Chapter 7 of 14

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Updated 11 June 2026

NCERT Solutions

Binomial Theorem

Rajasthan Board · Class 11 · Mathematics

NCERT Solutions for Binomial Theorem — Rajasthan Board Class 11 Mathematics.

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20 Questions Solved · 2 Sections

Exercise 7.1

1Expand

(1-2x)^5

.Show solution

Given:

(1-2x)^5

Formula (Binomial Theorem):

(a+b)^n = \sum_{r=0}^{n} {}^nC_r\, a^{n-r}\, b^r

Here

a=1

b=-2x

n=5

(1-2x)^5 = {}^5C_0(1)^5 + {}^5C_1(1)^4(-2x) + {}^5C_2(1)^3(-2x)^2 + {}^5C_3(1)^2(-2x)^3 + {}^5C_4(1)(-2x)^4 + {}^5C_5(-2x)^5

= 1 + 5(-2x) + 10(4x^2) + 10(-8x^3) + 5(16x^4) + 1(-32x^5)

\boxed{= 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5}

2Expand

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

.Show solution

Given:

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

Here

a=\dfrac{2}{x}

b=-\dfrac{x}{2}

n=5

\left(\frac{2}{x}-\frac{x}{2}\right)^5 = \sum_{r=0}^{5}{}^5C_r\left(\frac{2}{x}\right)^{5-r}\left(-\frac{x}{2}\right)^r

Computing each term:

-

r=0

{}^5C_0\left(\dfrac{2}{x}\right)^5 = \dfrac{32}{x^5}

r=1

{}^5C_1\left(\dfrac{2}{x}\right)^4\left(-\dfrac{x}{2}\right) = 5\cdot\dfrac{16}{x^4}\cdot\left(-\dfrac{x}{2}\right) = -\dfrac{40}{x^3}

r=2

{}^5C_2\left(\dfrac{2}{x}\right)^3\left(-\dfrac{x}{2}\right)^2 = 10\cdot\dfrac{8}{x^3}\cdot\dfrac{x^2}{4} = \dfrac{20}{x}

r=3

{}^5C_3\left(\dfrac{2}{x}\right)^2\left(-\dfrac{x}{2}\right)^3 = 10\cdot\dfrac{4}{x^2}\cdot\left(-\dfrac{x^3}{8}\right) = -5x

r=4

{}^5C_4\left(\dfrac{2}{x}\right)\left(-\dfrac{x}{2}\right)^4 = 5\cdot\dfrac{2}{x}\cdot\dfrac{x^4}{16} = \dfrac{5x^3}{8}

r=5

{}^5C_5\left(-\dfrac{x}{2}\right)^5 = -\dfrac{x^5}{32}

\boxed{\left(\frac{2}{x}-\frac{x}{2}\right)^5 = \frac{32}{x^5} - \frac{40}{x^3} + \frac{20}{x} - 5x + \frac{5x^3}{8} - \frac{x^5}{32}}

3Expand

(2x-3)^6

.Show solution

Given:

(2x-3)^6

Here

a=2x

b=-3

n=6

(2x-3)^6 = \sum_{r=0}^{6}{}^6C_r(2x)^{6-r}(-3)^r

Computing each term:

-

r=0

{}^6C_0(2x)^6 = 64x^6

r=1

{}^6C_1(2x)^5(-3) = 6\cdot32x^5\cdot(-3) = -576x^5

r=2

{}^6C_2(2x)^4(9) = 15\cdot16x^4\cdot9 = 2160x^4

r=3

{}^6C_3(2x)^3(-27) = 20\cdot8x^3\cdot(-27) = -4320x^3

r=4

{}^6C_4(2x)^2(81) = 15\cdot4x^2\cdot81 = 4860x^2

r=5

{}^6C_5(2x)(-243) = 6\cdot2x\cdot(-243) = -2916x

r=6

{}^6C_6(-3)^6 = 729

\boxed{(2x-3)^6 = 64x^6 - 576x^5 + 2160x^4 - 4320x^3 + 4860x^2 - 2916x + 729}

4Expand

\left(\dfrac{x}{3}+\dfrac{1}{x}\right)^5

.Show solution

Given:

\left(\dfrac{x}{3}+\dfrac{1}{x}\right)^5

Here

a=\dfrac{x}{3}

b=\dfrac{1}{x}

n=5

\left(\frac{x}{3}+\frac{1}{x}\right)^5 = \sum_{r=0}^{5}{}^5C_r\left(\frac{x}{3}\right)^{5-r}\left(\frac{1}{x}\right)^r

Computing each term:

-

r=0

{}^5C_0\left(\dfrac{x}{3}\right)^5 = \dfrac{x^5}{243}

r=1

{}^5C_1\left(\dfrac{x}{3}\right)^4\left(\dfrac{1}{x}\right) = 5\cdot\dfrac{x^4}{81}\cdot\dfrac{1}{x} = \dfrac{5x^3}{81}

r=2

{}^5C_2\left(\dfrac{x}{3}\right)^3\left(\dfrac{1}{x}\right)^2 = 10\cdot\dfrac{x^3}{27}\cdot\dfrac{1}{x^2} = \dfrac{10x}{27}

r=3

{}^5C_3\left(\dfrac{x}{3}\right)^2\left(\dfrac{1}{x}\right)^3 = 10\cdot\dfrac{x^2}{9}\cdot\dfrac{1}{x^3} = \dfrac{10}{9x}

r=4

{}^5C_4\left(\dfrac{x}{3}\right)\left(\dfrac{1}{x}\right)^4 = 5\cdot\dfrac{x}{3}\cdot\dfrac{1}{x^4} = \dfrac{5}{3x^3}

r=5

{}^5C_5\left(\dfrac{1}{x}\right)^5 = \dfrac{1}{x^5}

\boxed{\left(\frac{x}{3}+\frac{1}{x}\right)^5 = \frac{x^5}{243}+\frac{5x^3}{81}+\frac{10x}{27}+\frac{10}{9x}+\frac{5}{3x^3}+\frac{1}{x^5}}

5Expand

\left(x+\dfrac{1}{x}\right)^6

.Show solution

Given:

\left(x+\dfrac{1}{x}\right)^6

Here

a=x

b=\dfrac{1}{x}

n=6

\left(x+\frac{1}{x}\right)^6 = \sum_{r=0}^{6}{}^6C_r\, x^{6-r}\left(\frac{1}{x}\right)^r = \sum_{r=0}^{6}{}^6C_r\, x^{6-2r}

Computing each term:

-

r=0

{}^6C_0\, x^6 = x^6

r=1

{}^6C_1\, x^4 = 6x^4

r=2

{}^6C_2\, x^2 = 15x^2

r=3

{}^6C_3\, x^0 = 20

r=4

{}^6C_4\, x^{-2} = \dfrac{15}{x^2}

r=5

{}^6C_5\, x^{-4} = \dfrac{6}{x^4}

r=6

{}^6C_6\, x^{-6} = \dfrac{1}{x^6}

\boxed{\left(x+\frac{1}{x}\right)^6 = x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6}}

6Using binomial theorem, evaluate

(96)^3

.Show solution

Write:

96 = 100 - 4

(96)^3 = (100-4)^3

= {}^3C_0(100)^3 - {}^3C_1(100)^2(4) + {}^3C_2(100)(4)^2 - {}^3C_3(4)^3

= 1000000 - 3\times10000\times4 + 3\times100\times16 - 64

= 1000000 - 120000 + 4800 - 64

\boxed{(96)^3 = 884736}

7Using binomial theorem, evaluate

(102)^5

.Show solution

Write:

102 = 100 + 2

(102)^5 = (100+2)^5

= {}^5C_0(100)^5 + {}^5C_1(100)^4(2) + {}^5C_2(100)^3(2)^2 + {}^5C_3(100)^2(2)^3 + {}^5C_4(100)(2)^4 + {}^5C_5(2)^5

= 10000000000 + 5\times100000000\times2 + 10\times1000000\times4 + 10\times10000\times8 + 5\times100\times16 + 32

= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32

\boxed{(102)^5 = 11040808032}

8Using binomial theorem, evaluate

(101)^4

.Show solution

Write:

101 = 100 + 1

(101)^4 = (100+1)^4

= {}^4C_0(100)^4 + {}^4C_1(100)^3(1) + {}^4C_2(100)^2(1)^2 + {}^4C_3(100)(1)^3 + {}^4C_4(1)^4

= 100000000 + 4\times1000000 + 6\times10000 + 4\times100 + 1

= 100000000 + 4000000 + 60000 + 400 + 1

\boxed{(101)^4 = 104060401}

9Using binomial theorem, evaluate

(99)^5

.Show solution

Write:

99 = 100 - 1

(99)^5 = (100-1)^5

= {}^5C_0(100)^5 - {}^5C_1(100)^4 + {}^5C_2(100)^3 - {}^5C_3(100)^2 + {}^5C_4(100) - {}^5C_5

= 10000000000 - 5\times100000000 + 10\times1000000 - 10\times10000 + 5\times100 - 1

= 10000000000 - 500000000 + 10000000 - 100000 + 500 - 1

\boxed{(99)^5 = 9509900499}

10Using Binomial Theorem, indicate which number is larger

(1.1)^{10000}

1000

.Show solution

We show that $(1.1)^{10000} > 1000$ .

Write

1.1 = 1 + 0.1

. By the Binomial Theorem:

(1.1)^{10000} = (1+0.1)^{10000}

= {}^{10000}C_0 + {}^{10000}C_1(0.1) + {}^{10000}C_2(0.1)^2 + \cdots

All terms are positive. Taking only the first two terms:

(1.1)^{10000} &gt; {}^{10000}C_0 + {}^{10000}C_1(0.1)

= 1 + 10000 \times 0.1

= 1 + 1000 = 1001 &gt; 1000

\therefore\quad \boxed{(1.1)^{10000} &gt; 1000}

11Find

(a+b)^4-(a-b)^4

. Hence, evaluate

(\sqrt{3}+\sqrt{2})^4-(\sqrt{3}-\sqrt{2})^4

.Show solution

Step 1: Expand $(a+b)^4$ and $(a-b)^4$ .

(a+b)^4 = {}^4C_0 a^4 + {}^4C_1 a^3b + {}^4C_2 a^2b^2 + {}^4C_3 ab^3 + {}^4C_4 b^4

= a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4

(a-b)^4 = a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4

Step 2: Subtract.

(a+b)^4-(a-b)^4 = 2(4a^3b + 4ab^3) = 8a^3b + 8ab^3 = 8ab(a^2+b^2)

Step 3: Substitute $a=\sqrt{3}$ , $b=\sqrt{2}$ .

(\sqrt{3}+\sqrt{2})^4-(\sqrt{3}-\sqrt{2})^4 = 8(\sqrt{3})(\sqrt{2})\left[(\sqrt{3})^2+(\sqrt{2})^2\right]

= 8\sqrt{6}\,(3+2) = 8\sqrt{6}\times 5

\boxed{= 40\sqrt{6}}

12Find

(x+1)^6+(x-1)^6

. Hence or otherwise evaluate

(\sqrt{2}+1)^6+(\sqrt{2}-1)^6

.Show solution

Step 1: Expand $(x+1)^6$ and $(x-1)^6$ .

(x+1)^6 = {}^6C_0 x^6 + {}^6C_1 x^5 + {}^6C_2 x^4 + {}^6C_3 x^3 + {}^6C_4 x^2 + {}^6C_5 x + {}^6C_6

(x-1)^6 = {}^6C_0 x^6 - {}^6C_1 x^5 + {}^6C_2 x^4 - {}^6C_3 x^3 + {}^6C_4 x^2 - {}^6C_5 x + {}^6C_6

Step 2: Add (odd-power terms cancel).

(x+1)^6+(x-1)^6 = 2\left[{}^6C_0 x^6 + {}^6C_2 x^4 + {}^6C_4 x^2 + {}^6C_6\right]

= 2\left[x^6 + 15x^4 + 15x^2 + 1\right]

Step 3: Substitute $x=\sqrt{2}$ .

(\sqrt{2}+1)^6+(\sqrt{2}-1)^6 = 2\left[(\sqrt{2})^6 + 15(\sqrt{2})^4 + 15(\sqrt{2})^2 + 1\right]

= 2\left[8 + 15\times4 + 15\times2 + 1\right]

= 2\left[8 + 60 + 30 + 1\right] = 2\times99

\boxed{= 198}

13Show that

9^{n+1}-8n-9

is divisible by 64, whenever

n

is a positive integer.Show solution

We need to show:

9^{n+1}-8n-9 = 64k

for some non-negative integer

k

.

Write

9 = 1+8

, so

9^{n+1} = (1+8)^{n+1}

.

By the Binomial Theorem:

(1+8)^{n+1} = {}^{n+1}C_0 + {}^{n+1}C_1(8) + {}^{n+1}C_2(8)^2 + {}^{n+1}C_3(8)^3 + \cdots + {}^{n+1}C_{n+1}(8)^{n+1}

= 1 + (n+1)\cdot8 + {}^{n+1}C_2\cdot64 + {}^{n+1}C_3\cdot8^3 + \cdots

= 1 + 8n + 8 + 64\left[{}^{n+1}C_2 + {}^{n+1}C_3\cdot8 + \cdots + 8^{n-1}\right]

Therefore:

9^{n+1} - 8n - 9 = 1 + 8n + 8 + 64M - 8n - 9

where

M = {}^{n+1}C_2 + 8\cdot{}^{n+1}C_3 + \cdots + 8^{n-1}

is a positive integer.

= 64M

Hence

9^{n+1}-8n-9

is divisible by 64 for every positive integer

n

\blacksquare

14Prove that

\displaystyle\sum_{r=0}^{n} 3^r\,{}^nC_r = 4^n

.Show solution

We use the Binomial Theorem:

(1+x)^n = \sum_{r=0}^{n}{}^nC_r\, x^r

Substitute $x = 3$ :

(1+3)^n = \sum_{r=0}^{n}{}^nC_r\, 3^r

4^n = \sum_{r=0}^{n} 3^r\,{}^nC_r

Hence proved.

\blacksquare

Miscellaneous Exercise on Chapter 7

1If

a

and

b

are distinct integers, prove that

a-b

is a factor of

a^n - b^n

, whenever

n

is a positive integer. [Hint: write

a^n = (a-b+b)^n

and expand]Show solution

Proof:

Write

a = (a-b)+b

. Then:

a^n = [(a-b)+b]^n = \sum_{r=0}^{n}{}^nC_r(a-b)^r\, b^{n-r}

= {}^nC_0\, b^n + {}^nC_1(a-b)b^{n-1} + {}^nC_2(a-b)^2 b^{n-2} + \cdots + {}^nC_n(a-b)^n

= b^n + (a-b)\left[{}^nC_1 b^{n-1} + {}^nC_2(a-b)b^{n-2} + \cdots + (a-b)^{n-1}\right]

Therefore:

a^n - b^n = (a-b)\left[{}^nC_1 b^{n-1} + {}^nC_2(a-b)b^{n-2} + \cdots + (a-b)^{n-1}\right]

The expression in brackets is an integer (since

a

and

b

are integers). Hence

(a-b)

is a factor of

a^n - b^n

\blacksquare

2Evaluate

(\sqrt{3}+\sqrt{2})^6-(\sqrt{3}-\sqrt{2})^6

.Show solution

Let

a=\sqrt{3}

b=\sqrt{2}

.

Expand

(a+b)^6

and

(a-b)^6

(a+b)^6 = a^6+6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6

(a-b)^6 = a^6-6a^5b+15a^4b^2-20a^3b^3+15a^2b^4-6ab^5+b^6

Subtracting:

(a+b)^6-(a-b)^6 = 2\left[6a^5b+20a^3b^3+6ab^5\right] = 2ab\left[6a^4+20a^2b^2+6b^4\right]

Now substitute

a=\sqrt{3}

b=\sqrt{2}

, so

ab=\sqrt{6}

a^2=3

b^2=2

= 2\sqrt{6}\left[6(9)+20(3)(2)+6(4)\right]

= 2\sqrt{6}\left[54+120+24\right]

= 2\sqrt{6}\times198 = 396\sqrt{6}

\boxed{(\sqrt{3}+\sqrt{2})^6-(\sqrt{3}-\sqrt{2})^6 = 396\sqrt{6}}

3Find the value of

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

.Show solution

Let

x = a^2

and

y = \sqrt{a^2-1}

.

Expand

(x+y)^4

and

(x-y)^4

(x+y)^4+(x-y)^4 = 2\left[{}^4C_0 x^4 + {}^4C_2 x^2 y^2 + {}^4C_4 y^4\right]

= 2\left[x^4 + 6x^2y^2 + y^4\right]

Substitute back

x=a^2

y^2 = a^2-1

= 2\left[(a^2)^4 + 6(a^2)^2(a^2-1) + (a^2-1)^2\right]

= 2\left[a^8 + 6a^4(a^2-1) + (a^4-2a^2+1)\right]

= 2\left[a^8 + 6a^6 - 6a^4 + a^4 - 2a^2 + 1\right]

= 2\left[a^8 + 6a^6 - 5a^4 - 2a^2 + 1\right]

\boxed{= 2a^8 + 12a^6 - 10a^4 - 4a^2 + 2}

4Find an approximation of

(0.99)^5

using the first three terms of its expansion.Show solution

Write:

0.99 = 1 - 0.01

(0.99)^5 = (1-0.01)^5

Using the first three terms of the binomial expansion:

\approx {}^5C_0(1)^5 + {}^5C_1(1)^4(-0.01) + {}^5C_2(1)^3(-0.01)^2

= 1 + 5\times(-0.01) + 10\times(0.0001)

= 1 - 0.05 + 0.001

\boxed{(0.99)^5 \approx 0.951}

5Expand using Binomial Theorem

\left(1+\dfrac{x}{2}-\dfrac{2}{x}\right)^4,\; x\neq 0

.Show solution

Strategy: Group as

\left[\left(1+\dfrac{x}{2}\right)-\dfrac{2}{x}\right]^4

and let

p = 1+\dfrac{x}{2}

q = \dfrac{2}{x}

(p-q)^4 = {}^4C_0 p^4 - {}^4C_1 p^3 q + {}^4C_2 p^2 q^2 - {}^4C_3 p q^3 + {}^4C_4 q^4

Compute each part:

p = 1+\dfrac{x}{2}

q = \dfrac{2}{x}

p^2 = 1 + x + \dfrac{x^2}{4}

p^3 = \left(1+\dfrac{x}{2}\right)^3 = 1 + \dfrac{3x}{2} + \dfrac{3x^2}{4} + \dfrac{x^3}{8}

p^4 = \left(1+\dfrac{x}{2}\right)^4 = 1 + 2x + \dfrac{3x^2}{2} + \dfrac{x^3}{2} + \dfrac{x^4}{16}

q^2 = \dfrac{4}{x^2}

q^3 = \dfrac{8}{x^3}

q^4 = \dfrac{16}{x^4}

Term 1:

p^4 = 1+2x+\dfrac{3x^2}{2}+\dfrac{x^3}{2}+\dfrac{x^4}{16}

Term 2:

-4p^3 q = -4\left(1+\dfrac{3x}{2}+\dfrac{3x^2}{4}+\dfrac{x^3}{8}\right)\cdot\dfrac{2}{x}

= -\dfrac{8}{x}\left(1+\dfrac{3x}{2}+\dfrac{3x^2}{4}+\dfrac{x^3}{8}\right)

= -\dfrac{8}{x} - 12 - 6x - x^2

Term 3:

6p^2 q^2 = 6\left(1+x+\dfrac{x^2}{4}\right)\cdot\dfrac{4}{x^2}

= \dfrac{24}{x^2}\left(1+x+\dfrac{x^2}{4}\right)

= \dfrac{24}{x^2}+\dfrac{24}{x}+6

Term 4:

-4p q^3 = -4\left(1+\dfrac{x}{2}\right)\cdot\dfrac{8}{x^3}

= -\dfrac{32}{x^3}\left(1+\dfrac{x}{2}\right)

= -\dfrac{32}{x^3}-\dfrac{16}{x^2}

Term 5:

q^4 = \dfrac{16}{x^4}

Adding all terms:

\left(1+\frac{x}{2}-\frac{2}{x}\right)^4 = \frac{16}{x^4} - \frac{32}{x^3} + \frac{24}{x^2} - \frac{32}{x^2} + \frac{24}{x} - \frac{8}{x} - 12 + 6 + 1 + 2x - x^2 + \frac{3x^2}{2} - 6x + \frac{x^3}{2} + \frac{x^4}{16}

Collecting like powers:

-

x^4

\dfrac{x^4}{16}

x^3

\dfrac{x^3}{2}

x^2

-x^2+\dfrac{3x^2}{2} = \dfrac{x^2}{2}

x^1

2x-6x = -4x

x^0

-12+6+1 = -5

x^{-1}

\dfrac{24}{x}-\dfrac{8}{x} = \dfrac{16}{x}

x^{-2}

\dfrac{24}{x^2}-\dfrac{16}{x^2} = \dfrac{8}{x^2}

x^{-3}

-\dfrac{32}{x^3}

x^{-4}

\dfrac{16}{x^4}

\boxed{\left(1+\frac{x}{2}-\frac{2}{x}\right)^4 = \frac{x^4}{16}+\frac{x^3}{2}+\frac{x^2}{2}-4x-5+\frac{16}{x}+\frac{8}{x^2}-\frac{32}{x^3}+\frac{16}{x^4}}

6Find the expansion of

(3x^2-2ax+3a^2)^3

using binomial theorem.Show solution

Strategy: Write

3x^2-2ax+3a^2 = (3x^2-2ax)+3a^2

and apply the Binomial Theorem with

p = 3x^2-2ax

and

q = 3a^2

(p+q)^3 = p^3 + 3p^2q + 3pq^2 + q^3

Compute each part:

p = 3x^2-2ax

q = 3a^2

p^2 = (3x^2-2ax)^2 = 9x^4 - 12ax^3 + 4a^2x^2

p^3 = (3x^2-2ax)^3

Expand

p^3

using Binomial Theorem:

(3x^2-2ax)^3 = (3x^2)^3 - 3(3x^2)^2(2ax) + 3(3x^2)(2ax)^2 - (2ax)^3

= 27x^6 - 3\cdot9x^4\cdot2ax + 3\cdot3x^2\cdot4a^2x^2 - 8a^3x^3

= 27x^6 - 54ax^5 + 36a^2x^4 - 8a^3x^3

3p^2q = 3(9x^4-12ax^3+4a^2x^2)(3a^2)

= 9a^2(9x^4-12ax^3+4a^2x^2)

= 81a^2x^4 - 108a^3x^3 + 36a^4x^2

3pq^2 = 3(3x^2-2ax)(9a^4)

= 27a^4(3x^2-2ax)

= 81a^4x^2 - 54a^5x

q^3 = (3a^2)^3 = 27a^6

Adding all terms:

= 27x^6 - 54ax^5 + (36a^2+81a^2)x^4 + (-8a^3-108a^3)x^3 + (36a^4+81a^4)x^2 - 54a^5x + 27a^6

\boxed{(3x^2-2ax+3a^2)^3 = 27x^6 - 54ax^5 + 117a^2x^4 - 116a^3x^3 + 117a^4x^2 - 54a^5x + 27a^6}

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