4 \cdot 6^{2} + 4 \cdot 6^{3} + \dots + 4 \cdot 6^{n} = \frac{4 ( 6 ^{n + 1} - 1 )}{5}

for all positive integers

n

We prove the proposition $P (n)$ by mathematical induction.
$P (n) : 4 + 4 \cdot 6 + 4 \cdot 6^{2} + \dots + 4 \cdot 6^{n} = \frac{4 ( 6 ^{n + 1} - 1 )}{5}$
Step 1 (basis step). The book starts from $n = 0$ , where the sum is just the first term $4$ .
$\frac{4 ( 6 ^{0 + 1} - 1 )}{5} = \frac{4 \times 5}{5} = 4$
So $P (0)$ is true. (For $n = 1$ also: $4 + 24 = 28$ and $\frac{4 ( 36 - 1 )}{5} = 28$ .)
Step 2 (inductive step). Assume $P (k)$ is true for $n = k$ .
$4 + 4 \cdot 6 + 4 \cdot 6^{2} + \dots + 4 \cdot 6^{k} = \frac{4 ( 6 ^{k + 1} - 1 )}{5} (1)$
Add the next term $4 \cdot 6^{k + 1}$ to both sides of (1).
$4 + 4 \cdot 6 + \dots + 4 \cdot 6^{k} + 4 \cdot 6^{k + 1} = \frac{4 ( 6 ^{k + 1} - 1 )}{5} + 4 \cdot 6^{k + 1}$
Put the right side over $5$ and take out the factor $4$ .
$= \frac{4 ( 6 ^{k + 1} - 1 + 5 \cdot 6 ^{k + 1} )}{5}$
Combine $6^{k + 1} + 5 \cdot 6^{k + 1} = 6 \cdot 6^{k + 1} = 6^{k + 2}$ .
$= \frac{4 ( 6 \cdot 6 ^{k + 1} - 1 )}{5} = \frac{4 ( 6 ^{k + 2} - 1 )}{5}$
Since $k + 2 = (k + 1) + 1$ , this is $P (k + 1)$ . So $P (k + 1)$ is true whenever $P (k)$ is true, and $P (n)$ is true for all positive integers $n$ .

4 + 4 \cdot 6 + \dots + 4 \cdot 6^{n} = \frac{4 ( 6 ^{n + 1} - 1 )}{5}

Worked example: an inequality that starts at n = 5

Prove by mathematical induction that

2^{n} > n^{2}

for all integers

n \geq 5

. Why can the basis step not be taken at

n = 1

or at

n = 2

First test small values to see where the statement begins to hold.
$n = 1 : n = 2 : n = 3 : n = 4 : n = 5 : 2 > 1 true 4 > 4 false 8 > 9 false 16 > 16 false 32 > 25 true$
Step 1 (basis step). The statement is meant for $n \geq 5$ , so the basis step is $n = 5$ . It is true, as shown above. (Starting at $n = 1$ or $n = 2$ would not help: $n = 2, 3, 4$ fail, so the chain of dominoes cannot pass through them.)
$2^{5} = 32 > 25 = 5^{2}$
Step 2 (inductive step). Assume $P (k)$ for some integer $k \geq 5$ .
$2^{k} > k^{2}$
Multiply both sides by $2$ (a positive number, so the inequality keeps its direction). This gives $2^{k + 1}$ on the left.
$2^{k + 1} = 2 \cdot 2^{k} > 2 k^{2}$
We want to reach $(k + 1)^{2}$ , so compare $2 k^{2}$ with $(k + 1)^{2}$ by subtracting.
$2 k^{2} - (k + 1)^{2} = k^{2} - 2 k - 1 = (k - 1)^{2} - 2$
For $k \geq 5$ we have $(k - 1)^{2} \geq 16$ , so the difference is at least $14$ , which is positive. Hence $2 k^{2} > (k + 1)^{2}$ .
$2 k^{2} - (k + 1)^{2} \geq 16 - 2 = 14 > 0$
Join the two inequalities. This is $P (k + 1)$ .
$2^{k + 1} > 2 k^{2} > (k + 1)^{2}$
So $P (k + 1)$ is true whenever $P (k)$ is true for $k \geq 5$ . By the principle of mathematical induction, $2^{n} > n^{2}$ for all integers $n \geq 5$ .

x

y

(2, 4)

(4, 16)

n = 5

y = 2^{x}

y = x^{2}

The curves

y = 2^{x}

and

y = x^{2}

meet at

x = 2

and

x = 4

. For

x > 4

the exponential stays above the parabola, which is why the statement is proved from

n = 5

2^{n} > n^{2} \forall n \geq 5

7.2 and 7.2.1Pascal's triangle

Bi means two and nominal means terms. So a binomial is an algebraic expression of two terms, for example (x+y), \left(\dfrac1x-\dfrac12\right) and \left(x^2+\dfrac1x\right).

Often we need positive integral powers of a binomial, like the square, the cube or even higher powers. The higher the power, the longer the expansion. To handle such problems we use the binomial theorem. The general form of a binomial expression raised to a power is (a+b)^n, where n is a positive integer, and we can expand (a+b)^n by the binomial theorem. Another way to expand (a+b)^n is to use Pascal's triangle.

Pascal's triangle (7.2.1)

Write 1 and 1 in the first row. Each next row begins and ends with 1, and every other number is the sum of the two numbers just above it (the boxed additions in the book's figure: 1+1=2, 1+2=3, 3+1=4, and so on). This triangle of positive integers is known as Pascal's triangle.

Pascal's triangle up to the 6th row. Each row begins and ends with 1, and every other number is the sum of the two numbers above it (the sum 4+6=10 is the accent).

Now see how the triangle helps in the expansion of (a+b)^n. Observe some positive integral powers of (a+b).

The binomial coefficients of the first four powers
Power	Expansion	Coefficients
n=1	(a+b)^1=a+b	1\quad1
n=2	(a+b)^2=a^2+2ab+b^2	1\quad2\quad1
n=3	(a+b)^3=a^3+3a^2b+3ab^2+b^3	1\quad3\quad3\quad1
n=4	(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4	1\quad4\quad6\quad4\quad1

The binomial coefficients in these expansions are the same as the first four rows of Pascal's triangle. So we can find the binomial coefficients of (a+b)^n from the nth row of Pascal's triangle.

Also note from the expansions that the expansion starts with a^n; in each next term the exponent of a decreases by 1 and the exponent of b increases by 1; and the expansion ends with the term b^n.

Example 6

Expand (1+2x)^6 with the help of Pascal's triangle.

Compare with (a+b)^n.
a=1,\quad b=2x,\quad n=6
For the binomial coefficients we need the 6th row of Pascal's triangle. Build the rows one by one, adding neighbours.
\begin{aligned}&1\ \ 1\\&1\ \ 2\ \ 1\\&1\ \ 3\ \ 3\ \ 1\\&1\ \ 4\ \ 6\ \ 4\ \ 1\\&1\ \ 5\ \ 10\ \ 10\ \ 5\ \ 1\\&1\ \ 6\ \ 15\ \ 20\ \ 15\ \ 6\ \ 1\end{aligned}
Multiply each coefficient by a^{6-r}b^r, with the power of a=1 falling from 6 to 0 and the power of b=2x rising from 0 to 6.
(1+2x)^6=1(1)^6+6(1)^5(2x)+15(1)^4(2x)^2+20(1)^3(2x)^3+15(1)^2(2x)^4+6(1)^1(2x)^5+1(2x)^6
Work out the powers of 2x: (2x)^2=4x^2, (2x)^3=8x^3, (2x)^4=16x^4, (2x)^5=32x^5, (2x)^6=64x^6.
=1(1)+6(2x)+15(4x^2)+20(8x^3)+15(16x^4)+6(32x^5)+64x^6
Multiply the numbers in each term.
=1+12x+60x^2+160x^3+240x^4+192x^5+64x^6

(1+2x)^6=1+12x+60x^2+160x^3+240x^4+192x^5+64x^6

Example 7

Expand \left(2-\dfrac1x\right)^5 with the help of Pascal's triangle.

Write the subtraction as an addition of a negative term, so that the form is (a+b)^n.
\left(2-\dfrac1x\right)^5=\left[2+\left(-\dfrac1x\right)\right]^5,\qquad a=2,\ \ b=-\dfrac1x,\ \ n=5
For the binomial coefficients we need the 5th row of Pascal's triangle.
1\ \ 5\ \ 10\ \ 10\ \ 5\ \ 1
Write out the six terms with a=2 falling in power and b=-\dfrac1x rising in power.
=1(2)^5+5(2)^4\left(-\dfrac1x\right)+10(2)^3\left(-\dfrac1x\right)^2+10(2)^2\left(-\dfrac1x\right)^3+5(2)^1\left(-\dfrac1x\right)^4+1\left(-\dfrac1x\right)^5
Evaluate the powers. An odd power of -\dfrac1x is negative and an even power is positive.
=1(32)+5(16)\left(-\dfrac1x\right)+10(8)\left(\dfrac1{x^2}\right)+10(4)\left(-\dfrac1{x^3}\right)+5(2)\left(\dfrac1{x^4}\right)+1\left(-\dfrac1{x^5}\right)
Multiply the numbers.
=32-\dfrac{80}{x}+\dfrac{80}{x^2}-\dfrac{40}{x^3}+\dfrac{10}{x^4}-\dfrac{1}{x^5}

\left(2-\dfrac1x\right)^5=32-\dfrac{80}{x}+\dfrac{80}{x^2}-\dfrac{40}{x^3}+\dfrac{10}{x^4}-\dfrac{1}{x^5}

7.2.2Binomial theoremOften examined

The binomial theorem gives the expansion of (a+b)^n for any positive integer n without building Pascal's triangle row by row.

Statement. If a and b are any two real numbers and n is a positive integer, then

(a+b)^n=\dbinom{n}{0}a^nb^0+\dbinom{n}{1}a^{n-1}b^1+\dbinom{n}{2}a^{n-2}b^2+\cdots+\dbinom{n}{n-1}a^1b^{n-1}+\dbinom{n}{n}a^0b^n

Binomial theorem

Here \dbinom{n}{r}=\dfrac{n!}{r!\,(n-r)!} is the binomial coefficient (the same number as ^{n}C_{r} from Unit 6), and 0!=1.

Derivation Proof of the binomial theorem by mathematical induction

Aim. Prove that (a+b)^n=\sum_{r=0}^{n}\dbinom{n}{r}a^{n-r}b^r for every positive integer n.

Let P(n) be the statement of the theorem. We prove it by mathematical induction, so there are two steps.
P(n):\ (a+b)^n=\dbinom{n}{0}a^nb^0+\dbinom{n}{1}a^{n-1}b^1+\cdots+\dbinom{n}{n}a^0b^n
Step 1 (basis step). For n=1 the right side has two terms with \dbinom10=\dbinom11=1.
(a+b)^1=\dbinom{1}{0}a^1b^0+\dbinom{1}{1}a^{1-1}b^1=(1)a(1)+(1)(1)b=a+b
This is true, so P(1) is true.
Step 2 (inductive step). Assume P(k) is true for n=k.
(a+b)^k=\dbinom{k}{0}a^kb^0+\dbinom{k}{1}a^{k-1}b^1+\dbinom{k}{2}a^{k-2}b^2+\cdots+\dbinom{k}{k-1}ab^{k-1}+\dbinom{k}{k}a^0b^k
We must prove P(k+1). Multiply both sides of the assumption by (a+b), since (a+b)(a+b)^k=(a+b)^{k+1}. Multiply the long bracket by a and by b separately.
(a+b)^{k+1}=a\left[\dbinom{k}{0}a^kb^0+\cdots+\dbinom{k}{k}a^0b^k\right]+b\left[\dbinom{k}{0}a^kb^0+\cdots+\dbinom{k}{k}a^0b^k\right]
Multiplying by a raises each power of a by 1; multiplying by b raises each power of b by 1. Write the two lines one below the other.
\begin{aligned}(a+b)^{k+1}&=\left[\dbinom{k}{0}a^{k+1}b^0+\dbinom{k}{1}a^kb^1+\dbinom{k}{2}a^{k-1}b^2+\cdots+\dbinom{k}{k-1}a^2b^{k-1}+\dbinom{k}{k}a^1b^k\right]\\&\quad+\left[\dbinom{k}{0}a^kb^1+\dbinom{k}{1}a^{k-1}b^2+\dbinom{k}{2}a^{k-2}b^3+\cdots+\dbinom{k}{k-1}a^1b^k+\dbinom{k}{k}a^0b^{k+1}\right]\end{aligned}
Collect the like terms, that is, the terms with the same powers of a and b. The term a^{k+1}b^0 and the term a^0b^{k+1} occur once; every middle power a^{k+1-r}b^r occurs in both lines.
(a+b)^{k+1}=\dbinom{k}{0}a^{k+1}b^0+\left[\dbinom{k}{1}+\dbinom{k}{0}\right]a^kb+\left[\dbinom{k}{2}+\dbinom{k}{1}\right]a^{k-1}b^2+\cdots+\left[\dbinom{k}{k-1}+\dbinom{k}{k}\right]ab^k+\dbinom{k}{k}a^0b^{k+1}
Now use three facts about binomial coefficients: the end coefficients are all 1, and the sum of neighbouring coefficients follows Pascal's rule for 0\le r\le k.
\begin{aligned}\dbinom{k}{0}&=1=\dbinom{k+1}{0}\\ \dbinom{k}{k}&=1=\dbinom{k+1}{k+1}\\ \dbinom{k}{r-1}+\dbinom{k}{r}&=\dbinom{k+1}{r}\end{aligned}
Pascal's rule itself comes from putting the two fractions over a common denominator, because r!=r\,(r-1)! and (k-r+1)!=(k-r+1)(k-r)!.
\dfrac{k!}{(r-1)!\,(k-r+1)!}+\dfrac{k!}{r!\,(k-r)!}=\dfrac{k!\,[\,r+(k-r+1)\,]}{r!\,(k-r+1)!}=\dfrac{(k+1)!}{r!\,(k+1-r)!}=\dbinom{k+1}{r}
Replace each bracket by the single coefficient.
(a+b)^{k+1}=\dbinom{k+1}{0}a^{k+1}b^0+\dbinom{k+1}{1}a^kb+\dbinom{k+1}{2}a^{k-1}b^2+\cdots+\dbinom{k+1}{k}ab^k+\dbinom{k+1}{k+1}a^0b^{k+1}
This is exactly P(k+1). So the theorem is true for n=k+1 whenever it is true for n=k. By mathematical induction it is true for every positive integer n.

Result(a+b)^n=\sum_{r=0}^{n}\dbinom{n}{r}a^{n-r}b^r

The coefficients of the expansion are the binomial coefficients \dbinom{n}{r}, which are the numbers in the nth row of Pascal's triangle. Pascal's rule used in the last step is exactly the rule that each number of the triangle is the sum of the two above it.

Some properties of the binomial expansion

The number of terms in the expansion of (a+b)^n is one more than the index n, that is, n+1 (the values r=0,1,\dots,n).
The sum of the exponents of a and b in each term of the expansion of (a+b)^n is n.
The coefficients of the terms equidistant from the beginning and the end are the same, because \dbinom{n}{r}=\dbinom{n}{n-r}.
If n is even, there is an odd number of terms in the expansion of (a+b)^n, so there is one middle term. It is the \left(\dfrac n2+1\right)th term, which is the \left(\dfrac{n+2}{2}\right)th term.
If n is odd, there is an even number of terms in the expansion of (a+b)^n, so there are two middle terms. They are the \left(\dfrac{n+1}{2}\right)th and the \left(\dfrac{n+3}{2}\right)th terms.
In the expansion of (a+b)^n the exponent of a is n and the exponent of b is zero in the first term. In each next term the exponent of a decreases by 1 and the exponent of b increases by 1. In the last term the exponent of a becomes zero and the exponent of b reaches n.
Any particular (r+1)th term from the beginning, also known as the general term, in the expansion of (a+b)^n is given by T_{r+1}=\dbinom{n}{r}a^{n-r}b^r.
A term which is at the rth position from the end in the expansion of (a+b)^n is at the (n+2-r)th position from the beginning.

T_{r+1}=\dbinom{n}{r}a^{n-r}b^{r}

General term

Example 8

Expand (x-y)^5 using the binomial theorem.

Write the difference as a sum so that the form is (a+b)^n with b=-y.
(x-y)^5=[x+(-y)]^5
Apply the binomial theorem with n=5.
=\dbinom50x^5(-y)^0+\dbinom51x^4(-y)^1+\dbinom52x^3(-y)^2+\dbinom53x^2(-y)^3+\dbinom54x^1(-y)^4+\dbinom55x^0(-y)^5\qquad(1)
Find the binomial coefficients from \dbinom nr=\dfrac{n!}{r!(n-r)!}.
\begin{aligned}\dbinom50&=\dfrac{5!}{0!\,5!}=1\\ \dbinom51&=\dfrac{5!}{1!\,4!}=\dfrac{5\times4!}{4!}=5\\ \dbinom52&=\dfrac{5!}{2!\,3!}=\dfrac{5\times4\times3!}{2\times1\times3!}=10\end{aligned}
The rest follow, since \dbinom{5}{r}=\dbinom{5}{5-r}.
\dbinom53=\dfrac{5!}{3!\,2!}=10,\quad \dbinom54=\dfrac{5!}{4!\,1!}=5,\quad \dbinom55=\dfrac{5!}{5!\,0!}=1
Substitute the values in (1).
=(1)x^5(-y)^0+(5)x^4(-y)^1+(10)x^3(-y)^2+(10)x^2(-y)^3+(5)x^1(-y)^4+(1)x^0(-y)^5
Simplify the signs: an odd power of -y is negative, an even power is positive.
=x^5-5x^4y+10x^3y^2-10x^2y^3+5xy^4-y^5

(x-y)^5=x^5-5x^4y+10x^3y^2-10x^2y^3+5xy^4-y^5

Example 9

Find the constant term in the expansion of \left(x+\dfrac2x\right)^{10}.

The constant term is the term that is independent of x. Here the binomial is (a+b)^n with:
a=x,\quad b=\dfrac2x,\quad n=10
Write the general term T_{r+1}=\dbinom nr a^{n-r}b^r with these values.
T_{r+1}=\dbinom{10}{r}x^{10-r}\left(\dfrac2x\right)^r=\dbinom{10}{r}x^{10-r}\dfrac{2^r}{x^r}
Combine the powers of x: x^{10-r}\cdot x^{-r}=x^{10-2r}.
T_{r+1}=\dbinom{10}{r}x^{10-2r}\,2^r\qquad(1)
The term is independent of x when the exponent of x is zero.
10-2r=0\ \Rightarrow\ 2r=10\ \Rightarrow\ r=5
Put r=5 in (1). This is the 6th term.
T_{5+1}=\dbinom{10}{5}x^02^5=\dfrac{10!}{5!\,5!}\cdot1\cdot32=\dfrac{10\cdot9\cdot8\cdot7\cdot6\cdot5!}{5\cdot4\cdot3\cdot2\cdot1\cdot5!}\,(32)=252\times32
Multiply.
T_6=8064

\text{constant term}=T_6=8064

Example 10

Find the 3rd term from the end in the expansion of \left(2-\dfrac{5}{\sqrt x}\right)^5.

Compare with (a+b)^n.
a=2,\quad b=\dfrac{-5}{\sqrt x},\quad n=5
The rth term from the end is the (n+2-r)th term from the beginning. For r=3:
5+2-3=4\ \Rightarrow\ \text{the 4th term from the beginning}
For the 4th term, r=3 in T_{r+1}=\dbinom nr a^{n-r}b^r. Substitute.
T_{3+1}=\dbinom53 2^{5-3}\left(\dfrac{-5}{\sqrt x}\right)^3=\dfrac{5!}{3!\,2!}\cdot2^2\cdot\dfrac{-125}{x^{3/2}}
Evaluate: \dfrac{5!}{3!\,2!}=\dfrac{5\cdot4}{2}=10, 2^2=4, and (-5)^3=-125.
T_4=10(4)\left(\dfrac{-125}{x^{3/2}}\right)=10\left(-500\,x^{-3/2}\right)
Multiply.
T_4=-5000\,x^{-3/2}

\text{3rd term from the end}=-5000\,x^{-3/2}=-\dfrac{5000}{x^{3/2}}

Worked example: a required term and a missing constant term

In the expansion of \left(x^2-\dfrac2x\right)^8, (i) find the term containing x^4; (ii) show that there is no term independent of x.

Here a=x^2, b=-\dfrac2x and n=8. Write the general term T_{r+1}=\dbinom nr a^{n-r}b^r.
T_{r+1}=\dbinom8r(x^2)^{8-r}\left(-\dfrac2x\right)^r
Collect the numbers and the powers of x separately. Powers of x are x^{16-2r}\cdot x^{-r}.
T_{r+1}=\dbinom8r(-2)^r\,x^{16-3r}
(i) For x^4 the power must be 4. Solve for r.
16-3r=4\ \Rightarrow\ r=4
Put r=4 (this is the 5^{th} term). Remember (-2)^4=+16.
T_5=\dbinom84(-2)^4x^4=70\cdot16\,x^4=1120\,x^4
(ii) A term independent of x needs the power 0.
16-3r=0\ \Rightarrow\ r=\dfrac{16}{3}
But r must be a whole number from 0 to 8, and \dfrac{16}3 is not. So no term has power 0. As a check, the powers that occur are 16,13,10,7,4,1,-2,-5,-8, and 0 is not among them.

T_5=1120\,x^4;\ \text{no term independent of }x

Example 11

Find the remainder when 7^{101} is divided by 25.

Write the large power with a base close to a multiple of 25. Since 7^2=49=50-1, split 7^{101}=7\cdot7^{100}=7\,(7^2)^{50}.
\dfrac{7^{101}}{25}=\dfrac{7\cdot7^{100}}{25}=\dfrac{7\,(7^2)^{50}}{25}=\dfrac{7\,(49)^{50}}{25}=\dfrac{7}{25}(50-1)^{50}
Expand (50-1)^{50} by the binomial theorem.
(50-1)^{50}=\dbinom{50}{0}(50)^{50}(-1)^0+\dbinom{50}{1}(50)^{49}(-1)^1+\cdots+\dbinom{50}{49}(50)^1(-1)^{49}+\dbinom{50}{50}(50)^0(-1)^{50}
Every term except the last has a factor 50, so together they make a multiple of 50. The last term is (-1)^{50}=1. Take the first 50 terms together as 50M, where M is an integer.
(50-1)^{50}=50M+1
Multiply by \dfrac{7}{25} and separate the integer part.
\dfrac{7}{25}(50M+1)=\dfrac{350M}{25}+\dfrac{7}{25}=14M+\dfrac{7}{25}
14M is an integer, so the fractional part is \dfrac{7}{25}, and the numerator is the remainder.

\text{remainder}=7

Example (unnumbered in the book)

The fourth term in the expansion of \left(ax+\dfrac1x\right)^n is \dfrac52. Find the values of a and n.

The general term is T_{r+1}=\dbinom nr a^{n-r}b^r with first term ax and second term \dfrac1x.
T_{r+1}=\dbinom nr (ax)^{n-r}\left(\dfrac1x\right)^r
The fourth term has r=3.
T_{3+1}=\dbinom n3(ax)^{n-3}\left(\dfrac1x\right)^3=\dbinom n3a^{n-3}x^{n-3}x^{-3}=\dbinom n3a^{n-3}x^{n-6}
The fourth term is given to be \dfrac52.
\dbinom n3a^{n-3}x^{n-6}=\dfrac52\qquad(1)
The right side of (1) does not contain x, so the exponent of x on the left must be zero.
n-6=0\ \Rightarrow\ n=6
Put n=6 in (1), where x^0=1 and \dbinom63=\dfrac{6!}{3!\,3!}=\dfrac{6\cdot5\cdot4}{3\cdot2\cdot1}=20.
\dbinom63a^{3}=\dfrac52\ \Rightarrow\ 20a^3=\dfrac52
Solve for a^3 and take the cube root.
a^3=\dfrac{5}{40}=\dfrac18=\left(\dfrac12\right)^3\ \Rightarrow\ a=\dfrac12

n=6,\qquad a=\dfrac12

7.2.3Use of Pascal's triangle and binomial theorem in real world problems

Pascal's triangle and the binomial theorem are used in real-world problems such as cryptography, calculating the number of matches played in a tournament where n teams are playing, and calculating the possible number of protein structures and DNA sequences.

Example 12

Use Pascal's triangle to find the possible number of heads when three coins are tossed simultaneously.

List all the possible results when three coins are tossed together.
HHH\quad HHT\quad HTH\quad HTT\quad THH\quad THT\quad TTH\quad TTT
Count the results by the number of heads.
\begin{aligned}0\text{ heads}&=1\text{ result (TTT)}\\1\text{ head}&=3\text{ results (HTT, THT, TTH)}\\2\text{ heads}&=3\text{ results (HHT, HTH, THH)}\\3\text{ heads}&=1\text{ result (HHH)}\end{aligned}
Now take the 3rd row of Pascal's triangle, which is the row of coefficients of (a+b)^3.
1\quad3\quad3\quad1
The numbers of this row are exactly the numbers of results with 0,1,2,3 heads, in that order (figure below).

The third row of Pascal's triangle, 1,3,3,1, gives the number of results with 0, 1, 2 and 3 heads when three coins are tossed (the row and the arrows are the accent).

0\text{ heads}:1,\quad 1\text{ head}:3,\quad 2\text{ heads}:3,\quad 3\text{ heads}:1

7.3Binomial seriesOften examined

Expansion of (1+x)^n when n is a positive integer (7.3.1)

When the index n of (1+x)^n is a positive integer, we can use the binomial theorem directly. Put a=1 and b=x.

Derivation Binomial series for a positive integer index

Aim. Expand (1+x)^n in powers of x when n is a positive integer.

Apply the binomial theorem with a=1 and b=x. Every power of 1 is 1.
(1+x)^n=\dbinom n0(1)^nx^0+\dbinom n1(1)^{n-1}x^1+\dbinom n2(1)^{n-2}x^2+\cdots+\dbinom nn(1)^0x^n
Write each coefficient as \dbinom nr=\dfrac{n!}{r!\,(n-r)!}.
=\dfrac{n!}{0!\,(n-0)!}(1)(1)+\dfrac{n!}{1!\,(n-1)!}(1)x+\dfrac{n!}{2!\,(n-2)!}(1)x^2+\cdots+\dfrac{n!}{n!\,0!}x^n
Cancel the common factorials: \dfrac{n!}{(n-1)!}=n and \dfrac{n!}{(n-2)!}=n(n-1), and so on.
=\dfrac{n!}{n!}(1)+\dfrac{n\,(n-1)!}{(n-1)!}x+\dfrac{n(n-1)(n-2)!}{2!\,(n-2)!}x^2+\cdots+\dfrac{n!}{n!}x^n

Result(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\cdots+x^n

The series on the right is terminating and has (n+1) terms.

Expansion of (1+x)^n when n is not a positive integer (7.3.2)

When n is not a positive integer (for example a negative integer or a fraction), the expansion of (1+x)^n is non-terminating, that is, it goes on without end:

(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\dfrac{n(n-1)(n-2)}{3!}x^3+\cdots+\infty

The above series is convergent if |x|<1, that is, -1<x<1. A convergent series has a finite sum; otherwise the series is divergent. We focus only on those expansions of (1+x)^n which are convergent.

Binomial series: The series 1+nx+\dfrac{n(n-1)}{2!}x^2+\dfrac{n(n-1)(n-2)}{3!}x^3+\cdots is known as the binomial series.

The general term of the binomial series is given by

T_{r+1}=\dfrac{n(n-1)(n-2)\cdots(n-r+1)}{r!}\,x^{r}

General term of the binomial series

Worked example: approximating an expression for small x

If x is so small that its square and higher powers may be neglected, show that \dfrac{(8+x)^{1/3}}{(1-2x)^2}\approx2+\dfrac{97}{12}x.

The first bracket is not of the form (1+u)^n. Take 8 out: (8+x)^{1/3}=8^{1/3}\left(1+\dfrac x8\right)^{1/3} and 8^{1/3}=2.
(8+x)^{1/3}=2\left(1+\dfrac x8\right)^{1/3}
Expand with the binomial series, n=\dfrac13 and u=\dfrac x8, and keep terms up to x only.
2\left(1+\dfrac13\cdot\dfrac x8\right)=2\left(1+\dfrac x{24}\right)=2+\dfrac x{12}
Expand the denominator as (1+u)^n with n=-2 and u=-2x. Keep terms up to x.
(1-2x)^{-2}\approx1+(-2)(-2x)=1+4x
Multiply the two approximations.
\left(2+\dfrac x{12}\right)(1+4x)=2+8x+\dfrac x{12}+\dfrac{x^2}{3}
Neglect the term in x^2, as the question allows, and add the terms in x.
8x+\dfrac x{12}=\dfrac{96x+x}{12}=\dfrac{97}{12}x
Check with x=0.001: the expression is 2.008108 and 2+\dfrac{97}{12}(0.001)=2.008083, which agree to four decimal places.

\dfrac{(8+x)^{1/3}}{(1-2x)^2}\approx2+\dfrac{97}{12}x

Applications of binomial theorem

The binomial theorem has a wide range of applications in mathematics, like finding the remainder, finding the digits of a number, and so on. The most common applications of the binomial theorem are the following, each explained in its own section with examples.

Finding the last or unit place digit of an exponential number.
Finding the remainder using the binomial theorem.
Finding the digits of a number.
Finding the relation between two numbers (which is larger).
The divisibility test.

Finding the last or unit place digit of an exponential number

Consider the table below. The numbers (their unit digits 0 to 9) are written in the first column and their exponents in the first row. The table shows only the unit place digits of the powers.

Unit place digit of a number raised to a power
Number, power	1	2	3	4	5	6	7	8	9
0	0	0	0	0	0	0	0	0	0
1	1	1	1	1	1	1	1	1	1
2	2	4	8	6	2	4	8	6	2
3	3	9	7	1	3	9	7	1	3
4	4	6	4	6	4	6	4	6	4
5	5	5	5	5	5	5	5	5	5
6	6	6	6	6	6	6	6	6	6
7	7	9	3	1	7	9	3	1	7
8	8	4	2	6	8	4	2	6	8
9	9	1	9	1	9	1	9	1	9

From the table, the unit place digits of the powers of 2 are 2^1=2, 2^2=4, 2^3=8, 2^4=16 and 2^5=32. The digit 2 has come back at the fifth power, so the unit digits repeat after four powers. Therefore the cyclicity of 2 is 5-1=4.

In the same way the cyclicity of 3, 7 and 8 is also 4. For 1, 5 and 6 the unit digit remains the same for all exponents. For 4 the unit digit is either 4 or 6, and for 9 it is 9 or 1 only.

Hence, if the exponent is of the form 4n, 4n+1, 4n+2 or 4n+3, we can easily find the unit place digit of the power of any number.

Example 13

Find the unit digit of (i) 17^{203} (ii) 29^{26} (iii) 36^{307}.

(i) 17^{203}

The unit digit of 17^{203} depends only on the unit digit 7 of the base, whose cyclicity is 4. Divide the exponent by 4.
203=4\times50+3
The remainder is 3, so the unit digit is that of 7^3.
7^3=343

Hence the unit place digit of 17^{203} is 3.

(ii) 29^{26}

The unit digit of the base is 9. From the table, the powers of 9 end only in 9 and 1. Divide the exponent by 4 in the same way and note the remainder.
26=4\times6+2
So the unit digit is that of 9^2.
9^2=81

Hence the unit place digit of 29^{26} is 1.

(iii) 36^{307}

The unit digit of the base is 6, which always remains 6 at the unit place for every power. So 36^{307} has 6 at the unit place.

\text{(i) }3,\quad\text{(ii) }1,\quad\text{(iii) }6

Finding remainder using binomial theorem

The method is explained with the help of examples. To find the remainder when a large power is divided by n, rewrite the power as (n\cdot m\pm1)^{k} or a multiple of it, expand with the binomial theorem, and note that all terms except the last are multiples of n. The remainder comes from the last term.

Example 14

Find the remainder when 7^{103} is divided by 25.

Make the base 50\pm1: since 7^2=49, write 7^{103}=7\cdot7^{102}=7\,(7^2)^{51}.
\dfrac{7^{103}}{25}=\dfrac{7\,(49)^{51}}{25}=\dfrac{7\,(50-1)^{51}}{25}
Expand (50-1)^{51} by the binomial theorem. Every term except the last, (-1)^{51}=-1, has the factor 50, which is a multiple of 25. So (50-1)^{51}=25k-1 for some integer k.
(50-1)^{51}=25k-1
Substitute and multiply out.
\dfrac{7\,(25k-1)}{25}=\dfrac{175k-7}{25}
Add and subtract 25 in the numerator so that the part divisible by 25 shows.
=\dfrac{175k-25+25-7}{25}=\dfrac{25(7k-1)+18}{25}=(7k-1)+\dfrac{18}{25}
The first part is an integer, so the numerator of the fractional part is the remainder.

\text{remainder}=18

Example 15

If the fractional part of the number \dfrac{2^{403}}{15} is \dfrac{k}{15}, find k.

Write 2^{403} with the power 2^4=16=15+1. Since 403=3+4\times100:
\dfrac{2^{403}}{15}=\dfrac{2^3\,(2^4)^{100}}{15}=\dfrac{8}{15}(16)^{100}=\dfrac{8}{15}(15+1)^{100}
Expand (15+1)^{100}. Every term except the last, 1^{100}=1, has the factor 15, so (15+1)^{100}=15\lambda+1 for an integer \lambda.
\dfrac{8}{15}(15\lambda+1)=8\lambda+\dfrac{8}{15}
8\lambda is an integer, so the fractional part is \dfrac{8}{15}. Compare with \dfrac{k}{15}.

k=8

Worked example: a remainder from a number just below a multiple

Find the remainder when 8^{77} is divided by 9.

Write the base as a multiple of the divisor plus or minus a small number. Here 8 is just below 9.
8^{77}=(9-1)^{77}
Expand by the binomial theorem. Every term except the last contains a positive power of 9.
(9-1)^{77}=\dbinom{77}{0}9^{77}-\dbinom{77}{1}9^{76}+\cdots+\dbinom{77}{76}9(-1)^{76}+(-1)^{77}
Take 9 out of all terms except the last. The last term is (-1)^{77}=-1 because 77 is odd.
8^{77}=9m-1
A remainder cannot be negative, so borrow one 9: write 9m-1=9(m-1)+8.
8^{77}=9(m-1)+8
The number is a multiple of 9 plus 8, and 0\le8<9, so the remainder is 8.

\text{remainder}=8

Example 16

Find the unit digit of (i) 17^{203} (ii) 29^{26} (iii) 36^{307}.

The book prints this question a second time, without a solution, after Example 15. It is the same question as Example 13, so the same method is used here in short form.

(i) The unit digit of the base is 7 and the cyclicity is 4. Write 203=4\times50+3, so the digit is that of 7^3=343, which is 3.
(ii) The unit digit of the base is 9. Write 26=4\times6+2, so the digit is that of 9^2=81, which is 1.
(iii) The unit digit of the base is 6, and every power of a number ending in 6 ends in 6. The digit is 6.

\text{(i) }3,\quad\text{(ii) }1,\quad\text{(iii) }6

Finding digits of a number

The method is explained with the help of an example. To find the last two digits of a number, write it as (100m\pm r)^k or as a binomial whose terms from a certain point are all multiples of 100. Then only the first few terms of the expansion can change the last two digits.

Example 17

Find the last two digits of the number (13)^{10}.

Write the number with a base near a multiple of 100. Since 13^2=169:
13^{10}=(13^2)^5=(169)^5=(170-1)^5
Expand by the binomial theorem.
(170-1)^5=\dbinom50(170)^5-\dbinom51(170)^4+\dbinom52(170)^3-\dbinom53(170)^2+\dbinom54(170)-\dbinom55
The terms with 170^5,170^4,170^3,170^2 are all multiples of 100 (for example 170^2=28900). So they do not change the last two digits. Group them as 100k.
=100k+\dbinom54(170)-1
The remaining terms are \dbinom54=5 times 170, and -1.
5(170)-1=850-1=849
So the number equals 100k+849=100(k+8)+49. The last two digits are those of 49.

\text{last two digits}=49

Worked example: last two digits of a power

Find the last two digits of 7^{20}.

Find a small power of 7 that is 1 more than a multiple of 100.
7^2=49,\quad 7^4=2401=2400+1
Write the number as a power of 7^4.
7^{20}=(7^4)^5=(2400+1)^5
Expand by the binomial theorem. Each term except the last has a factor 2400, which is a multiple of 100.
(2400+1)^5=2400^5+5\cdot2400^4+\cdots+\dbinom54\cdot2400+1
So the number is a multiple of 100 plus 1.
7^{20}=100m+1
A multiple of 100 ends in 00. Adding 1 gives last two digits 01. Write the zero: the answer is 01, not 1.

\text{last two digits}=01

Relation between two numbers

The method is explained with the help of an example. To decide which of two large numbers is greater, write both as binomials with the same base and compare the expansions term by term.

Example 18

Which one is greater, 99^{50}+100^{50} or 101^{50}?

Write 101^{50} and 99^{50} as binomials with base 100 and expand.
101^{50}=(100+1)^{50}=100^{50}+50\cdot100^{49}+\dbinom{50}{2}100^{48}+\dbinom{50}{3}100^{47}+\cdots
The expansion of 99^{50} has the same terms, but the terms with an odd power of -1 are negative.
99^{50}=(100-1)^{50}=100^{50}-50\cdot100^{49}+\dbinom{50}{2}100^{48}-\dbinom{50}{3}100^{47}+\cdots
Subtract. The terms with even r cancel and the terms with odd r are doubled. Here \dbinom{50}{3}=\dfrac{50\cdot49\cdot48}{6}=25\cdot49\cdot16.
101^{50}-99^{50}=2\left[50\cdot100^{49}+25\cdot49\cdot16\cdot100^{47}+\cdots\right]
Multiply the first term by 2: 2\cdot50\cdot100^{49}=100\cdot100^{49}=100^{50}. The other terms are positive.
=100^{50}+50\cdot49\cdot16\cdot100^{47}+\cdots>100^{50}
Therefore the difference is greater than 100^{50}. Move 99^{50} to the other side.
101^{50}-99^{50}>100^{50}\ \Rightarrow\ 101^{50}>100^{50}+99^{50}

101^{50}\ \text{is greater than}\ 99^{50}+100^{50}

Divisibility test

The divisibility test means that we want to check whether a number is completely divisible by another number, that is, when a number is divided by another number the remainder is zero. If a number is too large to divide directly, we check the divisibility with the binomial theorem. The method is explained in the following example.

Example 19

Show that 11^9+9^{11} is divisible by 10.

Write 11=10+1 and 9=10-1.
11^9+9^{11}=(10+1)^9+(10-1)^{11}
Expand both brackets. Every term has a power of 10, except the last term of each expansion.
(10+1)^9=\dbinom90\,10^9+\dbinom91\,10^8+\cdots+\dbinom98\,10+\dbinom99
In the second expansion the signs alternate; its last term is \dbinom{11}{11}(-1)^{11}=-1.
(10-1)^{11}=\dbinom{11}0\,10^{11}-\dbinom{11}1\,10^{10}+\cdots+\dbinom{11}{10}\,10-1
In the sum, the last term \dbinom99=1 of the first expansion cancels the last term -1 of the second. Every other term has the factor 10, which can be taken out.
=10\left[\dbinom90\,10^8+\dbinom91\,10^7+\cdots+\dbinom98+\dbinom{11}0\,10^{10}-\dbinom{11}1\,10^9+\cdots+\dbinom{11}{10}\right]
The bracket is an integer k, so the sum is 10k, which is divisible by 10.

11^9+9^{11}=10k

Formula sheet

Principle of mathematical induction

P(1)\ \text{true},\qquad P(k)\Rightarrow P(k+1)\ \Longrightarrow\ P(n)\ \text{true for all}\ n\ge1

P(n): the statement; basis step: P(1) is true; inductive step: P(k+1) is true whenever P(k) is true

Pascal's rule

\dbinom{k}{r-1}+\dbinom{k}{r}=\dbinom{k+1}{r},\qquad 0\le r\le k

Each number of Pascal's triangle is the sum of the two numbers above it

Binomial coefficient

\dbinom{n}{r}=\dfrac{n!}{r!\,(n-r)!}

n: positive integer; r=0,1,\ldots,n; 0!=1

Binomial theorem

(a+b)^n=\dbinom{n}{0}a^nb^0+\dbinom{n}{1}a^{n-1}b^1+\cdots+\dbinom{n}{n}a^0b^n

a,b: real numbers; n: positive integer

General term of (a+b)^n

T_{r+1}=\dbinom{n}{r}a^{n-r}b^{r}

T_{r+1}: the (r+1)th term from the beginning, r=0,1,\ldots,n

Number of terms

n+1

The expansion of (a+b)^n has n+1 terms; the exponents of a and b in every term add up to n

Middle term, n even

\left(\dfrac{n}{2}+1\right)\text{th term}

One middle term when n is even

Middle terms, n odd

\left(\dfrac{n+1}{2}\right)\text{th and }\left(\dfrac{n+3}{2}\right)\text{th terms}

Two middle terms when n is odd

Term from the end

r\text{th term from the end}=(n+2-r)\text{th term from the beginning}

n: index of (a+b)^n

Binomial series, n a positive integer

(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\cdots+x^n

Terminating series of n+1 terms

Binomial series, n not a positive integer

(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\dfrac{n(n-1)(n-2)}{3!}x^3+\cdots

Non-terminating; convergent for |x|<1

General term of the binomial series

T_{r+1}=\dfrac{n(n-1)(n-2)\cdots(n-r+1)}{r!}\,x^{r}

r: one less than the term number

Key terms

Mathematical induction: A method of proving that a proposition P(n) is true for all positive integers n, using a basis step and an inductive step.
Basis step: The first step of an induction proof: show that P(1) is true (the first domino falls).
Inductive step: The second step: show that P(k+1) is true whenever P(k) is true.
Induction hypothesis: The assumption that P(k) is true for n=k, made in the inductive step.
Binomial: An algebraic expression of two terms, for example x+y or x^2+\dfrac1x.
Pascal's triangle: A triangle of positive integers in which each row starts and ends with 1 and every other number is the sum of the two numbers above it. Its nth row gives the coefficients of (a+b)^n.
Binomial theorem: The formula (a+b)^n=\sum_{r=0}^{n}\dbinom nr a^{n-r}b^r for a positive integer n.
Binomial coefficient: The number \dbinom nr=\dfrac{n!}{r!(n-r)!} that multiplies a^{n-r}b^r in the expansion of (a+b)^n.
General term: The (r+1)th term from the beginning of the expansion of (a+b)^n, T_{r+1}=\dbinom nr a^{n-r}b^r.
Middle term: The term at the centre of the expansion: one term when n is even, two terms when n is odd.
Binomial series: The series 1+nx+\dfrac{n(n-1)}{2!}x^2+\dfrac{n(n-1)(n-2)}{3!}x^3+\cdots, the expansion of (1+x)^n.
Convergent series: A series that has a finite sum. The binomial series is convergent when |x|<1.
Cyclicity: The number of powers after which the unit digit of the powers of a number repeats; for example the cyclicity of 2 is 5-1=4.
Divisibility test: A check whether a number is completely divisible by another, that is, whether the remainder is zero.

Exercise 7.1p. 179–180

Proof by mathematical induction: show P(n) for the first allowed value (basis step), assume P(k) (inductive hypothesis), then prove P(k+1) (inductive step). Questions 1 to 19 are identities, 20 to 24 are divisibility, 25 to 32 are inequalities. The book's answer key has no entry for this exercise because every question is a proof.

12026 Annual I, Set D, Q.2 (xii) OR

Prove by mathematical induction: 1+2+3+\cdots+n=\dfrac{n(n+1)}{2}\qquad\forall\,n\ge1.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
1+2+3+\cdots+n=\dfrac{n(n+1)}{2}
Basis step. Put n=1 in both sides and compare.
\text{LHS}=1,\qquad \text{RHS}=\dfrac{1(1+1)}{2}=1\;\Rightarrow\;P(1)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge 1.
1+2+3+\cdots+k=\dfrac{k(k+1)}{2}
To prove. P(k+1), that is, the same statement with k+1 in place of n.
1+2+3+\cdots+k+(k+1)=\dfrac{(k+1)(k+2)}{2}
Add (k+1) to both sides of P(k). The left side is now the left side of P(k+1).
1+2+\cdots+k+(k+1)=\dfrac{k(k+1)}{2}+(k+1)
Take the common factor (k+1) out of the right side and add inside the bracket.
=(k+1)\left(\dfrac{k}{2}+1\right)=(k+1)\cdot\dfrac{k+2}{2}
This is exactly the right side of P(k+1).
=\dfrac{(k+1)\big((k+1)+1\big)}{2}
So P(k+1) is true whenever P(k) is true. P(1) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge 1.

22025 Annual I, Set D, Q.2 (ix)

Prove by mathematical induction: 1^2+2^2+3^2+\cdots+n^2=\dfrac{n(n+1)(2n+1)}{6}\qquad\forall\,n\ge1.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
1^2+2^2+\cdots+n^2=\dfrac{n(n+1)(2n+1)}{6}
Basis step. Put n=1.
\text{LHS}=1^2=1,\qquad\text{RHS}=\dfrac{1\cdot2\cdot3}{6}=1\;\Rightarrow\;P(1)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge 1.
1^2+2^2+\cdots+k^2=\dfrac{k(k+1)(2k+1)}{6}
To prove. P(k+1), that is, the same statement with k+1 in place of n.
1^2+2^2+\cdots+k^2+(k+1)^2=\dfrac{(k+1)(k+2)(2k+3)}{6}
Add (k+1)^2 to both sides of P(k).
1^2+\cdots+k^2+(k+1)^2=\dfrac{k(k+1)(2k+1)}{6}+(k+1)^2
Take out the common factor (k+1) and put everything over the denominator 6.
=\dfrac{(k+1)\big[k(2k+1)+6(k+1)\big]}{6}
Simplify the bracket and factorise the quadratic 2k^2+7k+6=(k+2)(2k+3).
=\dfrac{(k+1)(2k^2+7k+6)}{6}=\dfrac{(k+1)(k+2)(2k+3)}{6}
Since 2k+3=2(k+1)+1, this is the right side of P(k+1).
=\dfrac{(k+1)\big((k+1)+1\big)\big(2(k+1)+1\big)}{6}
So P(k+1) is true whenever P(k) is true. P(1) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge 1.

32025 Annual I, Set B, Q.2 (ix)

Prove by mathematical induction: 1^3+2^3+3^3+\cdots+n^3=\dfrac{n^2(n+1)^2}{4}\qquad\forall\,n\ge1.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
1^3+2^3+\cdots+n^3=\dfrac{n^2(n+1)^2}{4}
Basis step. Put n=1.
\text{LHS}=1^3=1,\qquad\text{RHS}=\dfrac{1^2\cdot2^2}{4}=1\;\Rightarrow\;P(1)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge 1.
1^3+2^3+\cdots+k^3=\dfrac{k^2(k+1)^2}{4}
To prove. P(k+1), that is, the same statement with k+1 in place of n.
1^3+2^3+\cdots+k^3+(k+1)^3=\dfrac{(k+1)^2(k+2)^2}{4}
Add (k+1)^3 to both sides of P(k).
1^3+\cdots+k^3+(k+1)^3=\dfrac{k^2(k+1)^2}{4}+(k+1)^3
Take out the common factor (k+1)^2.
=\dfrac{(k+1)^2\big[k^2+4(k+1)\big]}{4}
The bracket is a perfect square: k^2+4k+4=(k+2)^2.
=\dfrac{(k+1)^2(k+2)^2}{4}=\dfrac{(k+1)^2\big((k+1)+1\big)^2}{4}
So P(k+1) is true whenever P(k) is true. P(1) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge 1.

Prove by mathematical induction: \dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+\cdots+\dfrac{1}{(3n-2)(3n+1)}=\dfrac{n}{3n+1}\qquad\forall\,n\ge1.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\cdots+\dfrac{1}{(3n-2)(3n+1)}=\dfrac{n}{3n+1}
Basis step. Put n=1. The sum has one term.
\text{LHS}=\dfrac{1}{1\cdot4}=\dfrac14,\qquad\text{RHS}=\dfrac{1}{3+1}=\dfrac14\;\Rightarrow\;P(1)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge 1.
\dfrac{1}{1\cdot4}+\cdots+\dfrac{1}{(3k-2)(3k+1)}=\dfrac{k}{3k+1}
To prove. P(k+1), that is, the same statement with k+1 in place of n.
\dfrac{1}{1\cdot4}+\cdots+\dfrac{1}{(3k-2)(3k+1)}+\dfrac{1}{(3k+1)(3k+4)}=\dfrac{k+1}{3k+4}
The next term is found by putting n=k+1 in \dfrac{1}{(3n-2)(3n+1)}: 3(k+1)-2=3k+1 and 3(k+1)+1=3k+4. Add it to both sides of P(k).
\text{LHS of }P(k+1)=\dfrac{k}{3k+1}+\dfrac{1}{(3k+1)(3k+4)}
Take the common denominator (3k+1)(3k+4).
=\dfrac{k(3k+4)+1}{(3k+1)(3k+4)}=\dfrac{3k^2+4k+1}{(3k+1)(3k+4)}
Factorise the numerator, 3k^2+4k+1=(3k+1)(k+1), and cancel the common factor (3k+1).
=\dfrac{(3k+1)(k+1)}{(3k+1)(3k+4)}=\dfrac{k+1}{3k+4}=\dfrac{k+1}{3(k+1)+1}
So P(k+1) is true whenever P(k) is true. P(1) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge 1.

Prove by mathematical induction: 1^2+3^2+5^2+\cdots+(2n-1)^2=\dfrac{n(4n^2-1)}{3}\qquad\forall\,n\ge1.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
1^2+3^2+5^2+\cdots+(2n-1)^2=\dfrac{n(4n^2-1)}{3}
Basis step. Put n=1.
\text{LHS}=1^2=1,\qquad\text{RHS}=\dfrac{1\cdot(4-1)}{3}=1\;\Rightarrow\;P(1)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge 1.
1^2+3^2+\cdots+(2k-1)^2=\dfrac{k(4k^2-1)}{3}
To prove. P(k+1), that is, the same statement with k+1 in place of n.
1^2+3^2+\cdots+(2k-1)^2+(2k+1)^2=\dfrac{(k+1)\big(4(k+1)^2-1\big)}{3}
The term after (2k-1)^2 is \big(2(k+1)-1\big)^2=(2k+1)^2. Add it to both sides of P(k).
\text{LHS of }P(k+1)=\dfrac{k(4k^2-1)}{3}+(2k+1)^2
Factorise 4k^2-1=(2k-1)(2k+1) and take out the common factor (2k+1).
=\dfrac{(2k+1)\big[k(2k-1)+3(2k+1)\big]}{3}
Simplify the bracket: 2k^2-k+6k+3=2k^2+5k+3=(2k+3)(k+1).
=\dfrac{(2k+1)(2k+3)(k+1)}{3}
Now work out the right side of P(k+1) to compare: 4(k+1)^2-1=\big(2(k+1)-1\big)\big(2(k+1)+1\big)=(2k+1)(2k+3).
\dfrac{(k+1)\big(4(k+1)^2-1\big)}{3}=\dfrac{(k+1)(2k+1)(2k+3)}{3}
So P(k+1) is true whenever P(k) is true. P(1) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge 1.

Prove by mathematical induction: 4^3+4^4+4^5+\cdots+4^n=4^3\left(\dfrac{4^{n-2}-1}{3}\right)\qquad\forall\,n\ge3.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
4^3+4^4+\cdots+4^n=4^3\left(\dfrac{4^{n-2}-1}{3}\right)
Basis step. The first value allowed is n=3. The sum is the single term 4^3.
\text{LHS}=4^3,\qquad\text{RHS}=4^3\left(\dfrac{4^{1}-1}{3}\right)=4^3\cdot1=4^3\;\Rightarrow\;P(3)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge3.
4^3+4^4+\cdots+4^k=4^3\left(\dfrac{4^{k-2}-1}{3}\right)
To prove. P(k+1), that is, the same statement with k+1 in place of n.
4^3+4^4+\cdots+4^k+4^{k+1}=4^3\left(\dfrac{4^{k-1}-1}{3}\right)
Add the next term 4^{k+1} to both sides of P(k).
\text{LHS of }P(k+1)=4^3\left(\dfrac{4^{k-2}-1}{3}\right)+4^{k+1}
Write 4^{k+1}=4^3\cdot4^{k-2} so that 4^3 can be taken out, and put over the denominator 3.
=4^3\left(\dfrac{4^{k-2}-1}{3}+4^{k-2}\right)=4^3\cdot\dfrac{4^{k-2}-1+3\cdot4^{k-2}}{3}
Collect the powers of 4: 4^{k-2}+3\cdot4^{k-2}=4\cdot4^{k-2}=4^{k-1}.
=4^3\left(\dfrac{4^{k-1}-1}{3}\right)=4^3\left(\dfrac{4^{(k+1)-2}-1}{3}\right)
So P(k+1) is true whenever P(k) is true. P(3) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge3.

Prove by mathematical induction: \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\cdots+\dfrac{1}{n(n+1)}=\dfrac{n}{n+1}\qquad\forall\,n\ge1.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\cdots+\dfrac{1}{n(n+1)}=\dfrac{n}{n+1}
Basis step. Put n=1.
\text{LHS}=\dfrac{1}{1\cdot2}=\dfrac12,\qquad\text{RHS}=\dfrac{1}{1+1}=\dfrac12\;\Rightarrow\;P(1)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge 1.
\dfrac{1}{1\cdot2}+\cdots+\dfrac{1}{k(k+1)}=\dfrac{k}{k+1}
To prove. P(k+1), that is, the same statement with k+1 in place of n.
\dfrac{1}{1\cdot2}+\cdots+\dfrac{1}{k(k+1)}+\dfrac{1}{(k+1)(k+2)}=\dfrac{k+1}{k+2}
Add the next term \dfrac{1}{(k+1)(k+2)} to both sides of P(k).
\text{LHS of }P(k+1)=\dfrac{k}{k+1}+\dfrac{1}{(k+1)(k+2)}
Common denominator (k+1)(k+2).
=\dfrac{k(k+2)+1}{(k+1)(k+2)}=\dfrac{k^2+2k+1}{(k+1)(k+2)}
The numerator is a perfect square, (k+1)^2. Cancel one factor (k+1).
=\dfrac{(k+1)^2}{(k+1)(k+2)}=\dfrac{k+1}{k+2}
So P(k+1) is true whenever P(k) is true. P(1) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge 1.

Prove by mathematical induction: \dfrac{3}{1.2.2}+\dfrac{4}{2.3.2^2}+\dfrac{5}{3.4.2^3}+\cdots+\dfrac{n+2}{n(n+1).2^n}=1-\dfrac{1}{(n+1).2^n}\qquad\forall\,n\ge1.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
\dfrac{3}{1\cdot2\cdot2}+\dfrac{4}{2\cdot3\cdot2^2}+\cdots+\dfrac{n+2}{n(n+1)\,2^n}=1-\dfrac{1}{(n+1)\,2^n}
Basis step. Put n=1. The sum has the single term \dfrac{1+2}{1\cdot2\cdot2^1}.
\text{LHS}=\dfrac{3}{4},\qquad\text{RHS}=1-\dfrac{1}{2\cdot2}=\dfrac34\;\Rightarrow\;P(1)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge 1.
\dfrac{3}{1\cdot2\cdot2}+\cdots+\dfrac{k+2}{k(k+1)\,2^k}=1-\dfrac{1}{(k+1)\,2^k}
To prove. P(k+1), that is, the same statement with k+1 in place of n.
\dfrac{3}{1\cdot2\cdot2}+\cdots+\dfrac{k+2}{k(k+1)\,2^k}+\dfrac{k+3}{(k+1)(k+2)\,2^{k+1}}=1-\dfrac{1}{(k+2)\,2^{k+1}}
Add the next term (put n=k+1 in the general term) to both sides of P(k).
\text{LHS of }P(k+1)=1-\dfrac{1}{(k+1)\,2^k}+\dfrac{k+3}{(k+1)(k+2)\,2^{k+1}}
Write the subtracted fraction over the common denominator (k+1)(k+2)\,2^{k+1}: multiply top and bottom by 2(k+2).
\dfrac{1}{(k+1)\,2^k}=\dfrac{2(k+2)}{(k+1)(k+2)\,2^{k+1}}
Combine the two fractions.
=1-\dfrac{2(k+2)-(k+3)}{(k+1)(k+2)\,2^{k+1}}=1-\dfrac{k+1}{(k+1)(k+2)\,2^{k+1}}
Cancel the common factor (k+1).
=1-\dfrac{1}{(k+2)\,2^{k+1}}=1-\dfrac{1}{\big((k+1)+1\big)\,2^{k+1}}
So P(k+1) is true whenever P(k) is true. P(1) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge 1.

Prove by mathematical induction: \dfrac{5}{1.2.3}+\dfrac{6}{2.3.4}+\dfrac{7}{3.4.5}+\cdots+\dfrac{n+4}{n(n+1)(n+2)}=\dfrac{n(3n+7)}{2(n+1)(n+2)}\qquad\forall\,n\ge1.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
\dfrac{5}{1\cdot2\cdot3}+\dfrac{6}{2\cdot3\cdot4}+\cdots+\dfrac{n+4}{n(n+1)(n+2)}=\dfrac{n(3n+7)}{2(n+1)(n+2)}
Basis step. Put n=1.
\text{LHS}=\dfrac{5}{6},\qquad\text{RHS}=\dfrac{1\cdot10}{2\cdot2\cdot3}=\dfrac{10}{12}=\dfrac56\;\Rightarrow\;P(1)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge 1.
\dfrac{5}{1\cdot2\cdot3}+\cdots+\dfrac{k+4}{k(k+1)(k+2)}=\dfrac{k(3k+7)}{2(k+1)(k+2)}
To prove. P(k+1), that is, the same statement with k+1 in place of n.
\dfrac{5}{1\cdot2\cdot3}+\cdots+\dfrac{k+4}{k(k+1)(k+2)}+\dfrac{k+5}{(k+1)(k+2)(k+3)}=\dfrac{(k+1)(3k+10)}{2(k+2)(k+3)}
Add the next term (put n=k+1 in the general term) to both sides of P(k).
\text{LHS of }P(k+1)=\dfrac{k(3k+7)}{2(k+1)(k+2)}+\dfrac{k+5}{(k+1)(k+2)(k+3)}
Common denominator 2(k+1)(k+2)(k+3).
=\dfrac{k(3k+7)(k+3)+2(k+5)}{2(k+1)(k+2)(k+3)}
Expand the numerator: k(3k+7)(k+3)=3k^3+16k^2+21k, and add 2k+10.
=\dfrac{3k^3+16k^2+23k+10}{2(k+1)(k+2)(k+3)}
Factorise the numerator. Since k=-1 makes it zero, (k+1) is a factor; dividing gives (k+1)(3k^2+13k+10)=(k+1)^2(3k+10).
3k^3+16k^2+23k+10=(k+1)^2(3k+10)
Cancel one factor (k+1) from numerator and denominator.
=\dfrac{(k+1)(3k+10)}{2(k+2)(k+3)}
This is the right side of P(k+1), because 3(k+1)+7=3k+10.
\dfrac{(k+1)\big(3(k+1)+7\big)}{2\big((k+1)+1\big)\big((k+1)+2\big)}=\dfrac{(k+1)(3k+10)}{2(k+2)(k+3)}\ \checkmark
So P(k+1) is true whenever P(k) is true. P(1) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge 1.

Prove by mathematical induction: 7+77+777+\cdots+\underbrace{77\ldots7}_{n\text{ times}}=\dfrac{7}{81}\left(10^{n+1}-9n-10\right)\qquad\forall\,n\ge1.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
7+77+777+\cdots+\underbrace{77\ldots7}_{n\text{ times}}=\dfrac{7}{81}\left(10^{n+1}-9n-10\right)
Basis step. Put n=1. The sum is just 7.
\text{RHS}=\dfrac{7}{81}(10^2-9-10)=\dfrac{7}{81}\cdot81=7=\text{LHS}\;\Rightarrow\;P(1)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge 1.
7+77+\cdots+\underbrace{77\ldots7}_{k\text{ times}}=\dfrac{7}{81}\left(10^{k+1}-9k-10\right)
To prove. P(k+1), that is, the same statement with k+1 in place of n.
7+77+\cdots+\underbrace{77\ldots7}_{k\text{ times}}+\underbrace{77\ldots7}_{k+1\text{ times}}=\dfrac{7}{81}\left(10^{k+2}-9k-19\right)
First express the new term. A number with k+1 sevens is 7 times the number 11\ldots1 (k+1 ones), and 11\ldots1=\dfrac{10^{k+1}-1}{9}.
\underbrace{77\ldots7}_{k+1\text{ times}}=7\cdot\dfrac{10^{k+1}-1}{9}
Add this term to both sides of P(k).
\text{LHS of }P(k+1)=\dfrac{7}{81}\left(10^{k+1}-9k-10\right)+\dfrac{7}{9}\left(10^{k+1}-1\right)
Take out \dfrac{7}{81}; the second fraction becomes \dfrac{7}{81}\cdot9(10^{k+1}-1).
=\dfrac{7}{81}\left[10^{k+1}-9k-10+9\cdot10^{k+1}-9\right]
Collect: 10^{k+1}+9\cdot10^{k+1}=10\cdot10^{k+1}=10^{k+2} and -9k-19=-9(k+1)-10.
=\dfrac{7}{81}\left[10^{k+2}-9k-19\right]=\dfrac{7}{81}\left[10^{(k+1)+1}-9(k+1)-10\right]
So P(k+1) is true whenever P(k) is true. P(1) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge 1.

Prove by mathematical induction: 1^3+3^3+5^3+\cdots+(2n+1)^3=(n+1)^2(2n^2+4n+1)\qquad\forall\,n\ge0.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
1^3+3^3+5^3+\cdots+(2n+1)^3=(n+1)^2(2n^2+4n+1)
Basis step. The first value is n=0. The sum then has the single term (2\cdot0+1)^3=1.
\text{LHS}=1,\qquad\text{RHS}=(0+1)^2(0+0+1)=1\;\Rightarrow\;P(0)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge0.
1^3+3^3+\cdots+(2k+1)^3=(k+1)^2(2k^2+4k+1)
To prove. P(k+1), that is, the same statement with k+1 in place of n.
1^3+3^3+\cdots+(2k+1)^3+(2k+3)^3=(k+2)^2(2k^2+8k+7)
The right side of P(k+1) simplifies to this: 2(k+1)^2+4(k+1)+1=2k^2+8k+7.
(k+2)^2\left[2(k+1)^2+4(k+1)+1\right]=(k+2)^2(2k^2+8k+7)
Add the next term (2k+3)^3 to both sides of P(k) and expand each part.
(k+1)^2(2k^2+4k+1)=2k^4+8k^3+11k^2+6k+1,\qquad(2k+3)^3=8k^3+36k^2+54k+27
Add the two expansions.
\text{LHS of }P(k+1)=2k^4+16k^3+47k^2+60k+28
Expand the right side of P(k+1) in the same way.
(k+2)^2(2k^2+8k+7)=(k^2+4k+4)(2k^2+8k+7)=2k^4+16k^3+47k^2+60k+28
The two expressions are identical, so P(k+1) is true.
2k^4+16k^3+47k^2+60k+28=2k^4+16k^3+47k^2+60k+28\ \checkmark
So P(k+1) is true whenever P(k) is true. P(0) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge0.

Prove by mathematical induction: 1.2^0+2.2^1+3.2^2+\cdots+n.2^{n-1}=(n-1).2^n+1\qquad\forall\,n\ge1.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
1\cdot2^0+2\cdot2^1+3\cdot2^2+\cdots+n\cdot2^{n-1}=(n-1)\,2^n+1
Basis step. Put n=1.
\text{LHS}=1\cdot2^0=1,\qquad\text{RHS}=(1-1)\cdot2+1=1\;\Rightarrow\;P(1)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge 1.
1\cdot2^0+2\cdot2^1+\cdots+k\cdot2^{k-1}=(k-1)\,2^k+1
To prove. P(k+1), that is, the same statement with k+1 in place of n.
1\cdot2^0+2\cdot2^1+\cdots+k\cdot2^{k-1}+(k+1)\,2^k=k\cdot2^{k+1}+1
Add the next term (k+1)\,2^{k} to both sides of P(k).
\text{LHS of }P(k+1)=(k-1)\,2^k+1+(k+1)\,2^k
Combine the two terms in 2^k: (k-1)+(k+1)=2k.
=2k\cdot2^k+1
Use 2\cdot2^k=2^{k+1}. This is the right side of P(k+1), since (k+1)-1=k.
=k\cdot2^{k+1}+1=\big((k+1)-1\big)\,2^{k+1}+1
So P(k+1) is true whenever P(k) is true. P(1) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge 1.

Prove by mathematical induction: 1.1!+2.2!+3.3!+\cdots+n.n!=(n+1)!-1\qquad\forall\,n\ge1.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
1\cdot1!+2\cdot2!+3\cdot3!+\cdots+n\cdot n!=(n+1)!-1
Basis step. Put n=1.
\text{LHS}=1\cdot1!=1,\qquad\text{RHS}=2!-1=1\;\Rightarrow\;P(1)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge 1.
1\cdot1!+2\cdot2!+\cdots+k\cdot k!=(k+1)!-1
To prove. P(k+1), that is, the same statement with k+1 in place of n.
1\cdot1!+2\cdot2!+\cdots+k\cdot k!+(k+1)(k+1)!=(k+2)!-1
Add the next term (k+1)\,(k+1)! to both sides of P(k).
\text{LHS of }P(k+1)=(k+1)!-1+(k+1)(k+1)!
Take out the common factor (k+1)!.
=(k+1)!\,\big[1+(k+1)\big]-1=(k+2)\,(k+1)!-1
By the definition of factorial, (k+2)(k+1)!=(k+2)!.
=(k+2)!-1=\big((k+1)+1\big)!-1
So P(k+1) is true whenever P(k) is true. P(1) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge 1.

Prove by mathematical induction: \left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)\cdots\left(1-\dfrac{1}{n^2}\right)=\dfrac{n+1}{2n}\qquad\forall\,n\ge2.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\cdots\left(1-\dfrac{1}{n^2}\right)=\dfrac{n+1}{2n}
Basis step. The first value is n=2. The product has the single factor 1-\dfrac{1}{2^2}.
\text{LHS}=1-\dfrac14=\dfrac34,\qquad\text{RHS}=\dfrac{2+1}{2\cdot2}=\dfrac34\;\Rightarrow\;P(2)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge2.
\left(1-\dfrac{1}{2^2}\right)\cdots\left(1-\dfrac{1}{k^2}\right)=\dfrac{k+1}{2k}
To prove. P(k+1), that is, the same statement with k+1 in place of n.
\left(1-\dfrac{1}{2^2}\right)\cdots\left(1-\dfrac{1}{k^2}\right)\left(1-\dfrac{1}{(k+1)^2}\right)=\dfrac{k+2}{2(k+1)}
This is a product, so multiply both sides of P(k) by the next factor 1-\dfrac{1}{(k+1)^2}.
\text{LHS of }P(k+1)=\dfrac{k+1}{2k}\left(1-\dfrac{1}{(k+1)^2}\right)
Simplify the bracket: 1-\dfrac{1}{(k+1)^2}=\dfrac{(k+1)^2-1}{(k+1)^2}=\dfrac{k(k+2)}{(k+1)^2}.
=\dfrac{k+1}{2k}\cdot\dfrac{k(k+2)}{(k+1)^2}
Cancel k and one factor (k+1).
=\dfrac{k+2}{2(k+1)}=\dfrac{(k+1)+1}{2(k+1)}
So P(k+1) is true whenever P(k) is true. P(2) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge2.

Prove by mathematical induction: \left(\dfrac11.\dfrac12\right)\left(\dfrac13.\dfrac14\right)\left(\dfrac15.\dfrac16\right)\cdots\left(\dfrac{1}{2n+1}.\dfrac{1}{2n+2}\right)=\dfrac{1}{(2n+2)!}\qquad\forall\,n\ge0.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
\left(\dfrac11\cdot\dfrac12\right)\left(\dfrac13\cdot\dfrac14\right)\cdots\left(\dfrac{1}{2n+1}\cdot\dfrac{1}{2n+2}\right)=\dfrac{1}{(2n+2)!}
Basis step. The first value is n=0. The product then has the single factor \dfrac11\cdot\dfrac12.
\text{LHS}=\dfrac11\cdot\dfrac12=\dfrac12,\qquad\text{RHS}=\dfrac{1}{2!}=\dfrac12\;\Rightarrow\;P(0)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge0.
\left(\dfrac11\cdot\dfrac12\right)\cdots\left(\dfrac{1}{2k+1}\cdot\dfrac{1}{2k+2}\right)=\dfrac{1}{(2k+2)!}
To prove. P(k+1), that is, the same statement with k+1 in place of n.
\begin{aligned}&\left(\dfrac11\cdot\dfrac12\right)\cdots\left(\dfrac{1}{2k+1}\cdot\dfrac{1}{2k+2}\right)\\ &\quad\times\left(\dfrac{1}{2k+3}\cdot\dfrac{1}{2k+4}\right)=\dfrac{1}{(2k+4)!}\end{aligned}
Multiply both sides of P(k) by the next factor, found by putting n=k+1: \dfrac{1}{2k+3}\cdot\dfrac{1}{2k+4}.
\text{LHS of }P(k+1)=\dfrac{1}{(2k+2)!}\cdot\dfrac{1}{(2k+3)(2k+4)}
Combine the denominators using (2k+4)!=(2k+4)(2k+3)(2k+2)!.
=\dfrac{1}{(2k+2)!\,(2k+3)(2k+4)}=\dfrac{1}{(2k+4)!}
This is the right side of P(k+1), because 2(k+1)+2=2k+4.
\dfrac{1}{\big(2(k+1)+2\big)!}=\dfrac{1}{(2k+4)!}\ \checkmark
So P(k+1) is true whenever P(k) is true. P(0) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge0.

Prove by mathematical induction: 1-2+2^2-2^3+\cdots+(-1)^{n-1}2^{n-1}=\dfrac{1-(-2)^n}{3}\qquad\forall\,n\ge0.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
1-2+2^2-2^3+\cdots+(-1)^{n-1}2^{n-1}=\dfrac{1-(-2)^n}{3}
Basis step. The first value is n=0. A sum with no terms is 0.
\text{LHS}=0,\qquad\text{RHS}=\dfrac{1-(-2)^0}{3}=\dfrac{1-1}{3}=0\;\Rightarrow\;P(0)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge0.
1-2+2^2-\cdots+(-1)^{k-1}2^{k-1}=\dfrac{1-(-2)^k}{3}
To prove. P(k+1), that is, the same statement with k+1 in place of n.
1-2+2^2-\cdots+(-1)^{k-1}2^{k-1}+(-1)^{k}2^{k}=\dfrac{1-(-2)^{k+1}}{3}
The new term is (-1)^k2^k=(-2)^k. Add it to both sides of P(k).
\text{LHS of }P(k+1)=\dfrac{1-(-2)^k}{3}+(-2)^k
Put everything over the denominator 3.
=\dfrac{1-(-2)^k+3(-2)^k}{3}=\dfrac{1+2(-2)^k}{3}
Write 2(-2)^k=-\big[(-2)(-2)^k\big]=-(-2)^{k+1}.
=\dfrac{1-(-2)^{k+1}}{3}
So P(k+1) is true whenever P(k) is true. P(0) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge0.

Prove by mathematical induction: \dbinom{n}{0}+\dbinom{n}{1}+\dbinom{n}{2}+\cdots+\dbinom{n}{n}=2^n\qquad\forall\,n\ge0.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
\dbinom{n}{0}+\dbinom{n}{1}+\cdots+\dbinom{n}{n}=2^n
Basis step. The first value is n=0. The sum has the single term \dbinom00=1.
\text{LHS}=\dbinom00=1,\qquad\text{RHS}=2^0=1\;\Rightarrow\;P(0)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge0.
\dbinom{k}{0}+\dbinom{k}{1}+\cdots+\dbinom{k}{k}=2^k
To prove. P(k+1), that is, the same statement with k+1 in place of n.
\dbinom{k+1}{0}+\dbinom{k+1}{1}+\cdots+\dbinom{k+1}{k+1}=2^{k+1}
Use Pascal's rule \dbinom{k+1}{r}=\dbinom{k}{r}+\dbinom{k}{r-1} on the middle terms (1\le r\le k). The first and last terms are both 1.
\sum_{r=0}^{k+1}\dbinom{k+1}{r}=1+\sum_{r=1}^{k}\left[\dbinom{k}{r}+\dbinom{k}{r-1}\right]+1
Split into two groups. Group 1 is 1+\sum_{r=1}^{k}\dbinom{k}{r} and group 2 is \sum_{r=1}^{k}\dbinom{k}{r-1}+1.
=\left[\dbinom{k}{0}+\sum_{r=1}^{k}\dbinom{k}{r}\right]+\left[\sum_{r=1}^{k}\dbinom{k}{r-1}+\dbinom{k}{k}\right]
Each group is the whole sum \dbinom{k}{0}+\dbinom{k}{1}+\cdots+\dbinom{k}{k} (group 2 is the same terms shifted by one place, from \dbinom{k}{0} to \dbinom{k}{k}). By P(k) each equals 2^k.
=2^k+2^k=2\cdot2^k=2^{k+1}
So P(k+1) is true whenever P(k) is true. P(0) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge0.

Prove by mathematical induction: \dbinom{n}{1}+2\dbinom{n}{2}+3\dbinom{n}{3}+\cdots+n\dbinom{n}{n}=n.2^{n-1}\qquad\forall\,n\ge1.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
\dbinom{n}{1}+2\dbinom{n}{2}+3\dbinom{n}{3}+\cdots+n\dbinom{n}{n}=n\cdot2^{n-1}
Basis step. Put n=1.
\text{LHS}=\dbinom11=1,\qquad\text{RHS}=1\cdot2^0=1\;\Rightarrow\;P(1)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge 1.
\sum_{r=1}^{k}r\dbinom{k}{r}=k\cdot2^{k-1}
To prove. P(k+1), that is, the same statement with k+1 in place of n.
\sum_{r=1}^{k+1}r\dbinom{k+1}{r}=(k+1)\,2^{k}
Write the left side of P(k+1) and use Pascal's rule \dbinom{k+1}{r}=\dbinom{k}{r}+\dbinom{k}{r-1}.
\sum_{r=1}^{k+1}r\dbinom{k+1}{r}=\sum_{r=1}^{k+1}r\dbinom{k}{r}+\sum_{r=1}^{k+1}r\dbinom{k}{r-1}
In the first sum \dbinom{k}{k+1}=0, so the last term vanishes and the sum is exactly the left side of P(k), which is k\cdot2^{k-1}.
\sum_{r=1}^{k+1}r\dbinom{k}{r}=\sum_{r=1}^{k}r\dbinom{k}{r}=k\cdot2^{k-1}
In the second sum put s=r-1 (so r=s+1, and s runs from 0 to k), and split (s+1).
\sum_{r=1}^{k+1}r\dbinom{k}{r-1}=\sum_{s=0}^{k}(s+1)\dbinom{k}{s}=\sum_{s=0}^{k}s\dbinom{k}{s}+\sum_{s=0}^{k}\dbinom{k}{s}
The first part is k\cdot2^{k-1} by P(k) (the term s=0 is zero). The second part is 2^k by Question 17.
=k\cdot2^{k-1}+2^k
Add the two results and simplify, using 2\cdot k\cdot2^{k-1}=k\cdot2^k.
\text{LHS of }P(k+1)=k\cdot2^{k-1}+k\cdot2^{k-1}+2^k=k\cdot2^k+2^k=(k+1)\,2^k\ \checkmark
So P(k+1) is true whenever P(k) is true. P(1) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge 1.

Note. The proof uses the result of Question 17.

Prove by mathematical induction: \dbinom{n}{0}+\dfrac12\dbinom{n}{1}+\dfrac13\dbinom{n}{2}+\cdots+\dfrac{1}{n+1}\dbinom{n}{n}=\dfrac{2^{n+1}-1}{n+1}\qquad\forall\,n\ge0.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
\sum_{r=0}^{n}\dfrac{1}{r+1}\dbinom{n}{r}=\dfrac{2^{n+1}-1}{n+1}
Basis step. Put n=0. The sum has the single term \dfrac11\dbinom00=1.
\text{LHS}=1,\qquad\text{RHS}=\dfrac{2^1-1}{1}=1\;\Rightarrow\;P(0)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge0.
\sum_{r=0}^{k}\dfrac{1}{r+1}\dbinom{k}{r}=\dfrac{2^{k+1}-1}{k+1}
To prove. P(k+1).
\sum_{r=0}^{k+1}\dfrac{1}{r+1}\dbinom{k+1}{r}=\dfrac{2^{k+2}-1}{k+2}
Use the identity \dfrac{1}{r+1}\dbinom{k+1}{r}=\dfrac{1}{k+2}\dbinom{k+2}{r+1}. It follows from the factorial formula.
\dfrac{1}{r+1}\cdot\dfrac{(k+1)!}{r!\,(k+1-r)!}=\dfrac{(k+1)!}{(r+1)!\,(k+1-r)!}=\dfrac{1}{k+2}\cdot\dfrac{(k+2)!}{(r+1)!\,\big((k+2)-(r+1)\big)!}
Apply it to every term of the left side of P(k+1). The index r+1 now runs from 1 to k+2.
\text{LHS of }P(k+1)=\dfrac{1}{k+2}\sum_{r=0}^{k+1}\dbinom{k+2}{r+1}=\dfrac{1}{k+2}\left[\dbinom{k+2}{1}+\dbinom{k+2}{2}+\cdots+\dbinom{k+2}{k+2}\right]
The bracket is the full sum \dbinom{k+2}{0}+\cdots+\dbinom{k+2}{k+2}=2^{k+2} (Question 17) without its first term \dbinom{k+2}{0}=1.
=\dfrac{2^{k+2}-1}{k+2}\ \checkmark
So P(k+1) is true whenever P(k) is true. P(0) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge0.

Prove by mathematical induction: n^3+2n is divisible by 3 \qquad\forall\,n\ge1.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
n^3+2n\ \text{is divisible by }3
Basis step. Put n=1.
1^3+2\cdot1=3=3\times1\;\Rightarrow\;P(1)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge 1.
k^3+2k=3m\quad\text{for some integer }m
To prove. P(k+1), that is, the same statement with k+1 in place of n.
(k+1)^3+2(k+1)\ \text{is divisible by }3
Expand (k+1)^3 and collect.
(k+1)^3+2(k+1)=k^3+3k^2+3k+1+2k+2=k^3+3k^2+5k+3
Regroup so that the hypothesis expression k^3+2k appears; the rest is a multiple of 3.
=(k^3+2k)+3k^2+3k+3=(k^3+2k)+3(k^2+k+1)
By the hypothesis k^3+2k=3m, so the whole expression is 3 times an integer.
=3m+3(k^2+k+1)=3\left(m+k^2+k+1\right)
So P(k+1) is true whenever P(k) is true. P(1) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge 1.

Prove by mathematical induction: 6 is a factor of n(n^2+5) \qquad\forall\,n\ge1.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
6\ \text{is a factor of }n(n^2+5)
Basis step. Put n=1.
1\cdot(1+5)=6=6\times1\;\Rightarrow\;P(1)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge 1.
k(k^2+5)=6m\quad\text{for some integer }m
To prove. P(k+1), that is, the same statement with k+1 in place of n.
(k+1)\big[(k+1)^2+5\big]\ \text{is divisible by }6
Expand the expression for n=k+1.
(k+1)(k^2+2k+6)=k^3+3k^2+8k+6
Regroup so that k(k^2+5)=k^3+5k appears.
=(k^3+5k)+3k^2+3k+6=k(k^2+5)+3k(k+1)+6
k(k+1) is a product of two consecutive integers, so it is even: k(k+1)=2t. Then 3k(k+1)=6t is a multiple of 6. Using the hypothesis for the first part:
=6m+6t+6=6(m+t+1)
So P(k+1) is true whenever P(k) is true. P(1) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge 1.

Prove by mathematical induction: \dfrac{n(3n^4+5n^2+7)}{15} is a rational number.

Solution

A number is rational if it can be written as a fraction of two integers. We prove something stronger: n(3n^4+5n^2+7) is divisible by 15, so the given number is an integer, and every integer is rational (m=\frac m1).

Proof by mathematical induction

Let P(n) be the statement to be proved.
N(n)=n(3n^4+5n^2+7)=3n^5+5n^3+7n\ \text{is divisible by }15
Basis step. Put n=1.
N(1)=1\cdot(3+5+7)=15=15\times1\;\Rightarrow\;P(1)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge 1.
N(k)=3k^5+5k^3+7k=15m\quad\text{for some integer }m
To prove. P(k+1), that is, the same statement with k+1 in place of n.
N(k+1)=3(k+1)^5+5(k+1)^3+7(k+1)\ \text{is divisible by }15
Find the difference N(k+1)-N(k) by expanding (k+1)^5=k^5+5k^4+10k^3+10k^2+5k+1 and (k+1)^3=k^3+3k^2+3k+1.
N(k+1)-N(k)=3\left(5k^4+10k^3+10k^2+5k+1\right)+5\left(3k^2+3k+1\right)+7
Multiply out and collect like terms.
=15k^4+30k^3+30k^2+15k+3+15k^2+15k+5+7=15k^4+30k^3+45k^2+30k+15
Every coefficient is a multiple of 15, so the difference is 15\left(k^4+2k^3+3k^2+2k+1\right).
N(k+1)=N(k)+15\left(k^4+2k^3+3k^2+2k+1\right)
By the hypothesis N(k)=15m, so N(k+1) is 15 times an integer.
N(k+1)=15\left(m+k^4+2k^3+3k^2+2k+1\right)
So P(k+1) is true whenever P(k) is true. P(1) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge 1.

Hence \dfrac{n(3n^4+5n^2+7)}{15} is an integer for every n\ge1, and so it is a rational number.

Note. The statement asks only for a rational number, which is true for every n trivially (any integer divided by 15 is rational). The induction shows the stronger fact that the quotient is an integer.

Prove by mathematical induction: 4^n+15n-1 is divisible by 9 \qquad\forall\,n\ge1.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
4^n+15n-1\ \text{is divisible by }9
Basis step. Put n=1.
4+15-1=18=9\times2\;\Rightarrow\;P(1)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge 1.
4^k+15k-1=9m\quad\text{for some integer }m
To prove. P(k+1), that is, the same statement with k+1 in place of n.
4^{k+1}+15(k+1)-1\ \text{is divisible by }9
Write 4^{k+1}=4\cdot4^k and replace 4^k using the hypothesis: 4^k=9m-15k+1.
4^{k+1}+15(k+1)-1=4\left(9m-15k+1\right)+15k+14
Multiply out and collect.
=36m-60k+4+15k+14=36m-45k+18
Every term is a multiple of 9.
=9\left(4m-5k+2\right)
So P(k+1) is true whenever P(k) is true. P(1) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge 1.

Prove by mathematical induction: 7^n-2^n is divisible by 5 \qquad\forall\,n\ge0.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
7^n-2^n\ \text{is divisible by }5
Basis step. The first value is n=0.
7^0-2^0=1-1=0=5\times0\;\Rightarrow\;P(0)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge0.
7^k-2^k=5m\quad\text{for some integer }m
To prove. P(k+1), that is, the same statement with k+1 in place of n.
7^{k+1}-2^{k+1}\ \text{is divisible by }5
Write 7^{k+1}=7\cdot7^k and 2^{k+1}=2\cdot2^k. Add and subtract 7\cdot2^k so that 7^k-2^k appears.
7^{k+1}-2^{k+1}=7\cdot7^k-7\cdot2^k+7\cdot2^k-2\cdot2^k=7\left(7^k-2^k\right)+5\cdot2^k
By the hypothesis 7^k-2^k=5m. Both parts are multiples of 5.
=7\cdot5m+5\cdot2^k=5\left(7m+2^k\right)
So P(k+1) is true whenever P(k) is true. P(0) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge0.

Prove by mathematical induction: 2^n<(n+1)!\qquad\forall\,n\ge2.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
2^n<(n+1)!
Basis step. The first value is n=2.
2^2=4,\qquad3!=6,\qquad4<6\;\Rightarrow\;P(2)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge2.
2^k<(k+1)!
To prove. P(k+1), that is, the same statement with k+1 in place of n.
2^{k+1}<(k+2)!
Multiply both sides of the hypothesis by 2 (a positive number, so the inequality keeps its direction).
2^{k+1}=2\cdot2^k<2\,(k+1)!
Since k\ge2, we have 2<k+2, so 2\,(k+1)!<(k+2)\,(k+1)!.
2\,(k+1)!<(k+2)\,(k+1)!=(k+2)!
Join the two inequalities.
2^{k+1}<(k+2)!=\big((k+1)+1\big)!
So P(k+1) is true whenever P(k) is true. P(2) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge2.

Prove by mathematical induction: 5^n+9<6^n\qquad\forall\,n\ge2.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
5^n+9<6^n
Basis step. The first value is n=2.
5^2+9=34,\qquad6^2=36,\qquad34<36\;\Rightarrow\;P(2)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge2.
5^k+9<6^k
To prove. P(k+1), that is, the same statement with k+1 in place of n.
5^{k+1}+9<6^{k+1}
Write 5^{k+1}+9 so that the hypothesis expression 5^k+9 appears: 5(5^k+9)=5^{k+1}+45, so 5^{k+1}+9=5(5^k+9)-36.
5^{k+1}+9=5\left(5^k+9\right)-36
Multiply the hypothesis by 5 (positive), and subtract 36 from the left side only; this makes the left side smaller still.
5^{k+1}+9=5\left(5^k+9\right)-36<5\cdot6^k-36<5\cdot6^k
Finally 5\cdot6^k<6\cdot6^k=6^{k+1}.
5^{k+1}+9<6^{k+1}
So P(k+1) is true whenever P(k) is true. P(2) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge2.

Prove by mathematical induction: if h>-1, then 1+nh\le(1+h)^n\qquad\forall\,n\ge0.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
1+nh\le(1+h)^n\quad(h>-1)
Basis step. The first value is n=0.
1+0\cdot h=1,\qquad(1+h)^0=1,\qquad1\le1\;\Rightarrow\;P(0)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge0.
1+kh\le(1+h)^k
To prove. P(k+1), that is, the same statement with k+1 in place of n.
1+(k+1)h\le(1+h)^{k+1}
Since h>-1, the number 1+h is positive. Multiplying both sides of the hypothesis by (1+h) keeps the inequality.
(1+h)^{k+1}=(1+h)(1+h)^k\ge(1+h)(1+kh)
Multiply out the right side.
(1+h)(1+kh)=1+kh+h+kh^2=1+(k+1)h+kh^2
kh^2\ge0 because it is k times a square, so we may drop it and the right side can only get smaller.
(1+h)^{k+1}\ge1+(k+1)h+kh^2\ge1+(k+1)h
So P(k+1) is true whenever P(k) is true. P(0) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge0.

Note. This is Bernoulli's inequality. The condition h>-1 is what makes 1+h positive.

Prove by mathematical induction: \dbinom{2n}{n}<2^{2n-2}\qquad\forall\,n\ge5.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
\dbinom{2n}{n}<2^{2n-2}
Basis step. The first value is n=5.
\dbinom{10}{5}=252,\qquad2^{8}=256,\qquad252<256\;\Rightarrow\;P(5)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge5.
\dbinom{2k}{k}<2^{2k-2}
To prove. P(k+1), that is, the same statement with k+1 in place of n.
\dbinom{2k+2}{k+1}<2^{2k}
Express \dbinom{2k+2}{k+1} in terms of \dbinom{2k}{k} using the factorial formula.
\dbinom{2k+2}{k+1}=\dfrac{(2k+2)!}{\big((k+1)!\big)^2}=\dfrac{(2k+2)(2k+1)\,(2k)!}{(k+1)^2\,(k!)^2}=\dfrac{2(2k+1)}{k+1}\dbinom{2k}{k}
The factor \dfrac{2(2k+1)}{k+1} is less than 4, because 4k+2<4k+4.
\dfrac{2(2k+1)}{k+1}<4
Use the hypothesis and this bound together (all quantities are positive).
\dbinom{2k+2}{k+1}=\dfrac{2(2k+1)}{k+1}\dbinom{2k}{k}<4\cdot\dbinom{2k}{k}<4\cdot2^{2k-2}
Simplify 4\cdot2^{2k-2}=2^2\cdot2^{2k-2}=2^{2k}, and note 2(k+1)-2=2k.
\dbinom{2k+2}{k+1}<2^{2k}=2^{2(k+1)-2}
So P(k+1) is true whenever P(k) is true. P(5) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge5.

Note. The condition n\ge5 is needed: \binom84=70>2^6=64, so the statement fails at n=4.

Prove by mathematical induction: \sqrt[n]{n}<2-\dfrac1n\qquad\forall\,n\ge2.

Solution

Both sides are positive for n\ge2 (the right side is at least \frac32), so raising both sides to the power n keeps the inequality. The statement is therefore the same as n<\left(2-\dfrac1n\right)^n. The induction is done on the simple inequality 2^{n-1}>n, and Bernoulli's inequality of Question 27 connects it to the given one.

Proof

Lemma. 2^{n-1}>n for all n\ge3. Basis n=3: 2^2=4>3. Assume 2^{k-1}>k for some k\ge3. Then multiply by 2:
2^{k}=2\cdot2^{k-1}>2k\ge k+1\quad(\text{since }k\ge1)
So 2^{(k+1)-1}>k+1 and the lemma holds for all n\ge3 by induction.
2^{n-1}>n\qquad(n\ge3)
Basis step of the main statement. n=2: the inequality \sqrt2<2-\frac12=\frac32 is the same as 2<\frac94.
\sqrt2<\dfrac32\iff2<\dfrac94\;\Rightarrow\;P(2)\ \text{is true}
**Step for n\ge3.** Take out the factor 2^n and write 2-\frac1n=2\left(1-\frac{1}{2n}\right).
\left(2-\dfrac1n\right)^n=2^n\left(1-\dfrac{1}{2n}\right)^n
Apply Bernoulli's inequality (Question 27) with h=-\dfrac{1}{2n}, which is greater than -1.
\left(1-\dfrac{1}{2n}\right)^n\ge1+n\left(-\dfrac{1}{2n}\right)=1-\dfrac12=\dfrac12
Multiply by 2^n, then use the lemma.
\left(2-\dfrac1n\right)^n\ge2^n\cdot\dfrac12=2^{n-1}>n
Take the nth root of both sides (both are positive).
\sqrt[n]{n}<2-\dfrac1n\ \checkmark

Note. The step from n=k to n=k+1 cannot be done directly on the root form, so the proof is arranged as an induction lemma plus Bernoulli's inequality (Question 27).

Prove by mathematical induction: 1+3n\le4^n\qquad\forall\,n\ge0.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
1+3n\le4^n
Basis step. The first value is n=0.
1+0=1,\qquad4^0=1,\qquad1\le1\;\Rightarrow\;P(0)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge0.
1+3k\le4^k
To prove. P(k+1), that is, the same statement with k+1 in place of n.
1+3(k+1)\le4^{k+1}
Add 3 to both sides of the hypothesis.
1+3(k+1)=(1+3k)+3\le4^k+3
Since k\ge0, we have 4^k\ge1, so 3\le3\cdot4^k.
4^k+3\le4^k+3\cdot4^k
Collect: 4^k+3\cdot4^k=4\cdot4^k=4^{k+1}.
1+3(k+1)\le4^{k+1}
So P(k+1) is true whenever P(k) is true. P(0) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge0.

Prove by mathematical induction: n^3>2n+1\qquad\forall\,n\ge2.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
n^3>2n+1
Basis step. The first value is n=2.
2^3=8,\qquad2\cdot2+1=5,\qquad8>5\;\Rightarrow\;P(2)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge2.
k^3>2k+1
To prove. P(k+1), that is, the same statement with k+1 in place of n.
(k+1)^3>2(k+1)+1=2k+3
Expand (k+1)^3 and apply the hypothesis to the first term.
(k+1)^3=k^3+3k^2+3k+1>(2k+1)+3k^2+3k+1=3k^2+5k+2
It remains to show 3k^2+5k+2>2k+3. Subtract 2k+3 from both sides.
3k^2+5k+2-(2k+3)=3k^2+3k-1
For k\ge2, 3k^2\ge12>1, so 3k^2+3k-1>0. Hence 3k^2+5k+2>2k+3.
(k+1)^3>3k^2+5k+2>2k+3=2(k+1)+1
So P(k+1) is true whenever P(k) is true. P(2) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge2.

Prove by mathematical induction: n!>n^2\qquad\forall\,n\ge4.

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
n!>n^2
Basis step. The first value is n=4.
4!=24,\qquad4^2=16,\qquad24>16\;\Rightarrow\;P(4)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge4.
k!>k^2
To prove. P(k+1), that is, the same statement with k+1 in place of n.
(k+1)!>(k+1)^2
Write (k+1)!=(k+1)\,k! and apply the hypothesis (multiply by the positive number k+1).
(k+1)!=(k+1)\,k!>(k+1)\,k^2
For k\ge4 we have k^2>k+1 (for example 16>5; in general k^2-k-1=k(k-1)-1\ge11).
k^2>k+1
Multiply this by (k+1) and join with the previous line.
(k+1)!>(k+1)k^2>(k+1)(k+1)=(k+1)^2
So P(k+1) is true whenever P(k) is true. P(4) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge4.

Exercise 7.2p. 187–188

Binomial expansion with Pascal's triangle (Question 1) and the binomial theorem T_{r+1}=\binom nr a^{n-r}b^r (the rest): expansions, middle term, specified term, term independent of x, and the coefficient identities. Questions 18 to 20 apply Pascal's triangle to tossing coins.

Expand the following with the help of Pascal's triangle.

Solution

(i) \left(2\sqrt x+\dfrac{1}{\sqrt x}\right)^5

Identify a, b and n in (a+b)^n (keep the sign with b). The coefficients come from the 5th row of Pascal's triangle.
a=2\sqrt x,\qquad b=\dfrac{1}{\sqrt x},\qquad n=5\qquad\text{row 5: }1, 5, 10, 10, 5, 1
Write the terms: the power of a falls from n to 0 while the power of b rises from 0 to n, each multiplied by its coefficient.
\left(2\sqrt x+\tfrac1{\sqrt x}\right)^5=1(2\sqrt x)^5+5(2\sqrt x)^4\!\left(\tfrac1{\sqrt x}\right)+10(2\sqrt x)^3\!\left(\tfrac1{\sqrt x}\right)^2+10(2\sqrt x)^2\!\left(\tfrac1{\sqrt x}\right)^3+5(2\sqrt x)\!\left(\tfrac1{\sqrt x}\right)^4+1\!\left(\tfrac1{\sqrt x}\right)^5
Simplify each term (powers of powers, products of fractions).
=32x^{5/2}+5(16x^2)\dfrac{1}{x^{1/2}}+10(8x^{3/2})\dfrac1x+10(4x)\dfrac{1}{x^{3/2}}+5(2x^{1/2})\dfrac{1}{x^2}+\dfrac{1}{x^{5/2}}

32x^{5/2}+80x^{3/2}+80\sqrt x+\dfrac{40}{\sqrt x}+\dfrac{10}{x^{3/2}}+\dfrac{1}{x^{5/2}}

(ii) \left(\dfrac3x+\dfrac y2\right)^6

Identify a, b and n in (a+b)^n (keep the sign with b). The coefficients come from the 6th row of Pascal's triangle.
a=\dfrac3x,\qquad b=\dfrac y2,\qquad n=6\qquad\text{row 6: }1, 6, 15, 20, 15, 6, 1
Write the terms: the power of a falls from n to 0 while the power of b rises from 0 to n, each multiplied by its coefficient.
\left(\tfrac3x+\tfrac y2\right)^6=1\left(\tfrac3x\right)^6+6\left(\tfrac3x\right)^5\!\left(\tfrac y2\right)+15\left(\tfrac3x\right)^4\!\left(\tfrac y2\right)^2+20\left(\tfrac3x\right)^3\!\left(\tfrac y2\right)^3+15\left(\tfrac3x\right)^2\!\left(\tfrac y2\right)^4+6\left(\tfrac3x\right)\!\left(\tfrac y2\right)^5+1\left(\tfrac y2\right)^6
Simplify each term (powers of powers, products of fractions).
=\dfrac{729}{x^6}+6\cdot\dfrac{243}{x^5}\cdot\dfrac y2+15\cdot\dfrac{81}{x^4}\cdot\dfrac{y^2}{4}+20\cdot\dfrac{27}{x^3}\cdot\dfrac{y^3}{8}+15\cdot\dfrac{9}{x^2}\cdot\dfrac{y^4}{16}+6\cdot\dfrac3x\cdot\dfrac{y^5}{32}+\dfrac{y^6}{64}

\dfrac{729}{x^6}+\dfrac{729y}{x^5}+\dfrac{1215y^2}{4x^4}+\dfrac{135y^3}{2x^3}+\dfrac{135y^4}{16x^2}+\dfrac{9y^5}{16x}+\dfrac{y^6}{64}

(iii) (2-x^{3/2})^7

Identify a, b and n in (a+b)^n (keep the sign with b). The coefficients come from the 7th row of Pascal's triangle.
a=2,\qquad b=-x^{3/2},\qquad n=7\qquad\text{row 7: }1, 7, 21, 35, 35, 21, 7, 1
Write the terms: the power of a falls from n to 0 while the power of b rises from 0 to n, each multiplied by its coefficient.
(2-x^{3/2})^7=1(2)^7+7(2)^6(-x^{3/2})+21(2)^5(-x^{3/2})^2+35(2)^4(-x^{3/2})^3+35(2)^3(-x^{3/2})^4+21(2)^2(-x^{3/2})^5+7(2)(-x^{3/2})^6+1(-x^{3/2})^7
Simplify each term (powers of powers, products of fractions).
=128-7(64)x^{3/2}+21(32)x^3-35(16)x^{9/2}+35(8)x^6-21(4)x^{15/2}+7(2)x^9-x^{21/2}

128-448x^{3/2}+672x^3-560x^{9/2}+280x^6-84x^{15/2}+14x^9-x^{21/2}

(iv) \left(\dfrac{x^2}{y^2}-\sqrt{\dfrac yx}\right)^5

Identify a, b and n in (a+b)^n (keep the sign with b). The coefficients come from the 5th row of Pascal's triangle.
a=\dfrac{x^2}{y^2},\qquad b=-\sqrt{\dfrac yx}=-\left(\dfrac yx\right)^{1/2},\qquad n=5\qquad\text{row 5: }1, 5, 10, 10, 5, 1
Write the terms: the power of a falls from n to 0 while the power of b rises from 0 to n, each multiplied by its coefficient.
=1a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5\quad\text{with }a=\dfrac{x^2}{y^2},\ b=-\left(\dfrac yx\right)^{1/2}
Simplify each term (powers of powers, products of fractions).
=\dfrac{x^{10}}{y^{10}}-5\dfrac{x^8}{y^8}\left(\dfrac yx\right)^{1/2}+10\dfrac{x^6}{y^6}\left(\dfrac yx\right)-10\dfrac{x^4}{y^4}\left(\dfrac yx\right)^{3/2}+5\dfrac{x^2}{y^2}\left(\dfrac yx\right)^{2}-\left(\dfrac yx\right)^{5/2}
Combine the powers of x and y in each term. For example \dfrac{x^8}{y^8}\cdot\dfrac{y^{1/2}}{x^{1/2}}=\dfrac{x^{15/2}}{y^{15/2}} and \dfrac{x^2}{y^2}\cdot\dfrac{y^2}{x^2}=1.
=\dfrac{x^{10}}{y^{10}}-5\dfrac{x^{15/2}}{y^{15/2}}+10\dfrac{x^5}{y^5}-10\dfrac{x^{5/2}}{y^{5/2}}+5-\dfrac{y^{5/2}}{x^{5/2}}

\dfrac{x^{10}}{y^{10}}-5\dfrac{x^{15/2}}{y^{15/2}}+10\dfrac{x^5}{y^5}-10\dfrac{x^{5/2}}{y^{5/2}}+5-\dfrac{y^{5/2}}{x^{5/2}}

Book answer. (i) 32x^{5/2}+80x^{3/2}+80\sqrt x+\dfrac{40}{\sqrt x}+\dfrac{10}{x^{3/2}}+\dfrac{1}{x^{5/2}}; (ii) \dfrac{729}{x^6}+\dfrac{729y}{x^5}+\dfrac{1215y^2}{4x^4}+\dfrac{135y^3}{2x^3}+\dfrac{135y^4}{16x^2}+\dfrac{9y^5}{16x}+\dfrac{y^6}{64}; (iii) 128-448x^{3/2}+672x^3-560x^{9/2}+280x^6-84x^{15/2}+14x^9-x^{21/2}; (iv) \dfrac{x^{10}}{y^{10}}-5\dfrac{x^{15/2}}{y^{15/2}}+10\dfrac{x^5}{y^5}-10\dfrac{x^{5/2}}{y^{5/2}}+5\dfrac{y^{5/2}}{x^{5/2}} (as printed)

Note. Parts (i) to (iii) agree with the key. For (iv) the key's last two terms read +5\dfrac{y^{5/2}}{x^{5/2}}, which is a misprint: the fifth term is 5ab^4=5 (a constant) and the last term is b^5=-\left(\dfrac yx\right)^{5/2}. Checked by substituting x=1.7, y=2.3 in both sides.

Expand the followings by using binomial theorem.

Solution

(i) \left(\dfrac{2x}{3}-\dfrac{3}{2x}\right)^5

Here a=\dfrac{2x}{3}, b=-\dfrac{3}{2x}, n=5. Use T_{r+1}=\dbinom5r a^{5-r}b^r for r=0,1,\ldots,5, with \dbinom50=1,\ \dbinom51=5,\ \dbinom52=10,\ \dbinom53=10,\ \dbinom54=5,\ \dbinom55=1.
\left(\tfrac{2x}{3}-\tfrac{3}{2x}\right)^5=\sum_{r=0}^{5}\dbinom5r\left(\tfrac{2x}{3}\right)^{5-r}\left(-\tfrac{3}{2x}\right)^r
Write out all six terms.
=\left(\tfrac{2x}{3}\right)^5+5\left(\tfrac{2x}{3}\right)^4\!\left(-\tfrac{3}{2x}\right)+10\left(\tfrac{2x}{3}\right)^3\!\left(\tfrac{3}{2x}\right)^2-10\left(\tfrac{2x}{3}\right)^2\!\left(\tfrac{3}{2x}\right)^3+5\left(\tfrac{2x}{3}\right)\!\left(\tfrac{3}{2x}\right)^4-\left(\tfrac{3}{2x}\right)^5
Simplify each term; the powers of x fall by 2 each time: x^5,x^3,x,x^{-1},x^{-3},x^{-5}.
=\dfrac{32x^5}{243}-5\cdot\dfrac{16x^4}{81}\cdot\dfrac{3}{2x}+10\cdot\dfrac{8x^3}{27}\cdot\dfrac{9}{4x^2}-10\cdot\dfrac{4x^2}{9}\cdot\dfrac{27}{8x^3}+5\cdot\dfrac{2x}{3}\cdot\dfrac{81}{16x^4}-\dfrac{243}{32x^5}

\dfrac{32x^5}{243}-\dfrac{40x^3}{27}+\dfrac{20x}{3}-\dfrac{15}{x}+\dfrac{135}{8x^3}-\dfrac{243}{32x^5}

(ii) (-x+y^{-1})^6

Here a=-x, b=y^{-1}=\dfrac1y, n=6. The coefficients are 1,6,15,20,15,6,1. An odd power of -x is negative, an even power is positive.
(-x+y^{-1})^6=\sum_{r=0}^{6}\dbinom6r(-x)^{6-r}\left(\tfrac1y\right)^r
Write out the seven terms and take care with the signs.
=x^6-6\dfrac{x^5}{y}+15\dfrac{x^4}{y^2}-20\dfrac{x^3}{y^3}+15\dfrac{x^2}{y^4}-6\dfrac{x}{y^5}+\dfrac{1}{y^6}

x^6-\dfrac{6x^5}{y}+\dfrac{15x^4}{y^2}-\dfrac{20x^3}{y^3}+\dfrac{15x^2}{y^4}-\dfrac{6x}{y^5}+\dfrac1{y^6}

(iii) (3u-1)^7

Here a=3u, b=-1, n=7. The coefficients are 1,7,21,35,35,21,7,1. The signs alternate because b=-1.
(3u-1)^7=\sum_{r=0}^{7}\dbinom7r(3u)^{7-r}(-1)^r
Write out the eight terms.
=(3u)^7-7(3u)^6+21(3u)^5-35(3u)^4+35(3u)^3-21(3u)^2+7(3u)-1
Use 3^7=2187,\ 3^6=729,\ 3^5=243,\ 3^4=81,\ 3^3=27,\ 3^2=9 and multiply by the coefficients.
=2187u^7-5103u^6+5103u^5-2835u^4+945u^3-189u^2+21u-1

2187u^7-5103u^6+5103u^5-2835u^4+945u^3-189u^2+21u-1

(iv) (a\sqrt2+b\sqrt3)^5

Here the first term is a\sqrt2, the second is b\sqrt3 and n=5. Use the powers (\sqrt2)^2=2,\ (\sqrt2)^3=2\sqrt2,\ (\sqrt2)^4=4,\ (\sqrt2)^5=4\sqrt2 and (\sqrt3)^2=3,\ (\sqrt3)^3=3\sqrt3,\ (\sqrt3)^4=9,\ (\sqrt3)^5=9\sqrt3.
(a\sqrt2+b\sqrt3)^5=\sum_{r=0}^{5}\dbinom5r(a\sqrt2)^{5-r}(b\sqrt3)^r
Write out the six terms.
=(a\sqrt2)^5+5(a\sqrt2)^4(b\sqrt3)+10(a\sqrt2)^3(b\sqrt3)^2+10(a\sqrt2)^2(b\sqrt3)^3+5(a\sqrt2)(b\sqrt3)^4+(b\sqrt3)^5
Replace each root power and multiply the numbers.
=4\sqrt2\,a^5+5(4)\sqrt3\,a^4b+10(2\sqrt2)(3)\,a^3b^2+10(2)(3\sqrt3)\,a^2b^3+5\sqrt2(9)\,ab^4+9\sqrt3\,b^5

4\sqrt2\,a^5+20\sqrt3\,a^4b+60\sqrt2\,a^3b^2+60\sqrt3\,a^2b^3+45\sqrt2\,ab^4+9\sqrt3\,b^5

(v) (1+2x-y)^4

Treat 1+2x as one term and -y as the other, so that the binomial theorem applies with n=4 (coefficients 1,4,6,4,1).
(1+2x-y)^4=(1+2x)^4-4(1+2x)^3y+6(1+2x)^2y^2-4(1+2x)y^3+y^4
Expand the powers of 1+2x with the binomial theorem.
(1+2x)^4=1+8x+24x^2+32x^3+16x^4,\qquad(1+2x)^3=1+6x+12x^2+8x^3,\qquad(1+2x)^2=1+4x+4x^2
Substitute and multiply out term by term.
=\left(1+8x+24x^2+32x^3+16x^4\right)-4y\left(1+6x+12x^2+8x^3\right)+6y^2\left(1+4x+4x^2\right)-4y^3(1+2x)+y^4
Remove the brackets and arrange by powers of y.
=1+8x+24x^2+32x^3+16x^4-4y-24xy-48x^2y-32x^3y+6y^2+24xy^2+24x^2y^2-4y^3-8xy^3+y^4

1+8x+24x^2+32x^3+16x^4-4y-24xy-48x^2y-32x^3y+6y^2+24xy^2+24x^2y^2-4y^3-8xy^3+y^4

(vi) \left(\dfrac1x+\dfrac2y+\dfrac3z\right)^4

Put u=\dfrac1x,\ v=\dfrac2y,\ w=\dfrac3z. Write the three-term power as \big[(u+v)+w\big]^4 and expand twice with n=4.
\big[(u+v)+w\big]^4=(u+v)^4+4(u+v)^3w+6(u+v)^2w^2+4(u+v)w^3+w^4
Expand the powers of u+v.
(u+v)^4=u^4+4u^3v+6u^2v^2+4uv^3+v^4,\quad(u+v)^3=u^3+3u^2v+3uv^2+v^3,\quad(u+v)^2=u^2+2uv+v^2
Collect the 15 terms; their coefficients are \dfrac{4!}{i!\,j!\,k!} for u^iv^jw^k with i+j+k=4.
=u^4+4u^3v+6u^2v^2+4uv^3+v^4+4u^3w+12u^2vw+12uv^2w+4v^3w+6u^2w^2+12uvw^2+6v^2w^2+4uw^3+4vw^3+w^4
Replace u,v,w by \dfrac1x,\dfrac2y,\dfrac3z and multiply the numbers, for example 4u^3v=4\cdot\dfrac1{x^3}\cdot\dfrac2y=\dfrac{8}{x^3y} and 12u^2vw=12\cdot\dfrac1{x^2}\cdot\dfrac2y\cdot\dfrac3z=\dfrac{72}{x^2yz}.
=\dfrac1{x^4}+\dfrac8{x^3y}+\dfrac{24}{x^2y^2}+\dfrac{32}{xy^3}+\dfrac{16}{y^4}+\dfrac{12}{x^3z}+\dfrac{72}{x^2yz}+\dfrac{144}{xy^2z}+\dfrac{96}{y^3z}+\dfrac{54}{x^2z^2}+\dfrac{216}{xyz^2}+\dfrac{216}{y^2z^2}+\dfrac{108}{xz^3}+\dfrac{216}{yz^3}+\dfrac{81}{z^4}

\dfrac1{x^4}+\dfrac8{x^3y}+\dfrac{24}{x^2y^2}+\dfrac{32}{xy^3}+\dfrac{16}{y^4}+\dfrac{12}{x^3z}+\dfrac{72}{x^2yz}+\dfrac{144}{xy^2z}+\dfrac{96}{y^3z}+\dfrac{54}{x^2z^2}+\dfrac{216}{xyz^2}+\dfrac{216}{y^2z^2}+\dfrac{108}{xz^3}+\dfrac{216}{yz^3}+\dfrac{81}{z^4}

Book answer. (i) \dfrac{32x^5}{243}-\dfrac{40x^3}{27}+\dfrac{20x}{3}-\dfrac{15}{x}+\dfrac{45}{x^2}-\dfrac{243}{32x^5}; (ii) x^6-6\dfrac{x^5}{y}+\dfrac{15x^4}{y^2}-\dfrac{20x^3}{y^3}+\dfrac{15x^2}{y^4}-\dfrac{6x}{y^5}+\dfrac1{y^6}; (iii) 2187u^7-5103u^6+5103u^5-2835u^4+945u^3-189u^2+21u-1; (iv) printed as a^5 2^{5/2}+20ab\sqrt3+30ab^2\sqrt2+20ab^3\sqrt3+45ab^4\sqrt2+b^53^{5/2}; (v) and (vi) as in our answers except where noted

Note. (ii), (iii) and (vi) agree with the key. The key has misprints in three parts. In (i) the fifth term is printed \dfrac{45}{x^2}; it must be \dfrac{135}{8x^3}, because the powers of x fall by 2 each time (x^5,x^3,x,x^{-1},x^{-3},x^{-5}). In (iv) the middle terms are printed with lost exponents and wrong coefficients (20ab\sqrt3, 30ab^2\sqrt2, 20ab^3\sqrt3); the correct terms are 20\sqrt3\,a^4b, 60\sqrt2\,a^3b^2, 60\sqrt3\,a^2b^3. In (v) the key prints -12x^2y and +6xy^3 where the expansion has -48x^2y and +24xy^2. Each of our results was checked by substituting numbers in the original expression and in the answer.

Expand and simplify.

Solution

(i) (1+10x)^4+(1-10x)^4

Expand each power with the binomial theorem (n=4, coefficients 1,4,6,4,1). The two expansions differ only in the sign of the odd powers of 10x.
(1+10x)^4=1+4(10x)+6(10x)^2+4(10x)^3+(10x)^4,\qquad(1-10x)^4=1-4(10x)+6(10x)^2-4(10x)^3+(10x)^4
Add. The odd-power terms cancel and the even-power terms double.
=2\left[1+6(100x^2)+10000x^4\right]=2\left[1+600x^2+10000x^4\right]

2+1200x^2+20000x^4

(ii) \left(2-\dfrac3x\right)(1+4x^2)

Neither factor needs the binomial theorem; multiply them term by term.
\left(2-\dfrac3x\right)(1+4x^2)=2(1)+2(4x^2)-\dfrac3x(1)-\dfrac3x(4x^2)
Simplify and arrange in decreasing powers of x.
=2+8x^2-\dfrac3x-12x=8x^2-12x+2-\dfrac3x

8x^2-12x+2-\dfrac3x

(iii) (1+2x+2x^2)(1-x)^5

Expand (1-x)^5 (coefficients 1,5,10,10,5,1, alternating signs).
(1-x)^5=1-5x+10x^2-10x^3+5x^4-x^5
Multiply the series by 1, by 2x and by 2x^2 in turn.
1\cdot(\ldots)=1-5x+10x^2-10x^3+5x^4-x^5
Second row: multiply by 2x (every power of x goes up by one).
2x\cdot(\ldots)=\ \ \ \,2x-10x^2+20x^3-20x^4+10x^5-2x^6
Third row: multiply by 2x^2 (every power of x goes up by two).
2x^2\cdot(\ldots)=\qquad\ \ 2x^2-10x^3+20x^4-20x^5+10x^6-2x^7
Add the like powers: x: -5+2=-3; x^2: 10-10+2=2; x^3: -10+20-10=0; x^4: 5-20+20=5; x^5: -1+10-20=-11; x^6: -2+10=8; x^7: -2.
=1-3x+2x^2+0\cdot x^3+5x^4-11x^5+8x^6-2x^7

1-3x+2x^2+5x^4-11x^5+8x^6-2x^7

(iv) (.99)^3+(1.01)^4

Write the numbers as 1\mp0.01 so that the binomial theorem can be used.
(0.99)^3=(1-0.01)^3,\qquad(1.01)^4=(1+0.01)^4
Expand (1-0.01)^3=1-3(0.01)+3(0.01)^2-(0.01)^3.
=1-0.03+0.0003-0.000001=0.970299
Expand (1+0.01)^4=1+4(0.01)+6(0.01)^2+4(0.01)^3+(0.01)^4.
=1+0.04+0.0006+0.000004+0.00000001=1.04060401
Add the two results.
0.970299+1.04060401=2.01090301

2.01090301

(v) \left(\dfrac2x-\dfrac x4\right)^4+\left(\dfrac2x+\dfrac x4\right)^4

Let a=\dfrac2x and b=\dfrac x4. The sum is (a-b)^4+(a+b)^4, in which the odd powers of b cancel and the even powers double.
(a+b)^4+(a-b)^4=2\left[a^4+6a^2b^2+b^4\right]
Compute each part: a^4=\dfrac{16}{x^4}, a^2b^2=\dfrac{4}{x^2}\cdot\dfrac{x^2}{16}=\dfrac14, b^4=\dfrac{x^4}{256}.
=2\left[\dfrac{16}{x^4}+6\cdot\dfrac14+\dfrac{x^4}{256}\right]=2\left[\dfrac{16}{x^4}+\dfrac32+\dfrac{x^4}{256}\right]
Multiply by 2.
=\dfrac{32}{x^4}+3+\dfrac{x^4}{128}

\dfrac{32}{x^4}+3+\dfrac{x^4}{128}

(vi) \left(a^2+\sqrt{a^2-1}\right)^4-\left(a^2-\sqrt{a^2-1}\right)^4

Let u=a^2 and s=\sqrt{a^2-1}. The expression is (u+s)^4-(u-s)^4, in which the even powers of s cancel and the odd powers double.
(u+s)^4-(u-s)^4=2\left[\dbinom41u^3s+\dbinom43us^3\right]=2\left[4u^3s+4us^3\right]
Simplify and substitute back u=a^2, s=\sqrt{a^2-1}.
=8u^3s+8us^3=8a^6\sqrt{a^2-1}+8a^2\left(\sqrt{a^2-1}\right)^3
Optional: since \left(\sqrt{a^2-1}\right)^2=a^2-1, this equals 8a^2\sqrt{a^2-1}\left(a^4+a^2-1\right).
8a^2\sqrt{a^2-1}\left[a^4+(a^2-1)\right]

8a^6\sqrt{a^2-1}+8a^2\left(\sqrt{a^2-1}\right)^3

Book answer. (i) 2+1200x^2+20000x^4; (ii) 8x^2-12x+2-\dfrac3x; (iii) 1-3x+2x^2+10x^3+5x^4-11x^5+8x^6-2x^7; (iv) 2.01090301; (v) \dfrac{32}{x^4}+2+\dfrac{x^4}{128}; (vi) 8a^6\sqrt{a^2-1}+8a^2\left(\sqrt{a^2-1}\right)^3

Note. Parts (i), (ii), (iv) and (vi) agree with the key. In (iii) the key prints +10x^3, but the x^3 terms are -10+20-10=0, so there is no x^3 term (checked by putting x=1.37 in both sides). In (v) the key prints the constant term as 2; the middle term of (a+b)^4+(a-b)^4 is 2\cdot6a^2b^2=12\cdot\dfrac14=3 (checked numerically at x=1.3).

Find the coefficient of the 8^{\text{th}} term in the expansion of \left(x^2+\dfrac y2\right)^{10}.

Solution

The (r+1)th term of (a+b)^n is T_{r+1}=\dbinom nr a^{n-r}b^r. For the 8th term, r+1=8, so r=7. Here a=x^2, b=\dfrac y2, n=10.
T_{8}=T_{7+1}=\dbinom{10}{7}(x^2)^{3}\left(\dfrac y2\right)^{7}
Evaluate \dbinom{10}{7}=\dbinom{10}{3}=\dfrac{10\cdot9\cdot8}{3\cdot2\cdot1}=120 and \left(\dfrac y2\right)^7=\dfrac{y^7}{128}.
T_8=120\cdot x^6\cdot\dfrac{y^7}{128}=\dfrac{15}{16}x^6y^7
The coefficient is the numerical factor of x^6y^7.
\text{coefficient}=\dfrac{120}{128}=\dfrac{15}{16}

\dfrac{15}{16}

Book answer. \dfrac{15}{16}x^6y^7

Note. The key prints the whole term; the coefficient asked for is \dfrac{15}{16}.

52026 Annual I, Set B, Q.5

Find the middle term in the expansion of the following.

Solution

(i) \left(3x^2-\dfrac{1}{2x}\right)^{10}

n=10 is even, so there is one middle term, the \left(\dfrac n2+1\right)th, that is the 6th term, T_6=T_{5+1}. Here a=3x^2, b=-\dfrac1{2x}.
T_6=\dbinom{10}{5}(3x^2)^5\left(-\dfrac1{2x}\right)^5
Evaluate: \dbinom{10}{5}=252, (3x^2)^5=243x^{10}, \left(-\dfrac1{2x}\right)^5=-\dfrac{1}{32x^5}.
T_6=252\cdot243x^{10}\cdot\left(-\dfrac1{32x^5}\right)
Multiply the numbers (252\times243=61236) and subtract the powers of x.
T_6=-\dfrac{61236}{32}x^5=-\dfrac{15309}{8}x^5

-\dfrac{15309}{8}x^5

(ii) \left(2x^2-\dfrac{1}{5x}\right)^{11}

n=11 is odd, so there are two middle terms: the \dfrac{n+1}{2}=6th and the \dfrac{n+3}{2}=7th. Here a=2x^2, b=-\dfrac1{5x}.
T_6=T_{5+1}=\dbinom{11}{5}(2x^2)^6\left(-\dfrac1{5x}\right)^5,\qquad T_7=T_{6+1}=\dbinom{11}{6}(2x^2)^5\left(-\dfrac1{5x}\right)^6
Evaluate with \dbinom{11}{5}=\dbinom{11}{6}=462.
T_6=462\cdot64x^{12}\cdot\left(-\dfrac1{3125x^5}\right)=-\dfrac{29568}{3125}x^7
The seventh term: (-1)^6=+1.
T_7=462\cdot32x^{10}\cdot\dfrac1{15625x^6}=\dfrac{14784}{15625}x^4=0.946176x^4

-\dfrac{29568}{3125}x^7\ \text{and}\ \dfrac{14784}{15625}x^4

(iii) \left(\dfrac{a}{\sqrt x}+\sqrt x\right)^8

n=8 is even, so the middle term is the \left(\dfrac82+1\right)=5th term, T_5=T_{4+1}. Here the first term is \dfrac a{\sqrt x}, the second is \sqrt x.
T_5=\dbinom84\left(\dfrac{a}{\sqrt x}\right)^4(\sqrt x)^4
\dbinom84=70; the powers of x cancel: \dfrac{a^4}{x^2}\cdot x^2=a^4.
T_5=70\cdot\dfrac{a^4}{x^2}\cdot x^2=70a^4

70a^4

(iv) \left(a-\dfrac{3}{x^2}\right)^{12}

n=12 is even, so the middle term is the \left(\dfrac{12}2+1\right)=7th term, T_7=T_{6+1}. Here b=-\dfrac3{x^2}.
T_7=\dbinom{12}{6}a^6\left(-\dfrac{3}{x^2}\right)^6
\dbinom{12}{6}=924, and the even power makes the sign positive: (-3)^6=729.
T_7=924\cdot729\cdot\dfrac{a^6}{x^{12}}=\dfrac{673596\,a^6}{x^{12}}

\dfrac{673596\,a^6}{x^{12}}

Book answer. (i) -\dfrac{15309}{8}x^5; (ii) 0.946176x^4; (iii) 70a^4; (iv) \dfrac{673596a^6}{x^{12}}

Note. All four agree with the key. In (ii) the exponent 11 is odd, so there are two middle terms; the key lists only the second one (0.946176x^4=\dfrac{14784}{15625}x^4).

Find the specified term in the following expansions.

Solution

(i) Term involving b^6 in the expansion of \left(\dfrac{a^2}{2}+2b^2\right)^{10}

The general term, with a-part \dfrac{a^2}{2} and b-part 2b^2, is T_{r+1}=\dbinom{10}{r}\left(\dfrac{a^2}2\right)^{10-r}(2b^2)^r. The power of b is 2r.
T_{r+1}=\dbinom{10}{r}\dfrac{a^{20-2r}}{2^{10-r}}\cdot2^rb^{2r}
We need b^6, so 2r=6 and r=3.
T_4=\dbinom{10}{3}\left(\dfrac{a^2}{2}\right)^7(2b^2)^3=120\cdot\dfrac{a^{14}}{128}\cdot8b^6
Simplify: 120\times8\div128=7.5.
T_4=\dfrac{15}{2}a^{14}b^6

\dfrac{15}{2}a^{14}b^6

(ii) Term involving q^8 in the expansion of \left(\dfrac{p^2}{2}+6q^2\right)^{12}

The general term is T_{r+1}=\dbinom{12}{r}\left(\dfrac{p^2}2\right)^{12-r}(6q^2)^r and the power of q is 2r. For q^8, 2r=8, so r=4.
T_5=\dbinom{12}{4}\left(\dfrac{p^2}{2}\right)^8(6q^2)^4
\dbinom{12}4=495, \left(\dfrac{p^2}2\right)^8=\dfrac{p^{16}}{256}, 6^4=1296.
T_5=495\cdot\dfrac{p^{16}}{256}\cdot1296\,q^8=\dfrac{641520}{256}p^{16}q^8
Divide numerator and denominator by 16.
T_5=\dfrac{40095}{16}p^{16}q^8

\dfrac{40095}{16}p^{16}q^8

(iii) Term involving x^4y^3 in the expansion of (3x^4-y)^4

The general term is T_{r+1}=\dbinom4r(3x^4)^{4-r}(-y)^r. The power of y is r, so y^3 needs r=3. Then the power of x is 4(4-3)=4, as required.
T_4=\dbinom43(3x^4)^1(-y)^3
\dbinom43=4 and (-y)^3=-y^3.
T_4=4\cdot3x^4\cdot(-y^3)=-12x^4y^3

-12x^4y^3

(iv) Term involving y^8x^3 in the expansion of (y^4-3x)^5

The general term is T_{r+1}=\dbinom5r(y^4)^{5-r}(-3x)^r. The power of x is r, so x^3 needs r=3. Then the power of y is 4(5-3)=8, as required.
T_4=\dbinom53(y^4)^2(-3x)^3
\dbinom53=10 and (-3x)^3=-27x^3.
T_4=10\cdot y^8\cdot(-27x^3)=-270\,y^8x^3

-270\,y^8x^3

Book answer. (i) \dfrac{15}{2}a^{14}b^6; (ii) \dfrac{40095}{16}p^{16}q^8; (iii) -12x^4y^3; (iv) -270y^8x^3

Note. All four agree with the key.

Find the term independent of x in the expansions of the following.

Solution

(i) \left(2x^2-\dfrac1x\right)^{12}

A term independent of x is the one in which the power of x is zero. Write the general term with a=2x^2, b=-\dfrac1x, n=12.
T_{r+1}=\dbinom{12}{r}(2x^2)^{12-r}\left(-\dfrac1x\right)^r=\dbinom{12}{r}2^{12-r}(-1)^r\,x^{24-2r}\,x^{-r}=\dbinom{12}{r}2^{12-r}(-1)^r\,x^{24-3r}
Put the power of x equal to zero.
24-3r=0\;\Rightarrow\;r=8
Substitute r=8 (\dbinom{12}8=\dbinom{12}4=495, and (-1)^8=1).
T_9=495\cdot2^{4}=495\cdot16=7920

7920

(ii) \left(\sqrt x+\dfrac{1}{3x^2}\right)^{10}

Write the general term with a=\sqrt x=x^{1/2}, b=\dfrac1{3x^2}, n=10.
T_{r+1}=\dbinom{10}{r}(x^{1/2})^{10-r}\left(\dfrac{1}{3x^2}\right)^r=\dbinom{10}{r}\dfrac{1}{3^r}\,x^{\frac{10-r}{2}-2r}
Put the power of x equal to zero.
\dfrac{10-r}{2}-2r=0\;\Rightarrow\;10-r=4r\;\Rightarrow\;r=2
Substitute r=2.
T_3=\dbinom{10}{2}\cdot\dfrac19=\dfrac{45}{9}=5

Book answer. (i) 1980; (ii) 5

Note. Part (ii) agrees with the key. For (i) the key prints 1980, but the term is \dbinom{12}{8}2^4=495\times16=7920 (1980 would be 495\times4, which needs 2^2). Substituting r=8 back and checking that the powers of x cancel confirms 7920.

Find the r^{th} term from the end in the expansion of (a+b)^n where 0\le r\le n.

Solution

The expansion of (a+b)^n has n+1 terms. By property 8 of the binomial expansion, the rth term from the end is the (n+2-r)th term from the beginning.
r\text{th from the end}=(n+2-r)\text{th from the beginning}
Write n+2-r=(n+1-r)+1, so in T_{k+1}=\dbinom nk a^{n-k}b^k we take k=n+1-r.
T_{n+2-r}=\dbinom{n}{n+1-r}a^{\,n-(n+1-r)}\,b^{\,n+1-r}
Simplify the power of a: n-(n+1-r)=r-1. Use \dbinom{n}{n+1-r}=\dbinom{n}{r-1} (the coefficients equidistant from the two ends are equal).
T_{n+2-r}=\dbinom{n}{r-1}a^{\,r-1}\,b^{\,n-r+1}
Check. For r=1 this gives \dbinom n0a^0b^n=b^n, the last term. For r=n+1 it gives \dbinom nn a^nb^0=a^n, the first term.
r=1:\ b^n,\qquad r=n+1:\ a^n

\dbinom{n}{r-1}\,a^{\,r-1}\,b^{\,n-r+1}

Book answer. T_{r+2}=\dbinom{n}{r+1}a^{n-r-1}b^{r+1}

Note. The key's expression is the (r+2)th term from the beginning (which is the (n-r)th from the end), not the rth from the end. Our answer passes the check for the last term (r=1) and the first term (r=n+1). Strictly r runs from 1 to n+1 since there are n+1 terms; the book's 0\le r\le n would leave out the first term.

Prove that sum of all the binomial coefficients in the expansion of (a+b)^n is 2^n; hence or otherwise prove that sum of odd coefficients is 2^{n-1}.

Solution

By the binomial theorem the coefficients of (a+b)^n are \dbinom n0,\dbinom n1,\ldots,\dbinom nn. Put a=1 and b=1 in the theorem.
(1+1)^n=\dbinom n0+\dbinom n1+\dbinom n2+\cdots+\dbinom nn
The left side is 2^n, so the sum of all the binomial coefficients is 2^n.
\dbinom n0+\dbinom n1+\cdots+\dbinom nn=2^n\qquad(1)
For the odd coefficients, put a=1 and b=-1 (n\ge1). The left side becomes 0^n=0.
(1-1)^n=\dbinom n0-\dbinom n1+\dbinom n2-\dbinom n3+\cdots+(-1)^n\dbinom nn=0
Move the negative terms to the other side. Write E for the sum of the coefficients with even lower index (\dbinom n0+\dbinom n2+\cdots) and O for those with odd lower index (\dbinom n1+\dbinom n3+\cdots).
E-O=0\;\Rightarrow\;E=O\qquad(2)
Equation (1) says E+O=2^n. Using E=O from (2):
2O=2^n\;\Rightarrow\;O=2^{n-1}
So the sum of the coefficients \dbinom n1+\dbinom n3+\cdots is 2^{n-1} (and the even ones add up to 2^{n-1} too).
\dbinom n1+\dbinom n3+\dbinom n5+\cdots=2^{n-1}

The sum of coefficients of first three terms in the expansion of \left(a-\dfrac{3}{a^2}\right)^n is 559. Find the term involving a^3 in the expansion.

Solution

Write the first three terms with b=-\dfrac3{a^2}. Their coefficients (numbers multiplying powers of a) are 1, -3n and \dfrac{n(n-1)}{2}\cdot9.
T_1=a^n,\qquad T_2=\dbinom n1a^{n-1}\left(-\dfrac3{a^2}\right),\qquad T_3=\dbinom n2a^{n-2}\left(\dfrac9{a^4}\right)
Add the three coefficients and set the sum equal to 559.
1-3n+\dfrac{9n(n-1)}{2}=559
Multiply by 2 and collect.
2-6n+9n^2-9n=1118\;\Rightarrow\;9n^2-15n-1116=0\;\Rightarrow\;3n^2-5n-372=0
Solve: n=\dfrac{5\pm\sqrt{25+4464}}{6}=\dfrac{5\pm67}{6}. The positive integer root is n=12 (the other root -\dfrac{31}{3} is rejected).
n=12
The general term of \left(a-\dfrac3{a^2}\right)^{12} has the power of a equal to 12-r-2r=12-3r.
T_{r+1}=\dbinom{12}{r}a^{12-r}\left(-\dfrac{3}{a^2}\right)^r=\dbinom{12}{r}(-3)^r\,a^{12-3r}
For a^3: 12-3r=3, so r=3.
T_4=\dbinom{12}{3}(-3)^3a^3=220\cdot(-27)\,a^3=-5940\,a^3

-5940\,a^3

Book answer. -5940

Note. Agrees with the key (the key gives only the coefficient). Check: 1-36+\dfrac{9\cdot12\cdot11}{2}=1-36+594=559.

If the coefficients of (r-5)^{th} and (2r-1)^{th} term in the expansion of (1+a)^{34} are equal; then find the value of r.

Solution

The kth term of (1+a)^{34} is T_k=\dbinom{34}{k-1}a^{k-1}, so its coefficient is \dbinom{34}{k-1}.
\text{coeff of }T_{r-5}=\dbinom{34}{r-6},\qquad\text{coeff of }T_{2r-1}=\dbinom{34}{2r-2}
Equal coefficients: \dbinom{34}{p}=\dbinom{34}{q} happens either when p=q or when p+q=34.
\dbinom{34}{r-6}=\dbinom{34}{2r-2}
Case 1: r-6=2r-2 gives r=-4, which is impossible (a term number cannot be negative).
r-6=2r-2\;\Rightarrow\;r=-4\ \text{(rejected)}
Case 2: (r-6)+(2r-2)=34.
3r-8=34\;\Rightarrow\;3r=42\;\Rightarrow\;r=14
Check. The terms are the 9th and the 27th: \dbinom{34}{8}=\dbinom{34}{26} because 8+26=34.
\dbinom{34}{8}=\dbinom{34}{26}\ \checkmark

r=14

Book answer. 14

If the coefficients of 2^{nd}, 3^{rd} and 4^{th} terms in the expansion of (1+x)^{2m} are in A.P., show that 2m^2-9m+7=0.

Solution

The coefficients of the 2nd, 3rd and 4th terms of (1+x)^{2m} are \dbinom{2m}{1},\dbinom{2m}{2},\dbinom{2m}{3}. Let N=2m to keep the algebra short.
\dbinom N1=N,\qquad\dbinom N2=\dfrac{N(N-1)}{2},\qquad\dbinom N3=\dfrac{N(N-1)(N-2)}{6}
Three numbers are in A.P. when the middle one is the average of the other two: 2\times\text{middle}=\text{first}+\text{last}.
2\cdot\dfrac{N(N-1)}{2}=N+\dfrac{N(N-1)(N-2)}{6}
Divide by N (it is not zero) and multiply by 6.
6(N-1)=6+(N-1)(N-2)
Expand and collect.
6N-6=6+N^2-3N+2\;\Rightarrow\;N^2-9N+14=0
Put back N=2m.
4m^2-18m+14=0\;\Rightarrow\;2m^2-9m+7=0

2m^2-9m+7=0

If coefficients of three consecutive terms in the expansion of (1+x)^n are in the ratio 6:33:110, then find the value of n and the position of terms.

Solution

Let the three consecutive terms be T_r,T_{r+1},T_{r+2}. Their coefficients are \dbinom n{r-1},\dbinom nr,\dbinom n{r+1}.
\dbinom n{r-1}:\dbinom nr:\dbinom n{r+1}=6:33:110
Use \dbinom nr\Big/\dbinom n{r-1}=\dfrac{n-r+1}{r} for the first two and \dbinom n{r+1}\Big/\dbinom nr=\dfrac{n-r}{r+1} for the last two.
\dfrac{n-r+1}{r}=\dfrac{33}{6}=\dfrac{11}{2},\qquad\dfrac{n-r}{r+1}=\dfrac{110}{33}=\dfrac{10}{3}
Cross-multiply both equations.
\begin{aligned}2n-2r+2&=11r\\ \Rightarrow\ 2n&=13r-2\\ 3n-3r&=10r+10\\ \Rightarrow\ 3n&=13r+10\end{aligned}
From the first equation 13r=2n+2. Replace 13r by 2n+2 in the second equation.
3n=(2n+2)+10\;\Rightarrow\;n=12
Then 13r=2n+2=26, so r=2.
r=2
The three terms are T_2,T_3,T_4. Check: \dbinom{12}1:\dbinom{12}2:\dbinom{12}3=12:66:220=6:33:110.
12:66:220=6:33:110\ \checkmark

n=12;\ \text{the terms are the }2^{\text{nd}},3^{\text{rd}},4^{\text{th}}

Book answer. 12,\ 2

Note. Agrees with the key (n=12, and the position r=2 means the terms start at the 2nd).

Prove that \dbinom n0+\dfrac12\dbinom n1+\dfrac13\dbinom n2+\cdots+\dfrac1{n+1}\dbinom nn=\dfrac{2^{n+1}-1}{n+1}

Solution

The general term of the left side is \dfrac{1}{r+1}\dbinom nr. Change it using factorials.
\dfrac{1}{r+1}\dbinom nr=\dfrac{1}{r+1}\cdot\dfrac{n!}{r!\,(n-r)!}=\dfrac{n!}{(r+1)!\,(n-r)!}
Multiply and divide by (n+1) to turn the numerator into (n+1)!.
=\dfrac{1}{n+1}\cdot\dfrac{(n+1)!}{(r+1)!\,\big((n+1)-(r+1)\big)!}=\dfrac{1}{n+1}\dbinom{n+1}{r+1}
Add this over r=0,1,\ldots,n. The lower index r+1 runs from 1 to n+1.
\text{LHS}=\dfrac{1}{n+1}\left[\dbinom{n+1}{1}+\dbinom{n+1}{2}+\cdots+\dbinom{n+1}{n+1}\right]
The bracket is the sum of all binomial coefficients of (1+x)^{n+1}, which is 2^{n+1} (Question 9), without the first one, \dbinom{n+1}{0}=1.
=\dfrac{1}{n+1}\left(2^{n+1}-1\right)=\dfrac{2^{n+1}-1}{n+1}

Prove that \dbinom n0-\dfrac12\dbinom n1+\dfrac13\dbinom n2-\cdots+\dfrac{(-1)^n}{n+1}\dbinom nn=\dfrac1{n+1}

Solution

As in Question 14, each term is \dfrac{(-1)^r}{r+1}\dbinom nr=\dfrac{(-1)^r}{n+1}\dbinom{n+1}{r+1}.
\text{LHS}=\dfrac{1}{n+1}\left[\dbinom{n+1}{1}-\dbinom{n+1}{2}+\dbinom{n+1}{3}-\cdots+(-1)^n\dbinom{n+1}{n+1}\right]
Put a=1, b=-1 in the binomial theorem for (a+b)^{n+1} (n+1\ge1, so the left side is 0).
0=\dbinom{n+1}{0}-\dbinom{n+1}{1}+\dbinom{n+1}{2}-\dbinom{n+1}{3}+\cdots+(-1)^{n+1}\dbinom{n+1}{n+1}
Move the first term \dbinom{n+1}{0}=1 to the other side and multiply by -1. The bracket of the first line is exactly 1.
\dbinom{n+1}{1}-\dbinom{n+1}{2}+\cdots+(-1)^{n}\dbinom{n+1}{n+1}=\dbinom{n+1}{0}=1
Substitute into the left side.
\text{LHS}=\dfrac{1}{n+1}\cdot1=\dfrac1{n+1}

Prove that \dbinom n0+\dfrac12\dbinom n1+\dfrac1{2^2}\dbinom n2+\cdots+\dfrac1{2^n}\dbinom nn=\left(\dfrac32\right)^n

Solution

Write the general term as \dbinom nr\left(\dfrac12\right)^r and compare with the binomial theorem (1+x)^n=\dbinom n0+\dbinom n1x+\dbinom n2x^2+\cdots+\dbinom nnx^n.
\dbinom n0+\dbinom n1\left(\tfrac12\right)+\dbinom n2\left(\tfrac12\right)^2+\cdots+\dbinom nn\left(\tfrac12\right)^n
This is the expansion of (1+x)^n with x=\dfrac12.
=\left(1+\dfrac12\right)^n=\left(\dfrac32\right)^n

Prove that \dbinom n0^2+\dbinom n1^2+\dbinom n2^2+\cdots+\dbinom nn^2=\dbinom{2n}{n}

Solution

Write (1+x)^{2n}=(1+x)^n(1+x)^n and find the coefficient of x^n on each side. On the left, the coefficient of x^n in (1+x)^{2n} is \dbinom{2n}{n}.
\text{coeff of }x^n\text{ in }(1+x)^{2n}=\dbinom{2n}{n}
On the right, write both factors as binomial expansions, one of them with the terms in reverse order.
(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom nnx^n,\qquad(1+x)^n=\dbinom nn x^n+\dbinom n{n-1}x^{n-1}+\cdots+\dbinom n0
Multiply the two expansions. x^n arises from \dbinom n0x^0\cdot\dbinom nnx^n, \dbinom n1x\cdot\dbinom n{n-1}x^{n-1}, and so on.
\text{coeff of }x^n=\dbinom n0\dbinom nn+\dbinom n1\dbinom n{n-1}+\dbinom n2\dbinom n{n-2}+\cdots+\dbinom nn\dbinom n0
Use \dbinom nr=\dbinom n{n-r} (the coefficients equidistant from the ends are equal), so every product is a square.
=\dbinom n0^2+\dbinom n1^2+\dbinom n2^2+\cdots+\dbinom nn^2
The two ways of finding the coefficient of x^n must agree.
\dbinom n0^2+\dbinom n1^2+\cdots+\dbinom nn^2=\dbinom{2n}{n}

Use pascals triangle to find the number of heads when six coins are tossed simultaneously.

Solution

As in the book's example with three coins, the row of Pascal's triangle that has n+1 entries lists the number of outcomes with 0,1,2,\ldots,n heads when n coins are tossed. For six coins we need the row n=6 (the 6th row in the book's counting, which starts with 1\ 1).

The entries of the row n=6 (outlined) give the number of ways of getting 0, 1, 2, 3, 4, 5, 6 heads with six coins.

Build the rows by adding the two numbers above each entry until the row n=6 is reached.
1;\quad1\ 1;\quad1\ 2\ 1;\quad1\ 3\ 3\ 1;\quad1\ 4\ 6\ 4\ 1;\quad1\ 5\ 10\ 10\ 5\ 1;\quad1\ 6\ 15\ 20\ 15\ 6\ 1
Read the last row: the entries are the numbers of outcomes with 0,1,2,3,4,5,6 heads.
\begin{array}{c|ccccccc}\text{heads}&0&1&2&3&4&5&6\\\hline\text{ways}&1&6&15&20&15&6&1\end{array}
Check: the total number of outcomes is 2^6=64.
1+6+15+20+15+6+1=64=2^6

1,\ 6,\ 15,\ 20,\ 15,\ 6,\ 1

Book answer. 1,6,15,20,15,6,1

Note. Agrees with the key. The book's example for three coins reads the row 1,3,3,1 in the same way.

If 7 coins are tossed how many times 5 heads will appear.

Solution

The number of outcomes with exactly 5 heads when 7 coins are tossed is the entry for 5 heads in the row n=7 of Pascal's triangle.

In the row n=7 the entry for 5 heads (outlined) is 21.

Row n=7 of Pascal's triangle (built from the row 1,6,15,20,15,6,1 above it).
1,\ 7,\ 21,\ 35,\ 35,\ 21,\ 7,\ 1\qquad(\text{for }0,1,2,\ldots,7\text{ heads})
The entry for 5 heads is the sixth number in the row. It is also \dbinom75=\dbinom72.
\dbinom75=\dfrac{7\cdot6}{2\cdot1}=21
So 5 heads appear in 21 of the 2^7=128 equally likely outcomes.
21\text{ times}

21\ \text{times}

Book answer. 21 times

If a coin is tossed 8 times how many times 3 tails will appear.

Solution

Tossing one coin 8 times is the same as tossing 8 coins once, so the number of outcomes with exactly 3 tails is the entry for 3 tails in the row n=8 of Pascal's triangle.

In the row n=8 the entry for 3 tails (outlined) is 56.

Row n=8 of Pascal's triangle (built from the row 1,7,21,35,35,21,7,1 above it).
1,\ 8,\ 28,\ 56,\ 70,\ 56,\ 28,\ 8,\ 1\qquad(\text{for }0,1,\ldots,8\text{ tails})
The entry for 3 tails is the fourth number. It is also \dbinom83.
\dbinom83=\dfrac{8\cdot7\cdot6}{3\cdot2\cdot1}=56
So 3 tails appear in 56 of the 2^8=256 equally likely outcomes.
56\text{ times}

56\ \text{times}

Book answer. 56 times

Exercise 7.3p. 189–190

Binomial series (1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\cdots for |x|<1 when n is negative or a fraction: expand (taking out a factor to get a leading 1), state the range of convergence, approximate numbers, neglect higher powers, and sum a series by recognising n and x.

Expand the followings upto four terms and also find the values of x for which the series is convergent.

Solution

(i) (1-\sqrt x)^{-3}

Use (1+u)^n=1+nu+\dfrac{n(n-1)}{2!}u^2+\dfrac{n(n-1)(n-2)}{3!}u^3+\cdots with n=-3 and u=-\sqrt x.
(1+u)^{-3}=1+(-3)u+\dfrac{(-3)(-4)}{2!}u^2+\dfrac{(-3)(-4)(-5)}{3!}u^3=1-3u+6u^2-10u^3
Put u=-\sqrt x, so u^2=x and u^3=-x^{3/2}.
(1-\sqrt x)^{-3}=1+3\sqrt x+6x+10x^{3/2}+\cdots
The series converges when |u|<1, that is |\sqrt x|<1. Since \sqrt x must be real, x\ge0.
0\le x<1

1+3x^{1/2}+6x+10x^{3/2}+\cdots,\quad0\le x<1

(ii) \left(3+\dfrac2x\right)^{-1/3}

Take 3 out so that the bracket has the form 1+u (the series needs a leading 1): (3+\frac2x)^{-1/3}=3^{-1/3}\left(1+\frac{2}{3x}\right)^{-1/3}. Here u=\dfrac{2}{3x}, n=-\dfrac13.
(1+u)^{-1/3}=1-\dfrac13u+\dfrac{(-\frac13)(-\frac43)}{2!}u^2+\dfrac{(-\frac13)(-\frac43)(-\frac73)}{3!}u^3=1-\dfrac u3+\dfrac{2u^2}{9}-\dfrac{14u^3}{81}
Substitute u=\dfrac2{3x}: \dfrac u3=\dfrac{2}{9x}, \dfrac{2u^2}9=\dfrac{8}{81x^2}, \dfrac{14u^3}{81}=\dfrac{112}{2187x^3}.
\left(3+\dfrac2x\right)^{-1/3}=3^{-1/3}\left[1-\dfrac{2}{9x}+\dfrac{8}{81x^2}-\dfrac{112}{2187x^3}+\cdots\right]
Multiply by 3^{-1/3} (using 9=3^2, 81=3^4, 2187=3^7).
=\dfrac1{3^{1/3}}-\dfrac{2}{3^{7/3}x}+\dfrac{8}{3^{13/3}x^2}-\dfrac{112}{3^{22/3}x^3}+\cdots
Convergence needs |u|<1.
\left|\dfrac{2}{3x}\right|<1\;\Rightarrow\;|x|>\dfrac23

3^{-1/3}-\dfrac{2}{3^{7/3}x}+\dfrac{8}{3^{13/3}x^2}-\dfrac{112}{3^{22/3}x^3}+\cdots,\quad|x|>\tfrac23

(iii) \left(\dfrac52-\dfrac3{x^2}\right)^{1/2}

Take out \dfrac52 so that a leading 1 appears: \left(\frac52-\frac3{x^2}\right)^{1/2}=\left(\frac52\right)^{1/2}\left(1-\frac{6}{5x^2}\right)^{1/2}. Here u=-\dfrac{6}{5x^2} and n=\dfrac12.
(1+u)^{1/2}=1+\dfrac u2-\dfrac{u^2}{8}+\dfrac{u^3}{16}-\cdots
Substitute: \dfrac u2=-\dfrac{3}{5x^2}, \dfrac{u^2}8=\dfrac{36}{200x^4}=\dfrac{9}{50x^4}, \dfrac{u^3}{16}=-\dfrac{216}{2000x^6}=-\dfrac{27}{250x^6}.
\left(1-\dfrac{6}{5x^2}\right)^{1/2}=1-\dfrac{3}{5x^2}-\dfrac{9}{50x^4}-\dfrac{27}{250x^6}-\cdots
Multiply by \left(\dfrac52\right)^{1/2} and keep four terms.
\left(\dfrac52-\dfrac3{x^2}\right)^{1/2}=\sqrt{\dfrac52}\left[1-\dfrac{3}{5x^2}-\dfrac{9}{50x^4}-\dfrac{27}{250x^6}\right]-\cdots
Convergence needs |u|<1.
\dfrac{6}{5x^2}<1\;\Rightarrow\;x^2>\dfrac65

\sqrt{\tfrac52}\left(1-\dfrac{3}{5x^2}-\dfrac{9}{50x^4}-\dfrac{27}{250x^6}-\cdots\right),\quad x^2>\tfrac65

(iv) \dfrac{3+x}{3-x}

Write as (3+x)(3-x)^{-1} and take 3 out of each bracket.
\dfrac{3+x}{3-x}=\dfrac{3\left(1+\frac x3\right)}{3\left(1-\frac x3\right)}=\left(1+\dfrac x3\right)\left(1-\dfrac x3\right)^{-1}
Expand (1-u)^{-1}=1+u+u^2+u^3+\cdots with u=\dfrac x3.
\left(1-\dfrac x3\right)^{-1}=1+\dfrac x3+\dfrac{x^2}{9}+\dfrac{x^3}{27}+\cdots
Multiply by \left(1+\dfrac x3\right): each power of x gets the two contributions \dfrac1{3^r}+\dfrac1{3^r} for r\ge1.
\left(1+\dfrac x3\right)\left(1+\dfrac x3+\dfrac{x^2}{9}+\dfrac{x^3}{27}+\cdots\right)=1+\dfrac{2x}{3}+\dfrac{2x^2}{9}+\dfrac{2x^3}{27}+\cdots
Convergence needs \left|\dfrac x3\right|<1.
-3<x<3

1+\dfrac{2x}{3}+\dfrac{2x^2}{3^2}+\dfrac{2x^3}{3^3}+\cdots,\quad-3<x<3

(v) \dfrac{1-2x}{\sqrt{3+\frac x2}}

Write the root as a power and take 3 out: (3+\frac x2)^{-1/2}=3^{-1/2}\left(1+\frac x6\right)^{-1/2}. So the expression is \dfrac1{\sqrt3}(1-2x)\left(1+\dfrac x6\right)^{-1/2}.
(1+u)^{-1/2}=1-\dfrac u2+\dfrac{3u^2}{8}-\dfrac{5u^3}{16}+\cdots
Put u=\dfrac x6.
\left(1+\dfrac x6\right)^{-1/2}=1-\dfrac{x}{12}+\dfrac{x^2}{96}-\dfrac{5x^3}{3456}+\cdots
Multiply by (1-2x) and collect each power of x: x: -\frac1{12}-2; x^2: \frac1{96}+\frac16; x^3: -\frac5{3456}-\frac1{48}.
(1-2x)\left(\ldots\right)=1-\dfrac{25}{12}x+\dfrac{17}{96}x^2-\dfrac{77}{3456}x^3+\cdots
Multiply by \dfrac1{\sqrt3}. Convergence needs \left|\dfrac x6\right|<1.
|x|<6

\dfrac1{\sqrt3}\left[1-\dfrac{25}{12}x+\dfrac{17}{96}x^2-\dfrac{77}{3456}x^3+\cdots\right],\quad-6<x<6

(vi) \dfrac{\sqrt{x+1}}{\sqrt{1-x}}

Write as a product of two binomial series.
\dfrac{\sqrt{1+x}}{\sqrt{1-x}}=(1+x)^{1/2}(1-x)^{-1/2}
Expand each.
(1+x)^{1/2}=1+\dfrac x2-\dfrac{x^2}{8}+\dfrac{x^3}{16}-\cdots,\qquad(1-x)^{-1/2}=1+\dfrac x2+\dfrac{3x^2}{8}+\dfrac{5x^3}{16}+\cdots
Multiply and collect: x: \frac12+\frac12=1; x^2: \frac38+\frac14-\frac18=\frac12; x^3: \frac5{16}+\frac3{16}-\frac1{16}+\frac1{16}=\frac12.
=1+x+\dfrac{x^2}{2}+\dfrac{x^3}{2}+\cdots
Both series need |x|<1.
-1<x<1

1+x+\dfrac{x^2}{2}+\dfrac{x^3}{2}+\cdots,\quad-1<x<1

Book answer. (i) 1+3x^{1/2}+6x+10x^{3/2}, 0<x<1; (ii) \dfrac1{3^{1/3}}-\dfrac{2}{3^{7/3}x}+\dfrac{24}{3^{13/3}x^2}-\dfrac{80}{3^{22/3}x^3}; (iii) \left(\tfrac52\right)^{1/2}-\dots, x^2>\tfrac65; (iv) 1+\tfrac{2x}3+\tfrac{2x^2}{3^2}+\tfrac{2x^3}{3^3}+\tfrac{x^4}{3^4}, -3<x<3; (v) -6<x<6; (vi) 1 (as printed)

Note. Parts (i), (iv) and (v) agree in the first terms and in the range. Differences: in (i) the end value x=0 also works. In (ii) the key's third and fourth terms (24 and 80) are wrong: the coefficients are 8 and 112; checking at x=10 the exact value is 0.678604, our four terms give 0.678603 and the key's give 0.67998. The key also prints the range as -\frac23>x>\frac23, which should read |x|>\frac23. In (iii) the key's third term has the opposite sign to ours (the term is negative). In (iv) the extra term printed is \frac{x^4}{3^4}; the fifth term is \frac{2x^4}{3^4} and only four terms are asked. In (v) the key prints \frac13 for the first term (it is \frac1{\sqrt3}) and the exponents of x in its later terms are garbled. In (vi) the key prints only 1, the first term.

Approximate the value upto four places of decimal.

Solution

(i) \sqrt[6]{65}

Write 65 near a perfect sixth power: 64=2^6.
\sqrt[6]{65}=(64+1)^{1/6}=2\left(1+\dfrac1{64}\right)^{1/6}
Expand with n=\dfrac16, u=\dfrac1{64}: (1+u)^{1/6}=1+\dfrac u6+\dfrac{\frac16\left(-\frac56\right)}{2}u^2=1+\dfrac u6-\dfrac{5u^2}{72}.
1+\dfrac{1}{384}-\dfrac{5}{72\cdot4096}=1+0.0026042-0.0000170=1.0025872
Multiply by 2.
2\times1.0025872=2.0051744\approx2.0052

2.0052

(ii) \sqrt{\dfrac53}

Write \sqrt{\frac53}=\dfrac{\sqrt{15}}{3} and use 15=16-1, since 16 is a perfect square.
\sqrt{\dfrac53}=\dfrac{\sqrt{16-1}}{3}=\dfrac43\left(1-\dfrac1{16}\right)^{1/2}
Expand with u=\dfrac1{16}: (1-u)^{1/2}=1-\dfrac u2-\dfrac{u^2}{8}-\dfrac{u^3}{16}.
=1-0.03125-0.000488-0.000015=0.968246
Multiply by \dfrac43.
\dfrac43\times0.968246=1.29099\approx1.2910

1.2910

(iii) (1.03)^{1/3}

(1+0.03)^{1/3}: n=\dfrac13, u=0.03.
1+\dfrac{0.03}{3}+\dfrac{\frac13\left(-\frac23\right)}{2}(0.03)^2+\dfrac{\frac13\left(-\frac23\right)\left(-\frac53\right)}{6}(0.03)^3
Evaluate each term.
=1+0.01-0.0001+0.0000017=1.0099017\approx1.0099

1.0099

(iv) (.95)^{2/7}

(1-0.05)^{2/7}: n=\dfrac27, u=-0.05.
1+\dfrac27(-0.05)+\dfrac{\frac27\left(-\frac57\right)}{2}(0.05)^2+\dfrac{\frac27\left(-\frac57\right)\left(-\frac{12}7\right)}{6}(-0.05)^3
Evaluate each term.
=1-0.0142857-0.0002551-0.0000073=0.9854519
Round to four places.
\approx0.9855

0.9855

Book answer. (i) 2.0052; (ii) 1.2963; (iii) 1.0099; (iv) 0.9859

Note. (i) and (iii) agree. (ii) The key's 1.2963 is the sum of the first four terms of \left(1+\frac23\right)^{1/2}, a series that converges too slowly; the true value of \sqrt{5/3} is 1.2910, which is what the 16-1 method gives. (iv) The key's 0.9859 does not match: (0.95)^{2/7}=e^{(2/7)\ln0.95}=0.98545, so the answer is 0.9855.

Find the term involving x^{14} in the product of (1+x^2)\left(2+\sqrt3x^3\right)^{-\frac12}.

Solution

Take 2 out of the second factor so it has the form (1+u)^n with u=\dfrac{\sqrt3}{2}x^3 and n=-\dfrac12.
(2+\sqrt3x^3)^{-1/2}=2^{-1/2}\left(1+\dfrac{\sqrt3}2x^3\right)^{-1/2}
The general term of (1+u)^{-1/2} is \dfrac{n(n-1)\cdots(n-r+1)}{r!}u^r, which contains x^{3r}.
T_{r+1}=\dfrac{(-\frac12)(-\frac32)\cdots}{r!}\left(\dfrac{\sqrt3}{2}\right)^rx^{3r}
x^{14} in the product arises as 1\cdot x^{3r} or x^2\cdot x^{3r}. The first needs 3r=14 (impossible); the second needs 3r=12, so r=4.
x^2\cdot x^{12}=x^{14}\quad(r=4)
Coefficient of u^4: \dfrac{(-\frac12)(-\frac32)(-\frac52)(-\frac72)}{4!}=\dfrac{105/16}{24}=\dfrac{35}{128}, and \left(\dfrac{\sqrt3}2\right)^4=\dfrac9{16}.
\dfrac{35}{128}\cdot\dfrac{9}{16}=\dfrac{315}{2048}
Multiply by 2^{-1/2}=\dfrac1{\sqrt2} and rationalise.
\dfrac{315}{2048\sqrt2}=\dfrac{315\sqrt2}{4096}

\dfrac{315\sqrt2}{4096}x^{14}

Book answer. \dfrac{315\sqrt2}{4096}

Note. Agrees with the key.

If x is so small that its square and higher powers may be neglected then prove that:

Solution

(i) \dfrac{\sqrt{2+x}\,(1-x)^{3/2}}{3+x}\approx\dfrac{\sqrt2}{3}\left(1-\dfrac{19}{12}x\right)

Make each factor of the form (1+u)^n and keep only terms up to x (higher powers are neglected).
\sqrt{2+x}=\sqrt2\left(1+\dfrac x2\right)^{1/2}\approx\sqrt2\left(1+\dfrac x4\right)
Expand the other two factors in the same way.
(1-x)^{3/2}\approx1-\dfrac{3x}{2},\qquad\dfrac1{3+x}=\dfrac13\left(1+\dfrac x3\right)^{-1}\approx\dfrac13\left(1-\dfrac x3\right)
Multiply and drop x^2.
\text{LHS}\approx\dfrac{\sqrt2}{3}\left(1+\dfrac x4\right)\left(1-\dfrac{3x}2\right)\left(1-\dfrac x3\right)\approx\dfrac{\sqrt2}{3}\left[1+\left(\dfrac14-\dfrac32-\dfrac13\right)x\right]
\dfrac14-\dfrac32-\dfrac13=\dfrac{3-18-4}{12}=-\dfrac{19}{12}.
\text{LHS}\approx\dfrac{\sqrt2}{3}\left(1-\dfrac{19}{12}x\right)\ \checkmark

(ii) \dfrac{\left(1+\frac23x\right)^{-5}+\sqrt{4+2x}}{(4+x)^{3/2}}\approx\dfrac18\left(3-\dfrac{95}{24}x\right)

Expand the numerator terms up to x.
\left(1+\dfrac23x\right)^{-5}\approx1-\dfrac{10}{3}x,\qquad\sqrt{4+2x}=2\left(1+\dfrac x2\right)^{1/2}\approx2+\dfrac x2
Add them.
\text{numerator}\approx3-\dfrac{10}3x+\dfrac x2=3-\dfrac{17}{6}x
Expand the denominator: (4+x)^{3/2}=8\left(1+\frac x4\right)^{3/2}\approx8\left(1+\frac{3x}8\right), so its reciprocal is
\dfrac1{(4+x)^{3/2}}\approx\dfrac18\left(1-\dfrac{3x}8\right)
Multiply and drop x^2: -\frac{17}6-\frac98=-\frac{68+27}{24}=-\frac{95}{24}.
\text{LHS}\approx\dfrac18\left(3-\dfrac{17}6x\right)\left(1-\dfrac{3x}8\right)\approx\dfrac18\left(3-\dfrac{95}{24}x\right)\ \checkmark

(iii) \dfrac{\sqrt{9-x}+\left(1+\frac34x\right)^{-5}}{2+x}\approx2-\dfrac{71}{24}x

Expand each part up to x.
\sqrt{9-x}=3\left(1-\dfrac x9\right)^{1/2}\approx3-\dfrac x6,\qquad\left(1+\dfrac34x\right)^{-5}\approx1-\dfrac{15}{4}x
Add: -\frac16-\frac{15}4=-\frac{2+45}{12}.
\text{numerator}\approx4-\dfrac{47}{12}x
\dfrac1{2+x}=\dfrac12\left(1+\dfrac x2\right)^{-1}\approx\dfrac12\left(1-\dfrac x2\right). Multiply: -\frac{47}{12}-2=-\frac{71}{12}.
\text{LHS}\approx\dfrac12\left(4-\dfrac{47}{12}x\right)\left(1-\dfrac x2\right)\approx\dfrac12\left(4-\dfrac{71}{12}x\right)=2-\dfrac{71}{24}x\ \checkmark

Note. Each statement is proved by expanding to the first power of x.

If x is so small that its cube and higher powers may be neglected then show that:

Solution

(i) \dfrac{(1+x)^{3/2}-(1+x^2)^3}{\sqrt{1-x}}\approx\dfrac32x-\dfrac{15}8x^2

Expand each part up to x^2.
(1+x)^{3/2}\approx1+\dfrac32x+\dfrac38x^2,\qquad(1+x^2)^3\approx1+3x^2,\qquad\dfrac1{\sqrt{1-x}}=(1-x)^{-1/2}\approx1+\dfrac x2+\dfrac38x^2
Numerator: subtract.
(1+x)^{3/2}-(1+x^2)^3\approx\dfrac32x+\left(\dfrac38-3\right)x^2=\dfrac32x-\dfrac{21}8x^2
Multiply by the expansion of (1-x)^{-1/2} and drop x^3: the x^2 term is \frac32\cdot\frac12-\frac{21}8=\frac68-\frac{21}8.
\left(\dfrac32x-\dfrac{21}8x^2\right)\left(1+\dfrac x2\right)\approx\dfrac32x-\dfrac{15}{8}x^2\ \checkmark

(ii) \dfrac{(4+x)^{-2}+(1-2x)^{-5}}{(1+2x)^2}\approx\dfrac{17}{16}+\dfrac{183}{32}x+\dfrac{8419}{256}x^2

Expand up to x^2.
(4+x)^{-2}=\dfrac1{16}\left(1+\dfrac x4\right)^{-2}\approx\dfrac1{16}\left(1-\dfrac x2+\dfrac{3x^2}{16}\right),\qquad(1-2x)^{-5}\approx1+10x+60x^2
Add (the x term is -\frac1{32}+10, the x^2 term is \frac3{256}+60).
\text{numerator}\approx\dfrac{17}{16}+\dfrac{319}{32}x+\dfrac{15363}{256}x^2
The denominator: (1+2x)^{-2}\approx1-4x+12x^2.
\dfrac1{(1+2x)^2}\approx1-4x+12x^2
Multiply and collect: x: \frac{319}{32}-4\cdot\frac{17}{16}=\frac{183}{32}; x^2: \frac{15363}{256}-4\cdot\frac{319}{32}+12\cdot\frac{17}{16}=\frac{15363-10208+3264}{256}.
\approx\dfrac{17}{16}+\dfrac{183}{32}x+\dfrac{8419}{256}x^2\ \checkmark

If x is so large that \left(\dfrac1x\right)^2 and higher powers may be neglected then show that \sqrt{x^2+25}-\sqrt{x^2+9}\approx\dfrac8x

Solution

Take x^2 out of each root so that the bracket is 1+u with u small (since x is large, \frac{1}{x^2} is small).
\sqrt{x^2+25}=x\left(1+\dfrac{25}{x^2}\right)^{1/2},\qquad\sqrt{x^2+9}=x\left(1+\dfrac9{x^2}\right)^{1/2}
Expand to the first power of \dfrac1{x^2} and neglect the rest.
\sqrt{x^2+25}\approx x\left(1+\dfrac{25}{2x^2}\right),\qquad\sqrt{x^2+9}\approx x\left(1+\dfrac{9}{2x^2}\right)
Subtract.
\sqrt{x^2+25}-\sqrt{x^2+9}\approx x\cdot\dfrac{25-9}{2x^2}=\dfrac{16}{2x}=\dfrac8x\ \checkmark

Find the term involving x^n while simplifying \dfrac{(2+x)^2}{(1+x)^3}.

Solution

Write the fraction as (2+x)^2(1+x)^{-3} and expand (1+x)^{-3}.
(1+x)^{-3}=\sum_{r\ge0}(-1)^r\dfrac{(r+1)(r+2)}{2}x^r
Check the coefficient: \dfrac{(-3)(-4)\cdots(-3-r+1)}{r!}=(-1)^r\dfrac{3\cdot4\cdots(r+2)}{r!}=(-1)^r\dfrac{(r+1)(r+2)}2.
1-3x+6x^2-10x^3+\cdots
Expand (2+x)^2=4+4x+x^2. The term in x^n comes from 4\cdot x^n-term, 4x\cdot x^{n-1}-term and x^2\cdot x^{n-2}-term.
\text{coeff of }x^n=4(-1)^n\dfrac{(n+1)(n+2)}2+4(-1)^{n-1}\dfrac{n(n+1)}2+(-1)^{n-2}\dfrac{(n-1)n}2
Take out (-1)^n (note (-1)^{n-1}=-(-1)^n, (-1)^{n-2}=(-1)^n) and simplify the bracket.
=(-1)^n\left[2(n+1)(n+2)-2n(n+1)+\dfrac{n(n-1)}2\right]=(-1)^n\left[4(n+1)+\dfrac{n^2-n}{2}\right]
Combine over 2.
=\dfrac{(-1)^n}{2}\left(8n+8+n^2-n\right)=\dfrac{(-1)^n}2\left(n^2+7n+8\right)

\dfrac{(-1)^n}{2}\left(n^2+7n+8\right)x^n

Book answer. \dfrac{(-1)^n}{2}(n^2+7n+8)x^n

Note. Agrees with the key. Check: n=1 gives -8, and 4(-3)+4=-8 directly.

Identify as binomial series and find the sum of the following:

Solution

(i) 1-\dfrac37+4\left(\dfrac37\right)^2-8\left(\dfrac37\right)^3+\cdots

Compare with (1+x)^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\cdots. From the first two terms nx=-\dfrac37. From the third, \dfrac{n(n-1)}2x^2=4\cdot\dfrac9{49}, i.e. n(n-1)x^2=\dfrac{72}{49}.
\dfrac{n(n-1)x^2}{(nx)^2}=\dfrac{72/49}{9/49}=8=\dfrac{n-1}{n}\;\Rightarrow\;n=-\dfrac17,\qquad x=3
With n=-\dfrac17, x=3 the first three terms are 1-\dfrac37+\dfrac{4\cdot9}{49}, as printed. The series is the binomial series for (1+3)^{-1/7}.
\left(1+3\right)^{-1/7}=4^{-1/7}=\dfrac{1}{\sqrt[7]{4}}

\dfrac1{\sqrt[7]{4}}

(ii) 1-\dfrac1{15}+\dfrac{4}{2!\,15^2}-\dfrac{4\cdot7}{3!\,15^3}+\cdots

From nx=-\dfrac1{15} and \dfrac{n(n-1)}{2!}x^2=\dfrac{4}{2!\,15^2} we get \dfrac{n-1}{n}=4.
n-1=4n\;\Rightarrow\;n=-\dfrac13,\qquad x=\dfrac{-1/15}{-1/3}=\dfrac15
Check the third term: \dfrac{n(n-1)(n-2)}{3!}x^3=\dfrac{\left(-\frac13\right)\left(-\frac43\right)\left(-\frac73\right)}{3!}\cdot\dfrac{1}{125}=-\dfrac{4\cdot7}{3!\cdot27\cdot125}, and 27\cdot125=15^3.
-\dfrac{4\cdot7}{3!\,15^3}\ \checkmark
So the series is (1+x)^n with x=\dfrac15, n=-\dfrac13 (|x|<1).
\left(1+\dfrac15\right)^{-1/3}=\left(\dfrac65\right)^{-1/3}=\left(\dfrac56\right)^{1/3}

\left(\dfrac56\right)^{1/3}

(iii) 1-\dfrac{2\cdot4}{5}+\dfrac{3\cdot4^2}{5^2}-\dfrac{4\cdot4^3}{5^3}+\cdots

The coefficients 1,2,3,4,\ldots with alternating signs suggest (1+y)^{-2}=1-2y+3y^2-4y^3+\cdots. Compare: y=\dfrac45.
1-2\left(\dfrac45\right)+3\left(\dfrac45\right)^2-4\left(\dfrac45\right)^3+\cdots
This is the binomial series with n=-2 and x=\dfrac45 (|x|<1, so it converges).
\left(1+\dfrac45\right)^{-2}=\left(\dfrac95\right)^{-2}=\dfrac{25}{81}
Check with partial sums: 1-1.6+1.92-2.048+\cdots oscillates and settles near 0.3086=\dfrac{25}{81}.
\dfrac{25}{81}=0.3086

\dfrac{25}{81}

(iv) 3-\dfrac3{18}-\dfrac{3}{2!\,18^2}-\dfrac{3^2}{3!\,18^3}-\cdots

Take out 3 and look at the terms 1-\dfrac1{18}-\dfrac{1}{2!\,18^2}\cdots. Compare with (1+x)^{1/2}=1+\dfrac x2-\dfrac{x^2}{8}+\dfrac{x^3}{16}-\cdots with x=-\dfrac19: \dfrac x2=-\dfrac1{18}.
\dfrac x2=-\dfrac1{18}\;\Rightarrow\;x=-\dfrac19,\ n=\dfrac12
Check the next terms: -\dfrac{x^2}8=-\dfrac1{8\cdot81}=-\dfrac1{2!\,18^2} and \dfrac{x^3}{16}=-\dfrac1{16\cdot729}=-\dfrac{3}{3!\,18^3}. So with the factor 3 the series is 3\left(1-\dfrac19\right)^{1/2}.
3\left(1-\dfrac19\right)^{1/2}=3\sqrt{\dfrac89}
Simplify.
3\cdot\dfrac{2\sqrt2}{3}=2\sqrt2

2\sqrt2

Book answer. (i) \dfrac1{\sqrt[7]{4}}; (ii) \left(\dfrac56\right)^{1/3}; (iii) (-3)^{6/5} (as printed); (iv) 2\sqrt2

Note. (ii) and (iv) agree. (i) agrees with the key as the sum (1+3)^{-1/7}; the first three terms match the series as printed, but the next printed coefficient 8 should be 20 for the series to be exactly this one, and since x=3 lies outside |x|<1 the series does not actually converge, so 4^{-1/7} is the formal binomial sum. (iii) The key's (-3)^{6/5} is not a value that this series can have; the series sums to \left(1+\frac45\right)^{-2}=\frac{25}{81} (checked numerically).

Show that:

Solution

(i) If y=\dfrac3{2^2\cdot1!}+\dfrac3{2^4\cdot2!}+\dfrac3{2^6\cdot3!}+\cdots, then show that 8y^2+16y-19=0.

Add 1 to both sides and compare 1+y with (1+x)^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\cdots. From the first term nx=\dfrac34; from the second \dfrac{n(n-1)}{2}x^2=\dfrac3{32}, so n(n-1)x^2=\dfrac3{16}.
\dfrac{n-1}{n}=\dfrac{3/16}{9/16}=\dfrac13\;\Rightarrow\;n=\dfrac32,\qquad x=\dfrac{3/4}{3/2}=\dfrac12
So 1+y is the binomial series of \left(1+\dfrac12\right)^{3/2} (convergent, |x|<1).
1+y=\left(\dfrac32\right)^{3/2}
Square both sides to remove the root.
(1+y)^2=\left(\dfrac32\right)^3=\dfrac{27}8\;\Rightarrow\;8(1+2y+y^2)=27
Expand and collect.
8y^2+16y+8-27=0\;\Rightarrow\;8y^2+16y-19=0\ \checkmark

(ii) If y=\dfrac13+\dfrac{1\cdot3}{3\cdot6}+\dfrac{1\cdot3\cdot5}{3\cdot6\cdot9}+\cdots then show that y^2-2y-2=0.

Write each term with the factors of 2 separated: \dfrac{1\cdot3}{3\cdot6}=\dfrac{1\cdot3}{2!\cdot3^2} and \dfrac{1\cdot3\cdot5}{3\cdot6\cdot9}=\dfrac{1\cdot3\cdot5}{3!\cdot3^3}. Compare 1+y with (1-x)^{-n}=1+nx+\dfrac{n(n+1)}{2!}x^2+\cdots.
nx=\dfrac13,\qquad\dfrac{n(n+1)}{2}x^2=\dfrac{1\cdot3}{2\cdot9}\;\Rightarrow\;n(n+1)x^2=\dfrac13
Divide by (nx)^2=\dfrac19.
\dfrac{n+1}{n}=3\;\Rightarrow\;n=\dfrac12,\qquad x=\dfrac{1/3}{1/2}=\dfrac23
Check the third term: \dfrac{n(n+1)(n+2)}{3!}x^3=\dfrac{\frac12\cdot\frac32\cdot\frac52}{6}\cdot\dfrac8{27}=\dfrac{1\cdot3\cdot5}{3!\cdot27}\ \checkmark. So 1+y=(1-x)^{-n}.
1+y=\left(1-\dfrac23\right)^{-1/2}=3^{1/2}=\sqrt3
Square: (1+y)^2=3.
y^2+2y+1=3\;\Rightarrow\;y^2+2y-2=0

y^2+2y-2=0

Note. In (i) the third printed term has a plus sign, but the series of (1+\frac12)^{3/2} has -\dfrac{3}{2^6\cdot3!} there (the signs go +,+,-,+,\ldots); the stated result 8y^2+16y-19=0 is the one for that series. In (ii) the printed result y^2-2y-2=0 is a misprint: y=\sqrt3-1=0.732, which satisfies y^2+2y-2=0 (a numerical sum of the series gives 0.7321), whereas y^2-2y-2=0 would need y=1+\sqrt3.

Approximations for numbers very close together.

Solution

(i) If x is very nearly equal to 1 then show that \dfrac{ax^b-bx^a}{x^b-x^a}\approx\dfrac1{1-x}.

Put x=1+h where h is so small that h^2 and higher powers can be neglected. Then x^a=(1+h)^a\approx1+ah and x^b\approx1+bh.
x^a\approx1+ah,\qquad x^b\approx1+bh
Numerator.
ax^b-bx^a\approx a(1+bh)-b(1+ah)=a-b
Denominator.
x^b-x^a\approx(1+bh)-(1+ah)=(b-a)h
Divide, then use h=x-1.
\dfrac{a-b}{(b-a)h}=-\dfrac1h=-\dfrac1{x-1}=\dfrac1{1-x}\ \checkmark

(ii) If p and q are approximately equal; then prove that \dfrac{q+2p}{p+2q}\approx\left(\dfrac pq\right)^{1/3}. Hence approximate the value of \left(\dfrac{2.01}{2}\right)^{1/3}.

Put p=q(1+h) with h small (h^2 neglected).
\dfrac{q+2p}{p+2q}=\dfrac{1+2(1+h)}{(1+h)+2}=\dfrac{3+2h}{3+h}=\left(1+\dfrac{2h}{3}\right)\left(1+\dfrac h3\right)^{-1}
Expand \left(1+\frac h3\right)^{-1}\approx1-\frac h3 and multiply, dropping h^2.
\approx\left(1+\dfrac{2h}3\right)\left(1-\dfrac h3\right)\approx1+\dfrac h3
The right side: \left(\dfrac pq\right)^{1/3}=(1+h)^{1/3}\approx1+\dfrac h3. The two sides agree.
\dfrac{q+2p}{p+2q}\approx\left(\dfrac pq\right)^{1/3}\ \checkmark
Hence. Take p=2.01, q=2.
\left(\dfrac{2.01}{2}\right)^{1/3}\approx\dfrac{2+2(2.01)}{2.01+2(2)}=\dfrac{6.02}{6.01}
Divide.
\dfrac{6.02}{6.01}=1.001664\approx1.0017

\approx1.0017

Note. The book gives no key entry for this question. Check: (1.005)^{1/3}=1.001664.

Exercise 7.4p. 193

Applications of the binomial theorem: unit digits (cyclicity table), remainders (write the base as a multiple of the divisor plus or minus a small number), last two digits, comparing large numbers (expand around a round number) and divisibility tests.

Find the unit place digits in 27^{304},108^{33},54^{203} and 503^{43}.

Solution

Only the unit digit of the base matters. The unit digits of the powers of 7, 8, 4, 3 repeat (cyclicity) as in the table of the book: powers of 7 give 7,9,3,1; powers of 8 give 8,4,2,6; powers of 4 give 4,6; powers of 3 give 3,9,7,1.

27^{304} ends in the same digit as 7^{304}. The cycle of 7 has length 4. Write the exponent as 4n+r: 304=4\times76+0, so the digit is that of 7^4.
7^4=2401\;\Rightarrow\;\text{unit digit }1
108^{33} ends like 8^{33}. 33=4\times8+1, so the digit is that of 8^1.
8^1=8\;\Rightarrow\;\text{unit digit }8
54^{203} ends like 4^{203}. For 4 the digit is 4 for odd exponents and 6 for even exponents; 203 is odd.
\text{unit digit }4
503^{43} ends like 3^{43}. 43=4\times10+3, so the digit is that of 3^3.
3^3=27\;\Rightarrow\;\text{unit digit }7

1,\ 8,\ 4,\ 7

Book answer. 1,8,4,7

Note. Agrees with the key.

22026 Annual I, Set B, Q.2 (iv)2026 Annual I, Set D, Q.2 (xi)

Find the remainder when:

Solution

a. 9^{205} is divided by 31.

Write 9^{205}=3^{410}=(3^5)^{82} and 3^5=243=8\times31-5=248-5.
9^{205}=(243)^{82}=(248-5)^{82}
Expand with the binomial theorem. Every term except the last has the factor 248, a multiple of 31.
(248-5)^{82}=31k+(-5)^{82}=31k+5^{82}
Reduce 5^{82}: 5^3=125=124+1=4\times31+1, and 82=3\times27+1.
5^{82}=5\cdot(5^3)^{27}=5(124+1)^{27}=5(31m+1)=31(5m)+5
So 9^{205}=31(k+5m)+5.
\text{remainder}=5

b. 8^{205} is divided by 48.

48=16\times3. Write 8^{205}=2^{615}=2^5\cdot2^{610}=32\cdot4^{305} and 4=3+1.
8^{205}=32\,(3+1)^{305}
By the binomial theorem (3+1)^{305}=3t+1 for an integer t.
8^{205}=32(3t+1)=96t+32
96=2\times48, so 96t is a multiple of 48 and the remainder is 32 (which is less than 48).
8^{205}=48(2t)+32

Book answer. a. 5; b. 2

Note. a. agrees. b. The key's 2 is not the remainder on division by 48. Direct check: 8^{205}=2^{615}\equiv0\pmod{16} and \equiv2\pmod3, and the only number below 48 with both properties is 32; also 8^3=512=10\times48+32 and 8^{k}\equiv32\pmod{48} for odd k\ge3. (A remainder of 2 is what 8^{205} leaves on division by 6.)

If fractional part of number \dfrac{2^{510}}{31} is \dfrac{k}{31}, then find the value of k.

Solution

Write the numerator with the base 2^5=32=31+1.
2^{510}=(2^5)^{102}=32^{102}=(31+1)^{102}
Expand: all terms except the last contain 31.
(31+1)^{102}=31m+1
Divide by 31.
\dfrac{2^{510}}{31}=\dfrac{31m+1}{31}=m+\dfrac1{31}
m is an integer, so the fractional part is \dfrac{1}{31}.
k=1

k=1

Book answer. 1

Note. Agrees with the key.

Find the last one or two digits of the number where applicable.

Solution

a. 15^8

Last digit. The unit digit of any power of 5 is 5, so the unit digit is 5.
5
Last two digits. 15^8=(15^2)^4=225^4=(200+25)^4. By the binomial theorem every term except 25^4 contains a power of 200, a multiple of 100.
(200+25)^4=100k+25^4=100k+390625
390625=100\times3906+25, so the last two digits are 25.
25

5\ \text{and}\ 25

b. 37^7

Last digit. It is that of 7^7. The cycle of 7 is 7,9,3,1 and 7=4+3, so the digit is that of 7^3=343.
3
Last two digits. Write 37=40-3. Every term of (40-3)^7 containing 40^2=1600 is a multiple of 100, so only the last two terms matter.
(40-3)^7=100k+\dbinom76(40)(-3)^6+(-3)^7
Evaluate: 7\cdot40\cdot729=204120 and (-3)^7=-2187.
204120-2187=201933
The last two digits of 201933 are 33.
33

3\ \text{and}\ 33

c. 29^{10}

Last digit. 29 ends in 9, and 9^{10} ends in 1 (the powers of 9 end alternately in 9, 1).
1
Last two digits. Write 29=30-1. In (30-1)^{10} every term containing 30^2=900 is a multiple of 100. Only the last two terms remain.
(30-1)^{10}=100k+\dbinom{10}{9}(30)(-1)^9+(-1)^{10}=100k-300+1
-300 is a multiple of 100, so the last two digits are those of +1, that is 01.
01

1\ \text{and}\ 01

Book answer. a. 5,\ 25; b. 3,\ 33; c. 1,\ 01

Note. All three agree with the key.

Which of the following is a larger number?

Solution

a. 98^{50}+100^{50} or 102^{50}

Write the bases around 100.
102^{50}=(100+2)^{50},\qquad98^{50}=(100-2)^{50}
Subtract. The even-power terms cancel and the odd-power terms double.
102^{50}-98^{50}=2\left[\dbinom{50}1(100)^{49}(2)+\dbinom{50}3(100)^{47}(2)^3+\cdots\right]
The first term is 2\cdot50\cdot2\cdot100^{49}=200\cdot100^{49}=2\cdot100^{50}, and all other terms are positive.
102^{50}-98^{50}=2\cdot100^{50}+(\text{positive terms})>100^{50}
Rearrange.
102^{50}>98^{50}+100^{50}

102^{50}\ \text{is larger}

b. 47^{30}+50^{30} or 53^{30}

Write the bases around 50.
53^{30}=(50+3)^{30},\qquad47^{30}=(50-3)^{30}
Subtract; the odd-power terms double.
53^{30}-47^{30}=2\left[\dbinom{30}1(50)^{29}(3)+\dbinom{30}3(50)^{27}(3)^3+\cdots\right]
The first term is 2\cdot30\cdot3\cdot50^{29}=180\cdot50^{29}=3.6\cdot50^{30}, which already exceeds 50^{30}; the other terms are positive.
53^{30}-47^{30}>3.6\cdot50^{30}>50^{30}
Rearrange.
53^{30}>47^{30}+50^{30}

53^{30}\ \text{is larger}

Note. The key has no entry for this question.

Show that 12^{15}+8^{15} is divisible by 10.

Solution

Write 12=10+2 and 8=10-2.
12^{15}+8^{15}=(10+2)^{15}+(10-2)^{15}
Expand. The terms with an odd power of 2 have opposite signs and cancel; the even-power terms double.
=2\left[10^{15}+\dbinom{15}{2}10^{13}2^2+\dbinom{15}{4}10^{11}2^4+\cdots+\dbinom{15}{14}\,10\,\cdot2^{14}\right]
The last term of the bracket still contains the factor 10, so every term has a factor 10.
=10\times(\text{an integer})
Hence the number is divisible by 10.
12^{15}+8^{15}\ \text{is divisible by }10

Show that 22^{25}+18^{25} is divisible by 20.

Solution

Write 22=20+2 and 18=20-2.
22^{25}+18^{25}=(20+2)^{25}+(20-2)^{25}
Odd powers of 2 cancel; even powers double.
=2\left[20^{25}+\dbinom{25}{2}20^{23}2^2+\cdots+\dbinom{25}{24}\,20\cdot2^{24}\right]
Every term contains at least 20^1.
=20\times(\text{an integer})
Hence it is divisible by 20.
22^{25}+18^{25}\ \text{is divisible by }20

Use binomial theorem to find the remainder when 5^{103} is divided by 13.

Solution

Find a power of 5 close to a multiple of 13: 5^2=25=26-1.
5^{103}=5\cdot5^{102}=5\,(5^2)^{51}=5\,(26-1)^{51}
Expand. All terms except the last contain 26, a multiple of 13; the last is (-1)^{51}=-1.
(26-1)^{51}=26k-1
Multiply by 5.
5^{103}=5(26k-1)=130k-5
A remainder must be positive and less than 13: write -5=-13+8.
5^{103}=130k-13+8=13(10k-1)+8

Book answer. 8

Note. Agrees with the key.

What is the remainder when 17^{1717} is divided by 9.

Solution

Write 17=18-1, where 18 is a multiple of 9.
17^{1717}=(18-1)^{1717}
Expand. All terms except the last contain 18; the last is (-1)^{1717}=-1 because 1717 is odd.
=18k-1
Write -1=-9+8 to get a positive remainder less than 9.
=18k-9+8=9(2k-1)+8

Book answer. 8

Note. Agrees with the key.

Using Binomial Theorem, indicate which number is larger (1.1)^{10000} or 1000.

Solution

Write 1.1=1+0.1 and expand the first two terms.
(1.1)^{10000}=(1+0.1)^{10000}=1+\dbinom{10000}{1}(0.1)+\dbinom{10000}2(0.1)^2+\cdots
All terms are positive, so the sum is larger than the first two terms.
(1.1)^{10000}>1+10000(0.1)=1+1000=1001
Since 1001>1000:
(1.1)^{10000}>1000

(1.1)^{10000}\ \text{is larger}

Show that 9^{n+1}-8n-9 is divisible by 64, whenever n is a positive integer.

Solution

Write 9=1+8 and expand (1+8)^{n+1}.
9^{n+1}=(1+8)^{n+1}=1+(n+1)8+\dbinom{n+1}2 8^2+\dbinom{n+1}3 8^3+\cdots+8^{n+1}
Subtract 8n+9. The first two terms give 1+8n+8-8n-9=0.
9^{n+1}-8n-9=\dbinom{n+1}2 8^2+\dbinom{n+1}3 8^3+\cdots+8^{n+1}
Every remaining term has the factor 8^2=64.
=64\left[\dbinom{n+1}2+8\dbinom{n+1}3+\cdots+8^{n-1}\right]
The bracket is an integer, so the expression is divisible by 64.
9^{n+1}-8n-9=64\times(\text{integer})

If a and b are distinct integers, prove that a-b is a factor of a^n-b^n, whenever n is a positive integer.

Solution

Let d=a-b, an integer different from 0. Then a=b+d.
a^n=(b+d)^n
Expand with the binomial theorem.
(b+d)^n=b^n+\dbinom n1b^{n-1}d+\dbinom n2b^{n-2}d^2+\cdots+d^n
Subtract b^n. Every remaining term contains d.
a^n-b^n=d\left[\dbinom n1b^{n-1}+\dbinom n2b^{n-2}d+\cdots+d^{n-1}\right]
The bracket is an integer, so d=a-b divides a^n-b^n.
a^n-b^n=(a-b)\times(\text{integer})

Show that 6^{n+3}-8n-6 is divisible by 6.

Solution

Test small values first. For n=1: 6^4-8-6=1296-14=1282=6\times213+4, which is not divisible by 6. So the statement cannot be proved as printed.

6^{n+3} is always a multiple of 6 (it has the factor 6). So the remainder on division by 6 depends only on -8n-6.
6^{n+3}-8n-6\equiv-8n\equiv-2n\pmod 6
This is a multiple of 6 only when 2n is, that is when n is a multiple of 3.
\begin{aligned}n=1:&\ 1282\ (\text{remainder }4)\\ n=2:&\ 7776-22=7754\ (\text{remainder }2)\\ n=3:&\ 46656-30=46626=6\times7771\end{aligned}
So the expression is divisible by 6 exactly when 3\mid n; for other n the statement is false.
6\mid\big(6^{n+3}-8n-6\big)\iff3\mid n

Note. The key has no entry. Counter-example: n=1 gives 1282, which leaves remainder 4 on division by 6.

Review Exercise 7p. 194

Mixed review: multiple choice on induction and the binomial theorem (Question 1), one induction proof, ratio and middle-term problems, a term that does not exist, an approximation, and a remainder.

1 (i)

Mathematical induction is used to check a proposition for all n where n is a/an:

Areal number
Brational number
Cinteger
Dpositive integerAnswer

Why

The basis step starts at n=1 and the inductive step moves from k to k+1, so the principle reaches 1,2,3,\ldots, the positive integers. It cannot reach real, rational or negative numbers, which are not reached by repeatedly adding 1.

Book answer. d

1 (ii)

A mathematical statement which is true for all positive integers is also true for all:

Anegative integers
Bpositive numbers
Cwhole numbers
DnoneAnswer

Why

Truth for n=1,2,3,\ldots says nothing about numbers outside that list. For example the statement n>0 is true for every positive integer but false for the whole number 0 and for negative integers. So, in general, it is not guaranteed for any of the other sets.

Book answer. c

Note. The key gives (c) whole numbers. Mathematically that does not follow: the statement 'n>0' holds for all positive integers but fails at n=0. A statement proved by induction from n=1 needs a separate check at n=0 before it can be extended to the whole numbers (as in Exercise 7.1 Question 11, where the basis step is n=0). If you are marked against the book key, write (c).

1 (iii)

If n is even positive integer then the middle term in the expansion of (a+b)^n is:

A\left(\frac n2\right)^{th} term
B\left(\frac{n+1}{2}\right)^{th} term
C\left(\frac{n-1}{2}\right)^{th} term
D\left(\frac n2+1\right)^{th} termAnswer

Why

The expansion has n+1 terms, an odd number when n is even, so there is one middle term. It is the \dfrac{(n+1)+1}{2}=\dfrac n2+1 th term. For example, n=4 has 5 terms and the middle one is the 3rd, and \frac42+1=3. Options (b) and (c) are not whole numbers when n is even.

Book answer. d

1 (iv)

In the expansion of (a+b)^{20} a term is at the 11^{th} position. Its position from the end is:

A9^{th}
B10^{th}
C11^{th}Answer
D12^{th}

Why

The rth term from the beginning is the (n+2-r)th from the end. With n=20 and r=11:

n+2-r=20+2-11=11

The 11th term is the middle term of the 21 terms, so it is 11th from either end.

Book answer. d

Note. The key gives (d) 12^{th}, but 20+2-11=11. Check: there are 21 terms, and 10 terms come before the 11th and 10 after it, so it is 11th from the end as well.

1 (v)

The coefficient of the 3^{rd} last term in the expansion of (1+x)^{300} is:

A277
B44850Answer
C303
D4305600

Why

The 3rd term from the end is the (300+2-3)=299th from the beginning, T_{299}=T_{298+1}.

T_{299}=\dbinom{300}{298}x^{298},\qquad\dbinom{300}{298}=\dbinom{300}{2}=\dfrac{300\cdot299}{2}=44850

Book answer. b

1 (vi)

\dbinom{11}{0}+\dbinom{11}{2}+\dbinom{11}{4}+\cdots+\dbinom{11}{10} is equal to

A2^{11}
B2^{12}
C2^{10}Answer
D2^{11}-1

Why

The sum of all the coefficients of (1+x)^{11} is 2^{11}, and the coefficients with even lower index add up to the same as those with odd lower index (Exercise 7.2 Question 9). So each half is 2^{11}/2.

\dbinom{11}{0}+\dbinom{11}{2}+\cdots+\dbinom{11}{10}=2^{10}

Book answer. c

1 (vii)

If the third term in the expansion of (1+x)^p is -\frac18x^2 then the value of p is:

A2
B\frac12Answer
C4
D3

Why

The third term is \dfrac{p(p-1)}{2!}x^2. Set its coefficient equal to -\dfrac18.

\dfrac{p(p-1)}2=-\dfrac18\;\Rightarrow\;p^2-p+\dfrac14=0\;\Rightarrow\;\left(p-\dfrac12\right)^2=0\;\Rightarrow\;p=\dfrac12

Book answer. b

Note. A non-integer p is allowed here because the series is a binomial series, valid for |x|<1.

1 (viii)

The coefficient of x^n in the expansion of (1+x+x^2+\cdots)^{-n} where n is a even number:

A1Answer
B-1
Cn
D-n+1

Why

For |x|<1, 1+x+x^2+\cdots=\dfrac1{1-x}, so the expression is \left(\dfrac{1}{1-x}\right)^{-n}=(1-x)^n.

(1-x)^n=\dbinom n0-\dbinom n1x+\cdots+(-1)^n\dbinom nnx^n

The coefficient of x^n is (-1)^n, which is +1 when n is even.

Book answer. a

1 (ix)

The greatest coefficient in the expansion of (1+x)^{10} is:

A2^x
B\binom{10}{5}Answer
C\binom{10}{6}
D2^{10}

Why

The coefficients of (1+x)^{10} rise to the middle term and then fall, so the greatest is the middle one, \dbinom{10}{5}=252. Option (c) is smaller (\dbinom{10}{6}=210). Option (d) 2^{10}=1024 is the sum of all the coefficients, not one coefficient.

Book answer. b

1 (x)

Binomial series (2+3x)^{-1/2} is valid when:

A|x|\le1
B|x|<1
C|x|<\frac23Answer
D|x|<\frac32

Why

Take out the 2 so that the bracket starts with 1.

(2+3x)^{-1/2}=2^{-1/2}\left(1+\dfrac{3x}{2}\right)^{-1/2}

The series (1+u)^n converges for |u|<1, here u=\dfrac{3x}{2}.

\left|\dfrac{3x}2\right|<1\;\Rightarrow\;|x|<\dfrac23

Book answer. b

Note. The key gives (b) |x|<1, which is the condition for (1+x)^{-1/2}, not for (2+3x)^{-1/2}. Check: at x=0.9 we have \left|\frac{3x}2\right|=1.35>1, so the series diverges there although |x|<1.

Using principle of mathematical induction prove that for all positive integers n: \dfrac1{1.2.3}+\dfrac1{2.3.4}+\cdots+\dfrac1{n(n+1)(n+2)}=\dfrac{n(n+3)}{4(n+1)(n+2)}

Solution

Proof by mathematical induction

Let P(n) be the statement to be proved.
\dfrac1{1\cdot2\cdot3}+\dfrac1{2\cdot3\cdot4}+\cdots+\dfrac1{n(n+1)(n+2)}=\dfrac{n(n+3)}{4(n+1)(n+2)}
Basis step. Put n=1.
\text{LHS}=\dfrac1{1\cdot2\cdot3}=\dfrac16,\qquad\text{RHS}=\dfrac{1\cdot4}{4\cdot2\cdot3}=\dfrac16\;\Rightarrow\;P(1)\ \text{is true}
Inductive hypothesis. Assume P(k) is true for some integer k\ge 1.
\dfrac1{1\cdot2\cdot3}+\cdots+\dfrac1{k(k+1)(k+2)}=\dfrac{k(k+3)}{4(k+1)(k+2)}
To prove. P(k+1), that is, the same statement with k+1 in place of n.
\dfrac1{1\cdot2\cdot3}+\cdots+\dfrac1{k(k+1)(k+2)}+\dfrac1{(k+1)(k+2)(k+3)}=\dfrac{(k+1)(k+4)}{4(k+2)(k+3)}
Add the next term to both sides of P(k).
\text{LHS of }P(k+1)=\dfrac{k(k+3)}{4(k+1)(k+2)}+\dfrac{1}{(k+1)(k+2)(k+3)}
Common denominator 4(k+1)(k+2)(k+3).
=\dfrac{k(k+3)^2+4}{4(k+1)(k+2)(k+3)}
Expand the numerator: k(k^2+6k+9)+4=k^3+6k^2+9k+4. It vanishes at k=-1, so (k+1) is a factor: k^3+6k^2+9k+4=(k+1)^2(k+4).
=\dfrac{(k+1)^2(k+4)}{4(k+1)(k+2)(k+3)}
Cancel one (k+1). This is the right side of P(k+1).
=\dfrac{(k+1)(k+4)}{4(k+2)(k+3)}=\dfrac{(k+1)\big((k+1)+3\big)}{4\big((k+1)+1\big)\big((k+1)+2\big)}
So P(k+1) is true whenever P(k) is true. P(1) is true, so by the principle of mathematical induction P(n) is true for all integers n\ge 1.

The ratio of coefficients of three consecutive terms in the binomial expansion of (1+x)^n is 2:15:70. Find the average of the three coefficients.

Solution

Let the coefficients be \dbinom n{r-1},\dbinom nr,\dbinom n{r+1}.
\dbinom n{r-1}:\dbinom nr:\dbinom n{r+1}=2:15:70
Ratios of neighbours: \dfrac{n-r+1}{r}=\dfrac{15}2 and \dfrac{n-r}{r+1}=\dfrac{70}{15}=\dfrac{14}3.
\begin{aligned}2(n-r+1)&=15r\\ \Rightarrow\ 2n+2&=17r\\ 3(n-r)&=14(r+1)\\ \Rightarrow\ 3n&=17r+14\end{aligned}
Replace 17r by 2n+2 in the second equation.
3n=(2n+2)+14\;\Rightarrow\;n=16
Then 17r=34, so r=2. The coefficients are \dbinom{16}{1},\dbinom{16}{2},\dbinom{16}{3}.
16,\quad120,\quad560\qquad(16:120:560=2:15:70\ \checkmark)
Average.
\dfrac{16+120+560}{3}=\dfrac{696}{3}=232

232

Book answer. 232

Note. Agrees with the key.

Show that the expansion of \left(x^2+\dfrac1x\right)^{12} does not contain any term involving \dfrac1x.

Solution

General term with a=x^2, b=\dfrac1x, n=12.
T_{r+1}=\dbinom{12}{r}(x^2)^{12-r}\left(\dfrac1x\right)^r=\dbinom{12}{r}x^{24-2r-r}=\dbinom{12}rx^{24-3r}
A term in \dfrac1x=x^{-1} would need the power of x to be -1.
24-3r=-1\;\Rightarrow\;3r=25\;\Rightarrow\;r=\dfrac{25}3
But r must be a whole number from 0 to 12, and \dfrac{25}{3} is not. So no such term exists.
\text{No term involving }\dfrac1x

If \alpha and \beta are nearly equal then show that \left(\dfrac{3\beta}{5\alpha-2\beta}\right)^{-1/3}\approx\dfrac{\alpha}{\alpha+2\beta}+\dfrac{\alpha+\beta}{3\beta}

Solution

Put \alpha=\beta(1+h) with h small, so h^2 and higher powers can be neglected.
\alpha=\beta(1+h)
Left side: 5\alpha-2\beta=\beta(3+5h), so the bracket is \dfrac{3}{3+5h}=\left(1+\dfrac{5h}{3}\right)^{-1}. The power -\frac13 turns it into \left(1+\frac{5h}3\right)^{1/3}.
\text{LHS}=\left(1+\dfrac{5h}{3}\right)^{1/3}\approx1+\dfrac{5h}{9}
Right side, first fraction: \dfrac{\alpha}{\alpha+2\beta}=\dfrac{1+h}{3+h}=\dfrac13(1+h)\left(1+\dfrac h3\right)^{-1}\approx\dfrac13(1+h)\left(1-\dfrac h3\right)\approx\dfrac13\left(1+\dfrac{2h}{3}\right).
\dfrac{\alpha}{\alpha+2\beta}\approx\dfrac13+\dfrac{2h}{9}
Second fraction: \dfrac{\alpha+\beta}{3\beta}=\dfrac{2+h}{3}.
\dfrac{\alpha+\beta}{3\beta}=\dfrac23+\dfrac h3
Add the two.
\text{RHS}\approx1+\dfrac{2h}{9}+\dfrac{3h}{9}=1+\dfrac{5h}9=\text{LHS}\ \checkmark

If {}^{22}C_r is the largest coefficient in the expansion of (1+x)^{22} then find {}^{13}C_r.

Solution

The coefficients of (1+x)^{22} are {}^{22}C_0,{}^{22}C_1,\ldots,{}^{22}C_{22}. They increase up to the middle and then decrease. Since n=22 is even, the middle term is the \left(\dfrac{22}2+1\right)=12th, with coefficient {}^{22}C_{11}.
{}^{22}C_r\ \text{largest}\;\Rightarrow\;r=11
Substitute r=11 in {}^{13}C_r and use {}^{n}C_r={}^{n}C_{n-r}.
{}^{13}C_{11}={}^{13}C_{2}=\dfrac{13\cdot12}{2\cdot1}=78

Book answer. 78

Note. Agrees with the key.

Use binomial theorem to prove that 6^n-5^n leaves a remainder 1, when divided by 5.

Solution

Write 6=5+1 and expand.
6^n=(5+1)^n=5^n+\dbinom n15^{n-1}+\dbinom n25^{n-2}+\cdots+\dbinom n{n-1}5+1
Subtract 5^n. Every remaining term except the last, 1, contains a factor 5.
6^n-5^n=5\left[\dbinom n15^{n-2}+\dbinom n25^{n-3}+\cdots+\dbinom n{n-1}\right]+1
The bracket is an integer m, so 6^n-5^n=5m+1. The remainder on division by 5 is 1.
6^n-5^n=5m+1

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2025 Annual I, Set B, MCQ 91 mark

What is the middle term in the expansion of (x+y)^6?

A\dbinom{6}{2}x^3y^3
B\dbinom{6}{3}x^3y^3Answer
C\dbinom{6}{3}x^4y^2
D\dbinom{6}{2}x^4y^2

Why

n=6 is even, so there is one middle term, the \left(\dfrac62+1\right)=4th term, T_4=T_{3+1}:

T_4=\dbinom63x^{6-3}y^3=\dbinom63x^3y^3

Options A and D have the wrong coefficient \dbinom62, and options C and D have the wrong exponents x^4y^2 (the exponents must add up to 6 and be equal here).

2026 Annual I, Set B, MCQ 121 mark

Binomial expansion uses:

AFactorization
BRoots
CDivision
DPascal triangleAnswer

Why

The book gives two ways to expand (a+b)^n: the binomial theorem and Pascal's triangle, whose nth row supplies the binomial coefficients.

Factorization, roots and division belong to polynomials (Unit 5) and are not used to write out a binomial expansion.

2026 Annual I, Set B, MCQ 131 mark

In mathematical induction assume true for n=k is:

ABase case
BProof
CHypothesisAnswer
DResult

Why

In a proof by induction we first prove the basis step P(1), then assume that the statement is true for n=k. This assumption is called the inductive hypothesis.

The base case (A) is the check at the first value of n, and the proof (B) and the result (D) come at the end, after P(k+1) has been shown.

2026 Annual I, Set B, Q.2 (ii)4 marks

Find the last two digits of (12)^{11}.

Solution

The last two digits of a number are its remainder on division by 100. Write the base as 10+2 so that powers of 10 appear.
12^{11}=(10+2)^{11}
Expand by the binomial theorem. A term containing 10^2=100 or a higher power of 10 is a multiple of 100, so it does not change the last two digits. Group all such terms as 100k.
(10+2)^{11}=100k+\dbinom{11}{10}(10)(2)^{10}+\dbinom{11}{11}(2)^{11}
Evaluate the last two terms: \dbinom{11}{10}=11, 2^{10}=1024 and 2^{11}=2048.
11\cdot10\cdot1024+2048=112640+2048=114688
So 12^{11}=100k+114688=100(k+1146)+88. The last two digits are those of 88.
12^{11}=100m+88

Note. Check: 12^{11}=743\,008\,370\,688, which ends in 88.

2026 Annual I, Set B, Q.2 (iv)4 marks

Find the remainder when 9^{205} is divided by 31 using Binomial expansion.

Solution

Write 9^{205} as a power of 3 and then of 243, which is close to a multiple of 31 (31\times8=248).
9^{205}=3^{410}=(3^5)^{82}=243^{82}=(248-5)^{82}
Expand by the binomial theorem. Every term except the last contains 248, a multiple of 31.
(248-5)^{82}=31k+(-5)^{82}=31k+5^{82}
Reduce 5^{82} in the same way: 5^3=125=124+1=4\times31+1 and 82=3\times27+1.
5^{82}=5\cdot(5^3)^{27}=5(124+1)^{27}=5(31m+1)=31(5m)+5
Combine: 9^{205}=31(k+5m)+5 with 5<31, so 5 is the remainder.
9^{205}=31(k+5m)+5

2026 Annual I, Set B, Q.2 (v) OR4 marks

Prove by mathematical induction that: 2^n\ge n+1 for all n\ge1.

Solution

Let P(n) be the statement.
P(n):\ 2^n\ge n+1,\qquad n\ge1
Basis step. For n=1 both sides can be compared directly.
2^1=2,\qquad 1+1=2,\qquad 2\ge2
So P(1) is true.
Inductive step. Assume P(k) is true for n=k.
2^k\ge k+1\qquad(1)
We must show P(k+1): 2^{k+1}\ge k+2. Multiply both sides of (1) by 2 (a positive number, so the inequality keeps its direction).
2^{k+1}=2\cdot2^k\ge2(k+1)=2k+2
Since k\ge1, we have 2k+2=(k+2)+k>k+2, so
2^{k+1}\ge2k+2>k+2=(k+1)+1
This is P(k+1). Hence P(k)\Rightarrow P(k+1) and, with the basis step, 2^n\ge n+1 for all n\ge1 by the principle of mathematical induction.

Note. Compare the book's Example 4 (3^n>n^2) and Exercise 7.1, questions 25 to 32, which prove inequalities the same way.

2025 Annual I, Set B, Q.2 (v) OR4 marks

Find the term containing x^3 in the expansion of (x+2)^6.

Solution

Here a=x, b=2, n=6. The general term is T_{r+1}=\dbinom nr a^{n-r}b^r.
T_{r+1}=\dbinom6rx^{6-r}\,2^r
The power of x is 6-r. For x^3 we need 6-r=3, so r=3 (the 4th term).
T_4=\dbinom63x^3\,2^3
Evaluate: \dbinom63=20 and 2^3=8.
T_4=20\cdot8\,x^3=160x^3

160x^3

2025 Annual I, Set B, Q.2 (ix)4 marks

Use the principle of mathematical induction to prove that 1^3+2^3+3^3+\dots+n^3=\left(\dfrac{n(n+1)}{2}\right)^2 for all n\in N, n\ge 1.

Solution

Let P(n) be the statement.
P(n):\ 1^3+2^3+3^3+\cdots+n^3=\left(\dfrac{n(n+1)}{2}\right)^2
Basis step. n=1: left side 1^3=1, right side \left(\dfrac{1\cdot2}{2}\right)^2=1. So P(1) is true.
Inductive step. Assume P(k) is true.
1^3+2^3+\cdots+k^3=\left(\dfrac{k(k+1)}{2}\right)^2=\dfrac{k^2(k+1)^2}{4}\qquad(1)
Add the next term (k+1)^3 to both sides of (1).
1^3+\cdots+k^3+(k+1)^3=\dfrac{k^2(k+1)^2}{4}+(k+1)^3
Take out (k+1)^2 and combine over the denominator 4.
=(k+1)^2\left[\dfrac{k^2}{4}+(k+1)\right]=(k+1)^2\cdot\dfrac{k^2+4k+4}{4}
Since k^2+4k+4=(k+2)^2:
=\dfrac{(k+1)^2(k+2)^2}{4}=\left(\dfrac{(k+1)\big((k+1)+1\big)}{2}\right)^2
This is P(k+1). By mathematical induction the formula holds for all n\ge1.

2025 Annual I, Set B, Q.2 (xii) OR4 marks

Use binomial theorem to approximate the value of \sqrt{101}\times\sqrt{99} up to three places of decimal.

Solution

Combine the two roots: \sqrt{101}\times\sqrt{99}=\sqrt{101\times99}, and 101\times99=(100+1)(100-1)=10000-1.
\sqrt{101}\sqrt{99}=\sqrt{10000-1}=\sqrt{10000}\,\left(1-\dfrac1{10000}\right)^{1/2}=100\,(1-0.0001)^{1/2}
Use the binomial series (1+u)^{n}=1+nu+\dfrac{n(n-1)}{2!}u^2+\cdots with n=\dfrac12 and u=-0.0001 (convergent since |u|<1).
(1-0.0001)^{1/2}=1+\dfrac12(-0.0001)+\dfrac{\frac12\left(-\frac12\right)}{2}(0.0001)^2+\cdots=1-0.00005-0.00000000125-\cdots
The third term is smaller than 10^{-9} and can be dropped. Multiply by 100.
100(1-0.00005)=100-0.005=99.995

\approx99.995

Note. Check: \sqrt{9999}=99.99499987\ldots

2025 Annual I, Set B, Q.48 marks

If x is very small such that its square and higher powers can be neglected, then show that \dfrac{(16+4x)^{3/4}}{(4+x)\sqrt{9-6x}}\approx\dfrac{2}{3}+\dfrac{13x}{72}.

Solution

Make every factor of the form (1+u)^n and keep terms up to x only (higher powers of x are neglected). The numerator: 16^{3/4}=(2^4)^{3/4}=8.
(16+4x)^{3/4}=16^{3/4}\left(1+\dfrac x4\right)^{3/4}=8\left(1+\dfrac x4\right)^{3/4}\approx8\left(1+\dfrac34\cdot\dfrac x4\right)=8\left(1+\dfrac{3x}{16}\right)
The factor (4+x) goes in the denominator, so take its reciprocal.
\dfrac1{4+x}=\dfrac14\left(1+\dfrac x4\right)^{-1}\approx\dfrac14\left(1-\dfrac x4\right)
The square root in the denominator becomes a power -\dfrac12, and 9^{-1/2}=\dfrac13.
\dfrac1{\sqrt{9-6x}}=(9-6x)^{-1/2}=\dfrac13\left(1-\dfrac{2x}{3}\right)^{-1/2}\approx\dfrac13\left(1+\dfrac12\cdot\dfrac{2x}{3}\right)=\dfrac13\left(1+\dfrac x3\right)
Multiply the three results. The constants give 8\cdot\dfrac14\cdot\dfrac13=\dfrac23. In the product of the brackets drop x^2.
\text{LHS}\approx\dfrac23\left(1+\dfrac{3x}{16}\right)\left(1-\dfrac x4\right)\left(1+\dfrac x3\right)\approx\dfrac23\left[1+\left(\dfrac3{16}-\dfrac14+\dfrac13\right)x\right]
Add the fractions: \dfrac3{16}-\dfrac14+\dfrac13=\dfrac{9-12+16}{48}=\dfrac{13}{48}.
\text{LHS}\approx\dfrac23\left(1+\dfrac{13x}{48}\right)=\dfrac23+\dfrac{13x}{72}\ \checkmark

Note. The method is that of Exercise 7.3, question 4.

2026 Annual I, Set B, Q.58 marks

Find the middle term(s) in the expansion of \left(3x^2-\dfrac{1}{2x}\right)^{10}. Also find the 9^{th} term.

Solution

Here a=3x^2, b=-\dfrac1{2x} and n=10. The general term is T_{r+1}=\dbinom nr a^{n-r}b^r.
T_{r+1}=\dbinom{10}{r}(3x^2)^{10-r}\left(-\dfrac1{2x}\right)^r
Middle term. n=10 is even, so there is one middle term, the \left(\dfrac n2+1\right)th, that is the 6th term, T_6=T_{5+1} (r=5).
T_6=\dbinom{10}{5}(3x^2)^5\left(-\dfrac1{2x}\right)^5
Evaluate: \dbinom{10}5=252, (3x^2)^5=243x^{10} and \left(-\dfrac1{2x}\right)^5=-\dfrac1{32x^5}.
T_6=252\cdot243x^{10}\cdot\left(-\dfrac1{32x^5}\right)=-\dfrac{61236}{32}x^5=-\dfrac{15309}{8}x^5
Ninth term. T_9=T_{8+1}, so r=8.
T_9=\dbinom{10}{8}(3x^2)^2\left(-\dfrac1{2x}\right)^8
Evaluate: \dbinom{10}8=45, (3x^2)^2=9x^4 and \left(-\dfrac1{2x}\right)^8=\dfrac1{256x^8} (an even power, so the sign is +).
T_9=45\cdot9x^4\cdot\dfrac1{256x^8}=\dfrac{405}{256}x^{-4}

T_6=-\dfrac{15309}{8}x^5,\qquad T_9=\dfrac{405}{256x^4}

Note. The middle-term part is Exercise 7.2, question 5(i) with the same expansion; the paper adds the 9^{th} term.

2026 Annual I, Set D, MCQ 71 mark

Coefficient of middle term:

Ais always 1
Bis independent of n
CDepends on nAnswer
Dis always zero

Why

The coefficient of the middle term of (a+b)^n is a binomial coefficient: \dbinom{n}{n/2} for even n (or the two equal coefficients \dbinom{n}{(n-1)/2} for odd n). Its value changes with n, for example \dbinom{2}{1}=2, \dbinom{4}{2}=6, \dbinom{6}{3}=20.

So it is not always 1 (A), not independent of n (B), and never zero (D).

Note. The question is printed incomplete (it does not say which expansion); it is read as the coefficient of the middle term of a binomial expansion.

2026 Annual I, Set D, MCQ 81 mark

In mathematical induction first step is:

AHypothesis
BResult
CConclusion
DBase caseAnswer

Why

The first step of a proof by induction is the basis step: verify the statement for the first value, usually n=1, the base case.

The hypothesis (A) is assumed in the second step, and the conclusion (C) and result (B) come last.

2025 Annual I, Set D, MCQ 81 mark

The number of ways to climb a 10-step staircase (taking either 1 or 2 steps at a time) is 89. Which method justifies the best conclusion?

AEmpirical counting
BPascal's Triangle
CMathematical InductionAnswer
DBinomial Theorem

Why

A statement such as \"the number of ways to climb n steps is the nth term of a sequence\" is a claim about every positive integer n. Counting the 10-step case does not show it for all n; the general claim is proved by mathematical induction (the count for n+1 steps follows from the counts for n and n-1).

Empirical counting only checks particular cases, and Pascal's triangle and the binomial theorem are for expanding (a+b)^n.

2025 Annual I, Set D, MCQ 91 mark

What is the middle term in the expansion of (x+y)^6?

A\dbinom{6}{2}x^3y^3
B\dbinom{6}{3}x^3y^3Answer
C\dbinom{6}{3}x^4y^2
D\dbinom{6}{2}x^4y^2

Why

n=6 is even, so the middle term is the \left(\dfrac62+1\right)=4th term:

T_4=\dbinom63x^{3}y^3

Options A and D use the wrong coefficient \dbinom62, and C and D use the exponents x^4y^2.

2026 Annual I, Set D, Q.2 (ix)4 marks

Determine the last two digits of (13)^{10}.

Solution

Write the number with a base near a multiple of 100. Since 13^2=169:
13^{10}=(13^2)^5=169^5=(170-1)^5
Expand by the binomial theorem.
(170-1)^5=\dbinom50170^5-\dbinom51170^4+\dbinom52170^3-\dbinom53170^2+\dbinom54170-\dbinom55
The terms with 170^5, 170^4, 170^3 and 170^2 are multiples of 100 (for example 170^2=28900). Group them as 100k. Only the last two terms matter.
=100k+\dbinom54(170)-1=100k+5(170)-1=100k+849
Write 849=800+49, so the number is 100(k+8)+49 and the last two digits are 49.
13^{10}=100(k+8)+49

Note. Check: 13^{10}=137\,858\,491\,849.

2025 Annual I, Set D, Q.2 (ix)4 marks

Use principle of mathematical induction to prove that 1^2+2^2+3^2+\dots+n^2=\dfrac{n(n+1)(2n+1)}{6} for all n\in N, n\ge 1.

Solution

Let P(n) be the statement.
P(n):\ 1^2+2^2+3^2+\cdots+n^2=\dfrac{n(n+1)(2n+1)}{6}
Basis step. n=1: left side 1, right side \dfrac{1\cdot2\cdot3}{6}=1. So P(1) is true.
Inductive step. Assume P(k) is true.
1^2+2^2+\cdots+k^2=\dfrac{k(k+1)(2k+1)}{6}\qquad(1)
Add the next term (k+1)^2 to both sides.
1^2+\cdots+k^2+(k+1)^2=\dfrac{k(k+1)(2k+1)}{6}+(k+1)^2
Take out the common factor (k+1) and use the denominator 6.
=\dfrac{(k+1)\big[k(2k+1)+6(k+1)\big]}{6}=\dfrac{(k+1)(2k^2+7k+6)}{6}
Factorise 2k^2+7k+6=(k+2)(2k+3).
=\dfrac{(k+1)(k+2)(2k+3)}{6}=\dfrac{(k+1)\big((k+1)+1\big)\big(2(k+1)+1\big)}{6}
This is P(k+1). By mathematical induction the formula is true for all n\ge1.

2026 Annual I, Set D, Q.2 (xi)4 marks

Use Binomial Theorem to find the remainder when 8^{205} is divided by 48.

Solution

Write 8^{205} as a power of 2 and split off a factor that makes 48=16\times3 appear: 8^{205}=2^{615}=2^5\cdot2^{610}=32\cdot4^{305} and 4=3+1.
8^{205}=32\,(3+1)^{305}
By the binomial theorem every term of (3+1)^{305} except the last contains 3, so (3+1)^{305}=3t+1 for an integer t.
8^{205}=32(3t+1)=96t+32
96=2\times48, so 96t is a multiple of 48. The remaining 32 is less than 48.
8^{205}=48(2t)+32

2025 Annual I, Set D, Q.2 (xi)4 marks

Find the term containing x^4 in the expansion of (2x-3)^7.

Solution

Here a=2x, b=-3, n=7.
T_{r+1}=\dbinom7r(2x)^{7-r}(-3)^r
The power of x is 7-r. For x^4: 7-r=4, so r=3 (the 4th term).
T_4=\dbinom73(2x)^4(-3)^3
Evaluate: \dbinom73=35, (2x)^4=16x^4, (-3)^3=-27.
T_4=35\cdot16\cdot(-27)\,x^4=-15120\,x^4

-15120\,x^4

2026 Annual I, Set D, Q.2 (xii) OR4 marks

Use mathematical induction to prove: 1+2+3+\dots+n=\dfrac{n(n+1)}{2}.

Solution

Let P(n) be the statement.
P(n):\ 1+2+3+\cdots+n=\dfrac{n(n+1)}{2}
Basis step. For n=1 the left side is 1 and the right side is \dfrac{1\cdot2}{2}=1. So P(1) is true.
1=\dfrac{1(1+1)}{2}=1
Inductive step. Assume P(k) is true.
1+2+3+\cdots+k=\dfrac{k(k+1)}{2}\qquad(1)
Add the next term, k+1, to both sides of (1).
1+2+\cdots+k+(k+1)=\dfrac{k(k+1)}{2}+(k+1)
Take out the common factor (k+1) on the right.
=(k+1)\left(\dfrac k2+1\right)=(k+1)\cdot\dfrac{k+2}{2}
This is the formula with n=k+1, so P(k+1) is true.
1+2+\cdots+(k+1)=\dfrac{(k+1)\big((k+1)+1\big)}{2}
By the principle of mathematical induction, P(n) is true for all n\ge1.

2025 Annual I, Set D, Q.2 (xii) OR4 marks

Use binomial theorem to approximate the value of \dfrac{(1.03)^4}{(0.98)^3} up to three places of the decimal.

Solution

Write each bracket as (1+u)^n: 1.03=1+0.03 and 0.98=1-0.02, so \dfrac{(1.03)^4}{(0.98)^3}=(1+0.03)^4(1-0.02)^{-3}.
Expand (1+0.03)^4 (a positive integer index, so it stops).
(1+0.03)^4=1+4(0.03)+6(0.03)^2+4(0.03)^3+\cdots=1+0.12+0.0054+0.000108+\cdots=1.1255
Expand (1-0.02)^{-3} with the binomial series (1+u)^{-3}=1-3u+6u^2-10u^3+\cdots, u=-0.02 (convergent, |u|<1).
(1-0.02)^{-3}=1+3(0.02)+6(0.02)^2+10(0.02)^3+\cdots=1+0.06+0.0024+0.00008+\cdots=1.0625
Multiply.
1.1255\times1.0625\approx1.1958
Round to three decimal places.
\approx1.196

\approx1.196

2026 Annual I, Set D, Q.44+4 = 8 marks

Find the term independent of x in the expansion of \left(2x-\dfrac1x\right)^{12}. Also find the 10^{th} term.

Solution

Here a=2x, b=-\dfrac1x, n=12. Write the general term and collect the powers of x.
T_{r+1}=\dbinom{12}{r}(2x)^{12-r}\left(-\dfrac1x\right)^r=\dbinom{12}{r}2^{12-r}(-1)^r\,x^{12-r}\,x^{-r}=\dbinom{12}{r}2^{12-r}(-1)^r\,x^{12-2r}
**Term independent of x.** The power of x must be zero.
12-2r=0\ \Rightarrow\ r=6
Put r=6 (the 7th term): \dbinom{12}6=924, 2^6=64, (-1)^6=1.
T_7=924\cdot64=59136
Tenth term. T_{10}=T_{9+1}, so r=9.
T_{10}=\dbinom{12}{9}(2x)^3\left(-\dfrac1x\right)^9
Evaluate: \dbinom{12}9=220, (2x)^3=8x^3 and \left(-\dfrac1x\right)^9=-x^{-9}.
T_{10}=220\cdot8x^3\cdot(-x^{-9})=-1760\,x^{-6}

T_7=59136,\qquad T_{10}=-\dfrac{1760}{x^{6}}

Note. Exercise 7.2, question 7(i), asks the same for (2x^2-1/x)^{12}.

2025 Annual I, Set D, Q.48 marks

If x is very small such that its square and higher powers can be neglected, then show that \dfrac{(8+2x)^{2/3}}{(2+x)\left(\sqrt{4-2x}\right)}\approx 1-\dfrac{x}{12}.

Solution

Put each factor in the form (1+u)^n and keep terms up to x. The numerator: 8^{2/3}=4.
(8+2x)^{2/3}=8^{2/3}\left(1+\dfrac x4\right)^{2/3}\approx4\left(1+\dfrac23\cdot\dfrac x4\right)=4\left(1+\dfrac x6\right)
The factor (2+x) is in the denominator.
\dfrac1{2+x}=\dfrac12\left(1+\dfrac x2\right)^{-1}\approx\dfrac12\left(1-\dfrac x2\right)
The square root in the denominator, with 4^{-1/2}=\dfrac12:
\dfrac1{\sqrt{4-2x}}=\dfrac12\left(1-\dfrac x2\right)^{-1/2}\approx\dfrac12\left(1+\dfrac12\cdot\dfrac x2\right)=\dfrac12\left(1+\dfrac x4\right)
Multiply. The constants give 4\cdot\dfrac12\cdot\dfrac12=1. Drop x^2 in the product.
\text{LHS}\approx\left(1+\dfrac x6\right)\left(1-\dfrac x2\right)\left(1+\dfrac x4\right)\approx1+\left(\dfrac16-\dfrac12+\dfrac14\right)x
\dfrac16-\dfrac12+\dfrac14=\dfrac{2-6+3}{12}=-\dfrac1{12}.
\text{LHS}\approx1-\dfrac{x}{12}\ \checkmark

Note. The method is that of Exercise 7.3, question 4.

Exam focus

Show that an expression such as \dfrac{(16+4x)^{3/4}}{(4+x)\sqrt{9-6x}} is approximately a linear expression in x when x is very small. (long)
8-mark Q.4 in both 2025 sets (Set B and Set D, with different numbers). Expand each factor as (1+u)^n, keep terms up to x, multiply. The same method is Exercise 7.3, question 4.
Go to answer
Prove a sum formula by mathematical induction, such as 1+2+\cdots+n=\dfrac{n(n+1)}2, 1^2+\cdots+n^2 or 1^3+\cdots+n^3. (problem)
4-mark question in every set: 2026 Set D Q.2 (xii) OR, 2025 Set B Q.2 (ix) (cubes) and 2025 Set D Q.2 (ix) (squares). Write the basis step, the assumption P(k), the next term and P(k+1).
Go to answer
Find the middle term(s), a term independent of x, or a required term in a binomial expansion, for example \left(2x-\dfrac1x\right)^{12} or (2x-3)^7. (long)
2026 Set B Q.5 (middle terms and the 9^{th} term, 8 marks), 2026 Set D Q.4 (term independent of x and the 10^{th} term), 2025 Set B Q.2 (v) OR and Set D Q.2 (xi) (term containing x^3 or x^4). Use T_{r+1}=\dbinom nr a^{n-r}b^r.
Go to answer
Find the remainder when a large power is divided by a number, for example 9^{205} by 31 or 8^{205} by 48, using the binomial theorem. (problem)
2026 Set B Q.2 (iv) and 2026 Set D Q.2 (xi), both 4 marks and both Exercise 7.4 question 2. Write the base as a multiple of the divisor plus or minus a small number.
Go to answer
Find the last two digits of a power, for example 12^{11} or 13^{10}. (problem)
2026 Set B Q.2 (ii) and Set D Q.2 (ix), 4 marks each. Write the number as (10\pm a)^n or (170-1)^5 and keep only the last terms; 13^{10} is Example 17 of the book.
Go to answer
Approximate a number with the binomial series, for example \sqrt{101}\times\sqrt{99} or \dfrac{(1.03)^4}{(0.98)^3}, to three decimal places. (problem)
2025 Set B Q.2 (xii) OR and Set D Q.2 (xii) OR (4 marks each). Write the number as (1+u)^n with small u and add terms until the next is below the accuracy needed.
Go to answer
Name the steps of mathematical induction and the middle term of a binomial expansion (MCQ). (mcq)
MCQs on the inductive hypothesis (2026 Set B MCQ 13), the base case (2026 Set D MCQ 8), the induction method (2025 Set D MCQ 8) and the middle term of (x+y)^6 (2025 Set B and Set D MCQ 9).
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Prove an inequality by mathematical induction, for example 2^n\ge n+1. (problem)
2026 Set B Q.2 (v) OR. Exercise 7.1, questions 25 to 32, practise this type; build the larger side from the assumption P(k).
Go to answer
Use Pascal's triangle to count outcomes, such as the number of heads when coins are tossed. (short)
Not in the 2025 or 2026 paper, but Exercise 7.2, questions 18 to 20, and Example 12 of the book ask it, so it can still come as a short question.
Go to answer

Number, power	1	2	3	4	5	6	7	8	9
0	0	0	0	0	0	0	0	0	0
1	1	1	1	1	1	1	1	1	1
2	2	4	8	6	2	4	8	6	2
3	3	9	7	1	3	9	7	1	3
4	4	6	4	6	4	6	4	6	4
5	5	5	5	5	5	5	5	5	5
6	6	6	6	6	6	6	6	6	6
7	7	9	3	1	7	9	3	1	7
8	8	4	2	6	8	4	2	6	8
9	9	1	9	1	9	1	9	1	9

Number, power	1	2	3	4	5	6	7	8	9
0	0	0	0	0	0	0	0	0	0
1	1	1	1	1	1	1	1	1	1
2	2	4	8	6	2	4	8	6	2
3	3	9	7	1	3	9	7	1	3
4	4	6	4	6	4	6	4	6	4
5	5	5	5	5	5	5	5	5	5
6	6	6	6	6	6	6	6	6	6
7	7	9	3	1	7	9	3	1	7
8	8	4	2	6	8	4	2	6	8
9	9	1	9	1	9	1	9	1	9

Number, power	1	2	3	4	5	6	7	8	9
0	0	0	0	0	0	0	0	0	0
1	1	1	1	1	1	1	1	1	1
2	2	4	8	6	2	4	8	6	2
3	3	9	7	1	3	9	7	1	3
4	4	6	4	6	4	6	4	6	4
5	5	5	5	5	5	5	5	5	5
6	6	6	6	6	6	6	6	6	6
7	7	9	3	1	7	9	3	1	7
8	8	4	2	6	8	4	2	6	8
9	9	1	9	1	9	1	9	1	9