# NCERT Solutions for Class 11 Maths Chapter 7 - Binomial Theorem

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Exercise 8.1

Direction (Q. Nos. 1 to 5): Expand each of the expressions.

1. (1 – 2x)5.

Sol. We know that

(a + b)n =$$\displaystyle\sum_{k=0}^n\space^n\text{C}_{k}\space a^{n-k}b^{k}$$

= nC0an + nC1 an – 1 b′ + nC2 an – 2 b2 + …

Here a = 1, b = – 2n, n = 5

Therefore, expression of (1 – 2x)5 can be written as

(1 – 2x)5 = 5C0 (1)5 (– 2x)0 + 5C1 (1)5 – 1 (– 2x)1 + 5C2 (1)5 – 2 (– 2x)2 + 5C3 (1)5 – 3 (– 2x)3 + 5C4 (1)5 – 4 (– 2x)4 + 5C5 (1)5 – 5 (– 2x)5 …(i)

We know that

$$^n\text{C}_r=\frac{n!}{r!(n-r)!}\\\text{So, from equation (i)}\\\text{(1-2x)}^{5}=\frac{5!}{0!(5-0)!}(1)^{5}(-2x)^{0}\\+\frac{5!}{1!(5-1)!}(-2x)^1+\frac{5!}{2!(5-2)!}(-2x)^{2}\\+\frac{5!}{3!(5-3)!}(-2x)^{3}+\frac{5!}{2!(5-4)!}(-2x)^{4}\\+\frac{5!}{5!(5-5)!}(-2x)^{5}$$

= 1 + 5(– 2x) + 10(4x)2 + 10(– 8x3) + 5(– 16x4) + (– 32x5)

= 1 – 10x + 40x2 – 80x3 + 80x4 – 32x5.

$$\textbf{2.}\space\bigg(\frac{\textbf{2}}{\textbf{x}}-\frac{\textbf{x}}{\textbf{2}}\bigg)^{5}.$$

Sol. We know that

$$(a+b)^{n}=\displaystyle\sum_{k=0}^n\space^n\text{C}_ka^{n-k}b^{k}$$

= nC0 an + nC1 an – 1 b + nC2 an – 2 b2

$$\text{Here}\space a=\frac{2}{x},\space b=\frac{-x}{2},\\n=5\\\text{Therefore, expression of}\bigg(\frac{2}{x}-\frac{x}{2}\bigg)^{5}\\\text{can be written as}\\\bigg(\frac{2}{x}-\frac{x}{2}\bigg)^{5}=\space^5\text{C}_0\bigg(\frac{2}{x}\bigg)^{5}\bigg(\frac{-x}{2}\bigg)^{0}\\+^5\text{C}_1\bigg(\frac{2}{x}\bigg)^{4}\bigg(\frac{-x}{2}\bigg)+\space^5\text{C}_2\bigg(\frac{2}{x}\bigg)^{3}\bigg(\frac{-x}{2}\bigg)^{2}$$

$$+\space^5\text{C}_3\bigg(\frac{2}{x}\bigg)^{2}\bigg(\frac{-x}{2}\bigg)^{3}+\space^5\text{C}_4\bigg(\frac{2}{x}\bigg)^1\bigg(\frac{-x}{2}\bigg)^4\\+ \space^5\text{C}_5\bigg(\frac{2}{x}\bigg)^{0}\bigg(\frac{-x}{2}\bigg)^{5}\space\text{...(i)}\\\text{We know}\space ^n\text{C}_r=\frac{n!}{r!(n-r)!}$$

So from equation (i)

$$\bigg(\frac{2}{x}-\frac{x}{2}\bigg)^5=\frac{5!}{0!(5-0)!}\bigg(\frac{2}{x}\bigg)^5\bigg(\frac{-x}{2}\bigg)^0$$

$$+\space\frac{5!}{1!(5-1)!}\bigg(\frac{2}{x}\bigg)^{4}\bigg(\frac{-x}{2}\bigg)^{1}\\+\frac{5!}{2!(5-2)!}\bigg(\frac{2}{x}\bigg)^3\bigg(\frac{-x}{2}\bigg)^2\\+\frac{5!}{3!(5-3)!}\bigg(\frac{2}{x}\bigg)^2\bigg(\frac{-x}{2}\bigg)^3\\+\space\frac{5!}{4!(5-4)!}\bigg(\frac{2}{x}\bigg)^1\bigg(\frac{-x}{2}\bigg)^4\\+\frac{5!}{5!(5-5)!}\bigg(\frac{2}{x}\bigg)^0\bigg(\frac{-x}{2}\bigg)^5$$

$$=\space\frac{32}{x^5}-5×\frac{16}{x^4}×\frac{x}{2}+10×\frac{8}{x^3}×\frac{x^2}{4}\\+10\bigg(\frac{4}{x^2}\bigg)\bigg(\frac{-x^3}{8}\bigg)+5\bigg(\frac{2}{x}\bigg)\bigg(\frac{x^4}{16}\bigg)\\1.1\bigg(\frac{-x^6}{32}\bigg)\\=\frac{32}{x^5}-\frac{40}{x^3}+\frac{20}{x}-5x+\frac{5}{8}x^3-\frac{x^5}{32}.$$

3. (2x – 3)6.

Sol. We know that

$$(a+b)^n=\displaystyle\sum_{k=0}^n\space ^n\text{C}_k\space a^{n-k}b^k\\=\space^n\text{C}_0a^n+\space^n\text{C}_1a^{n-1}b+\space^n\text{C}_2a^{n-2}b^2+...$$

Here a = 2x, b = – 3, n = 6

Therefore, expansion of (2x – 3)6 can be written as

(2x – 3)6 = 6C0 (2x)6 (– 3)0 + 6C1 (2x)5 (– 3)1+ 6C2 (2x)4 (– 3)2 + 6C3 (2x)3 (– 3)3 + 6C4 (2x)2 (– 3)4 + 6C5 (2x) (– 3)5
+ 6C6 (2x)0 (– 3)6

We know that

$$^n\text{C}_r=\frac{n!}{r!(n-r)!}$$

So from equation (i),

$$(2x-3)^6=\frac{6!}{0!(6-0!)}(2x)^6(-3x)^0\\+\frac{6!}{1!(6-1)!}(2x)^5(-3)^1\\+\frac{6!}{2!(6-2)!}(2x^4)(-3)^2\\+\frac{6!}{3!(6-3)!}(2x)^3(-3)^3\\+\frac{6!}{4!(6-4)!}(2x)^2(-3)^4\\+\frac{6!}{5!(6-5)!}(2x)^1(-3)^5\\+\frac{6!}{6!(6-6)!}(2x)^0(-3)^6$$

= 64x6 + 6(32x5) (– 3) + (15) (16x4) (9) + 20(8x3) (27) + 15(4x2) (81) + 6(2x) (– 243) + 729

= 64x6 – 576x5 + 2160x4 – 4320x3 + 4860x2 – 2916x + 729.

$$\textbf{4.}\space\bigg(\frac{\textbf{x}}{\textbf{3}}+\frac{\textbf{1}}{\textbf{x}}\bigg)^\textbf{5}\textbf{.}$$

Sol. We know that

$$\text{(a+b)}^n=\displaystyle\sum_\text{k=0}^n\space^n\text{C}_k\space a^{n-k}b^k\\=\space^n\text{C}_0a^n+\space ^n\text{C}_1a^{n-1}b+\space^n\text{C}_2a^{n-2}b^2+...\\\text{Here}\space a=\frac{x}{3},\space b=\frac{1}{x},n=5$$

$$\text{Therefore, expansion of}\space\bigg(\frac{x}{3}-\frac{1}{3}\bigg)^5\\\text{can be written as}\\\bigg(\frac{x}{3}+\frac{1}{x}\bigg)^5=\space^5\text{C}_0\bigg(\frac{x}{3}\bigg)^5\bigg(\frac{1}{x}\bigg)^0+\\\space^5\text{C}_1\bigg(\frac{x}{3}\bigg)^4\bigg(\frac{1}{x}\bigg)^1$$

$$\text{Therefore, expansion of}\space\bigg(\frac{x}{3}-\frac{1}{3}\bigg)^5\\\text{can be written as}\\\bigg(\frac{x}{3}+\frac{1}{x}\bigg)^5=\space^5\text{C}_0\bigg(\frac{x}{3}\bigg)^5\bigg(\frac{1}{x}\bigg)^0+\\\space^5\text{C}_1\bigg(\frac{x}{3}\bigg)^4\bigg(\frac{1}{x}\bigg)^1\\+\space^5\text{C}_2\bigg(\frac{x}{3}\bigg)^3\bigg(\frac{1}{x}\bigg)^2\\+\space^5\text{C}_3\bigg(\frac{x}{3}\bigg)^{2}\bigg(\frac{1}{x}\bigg)^3\\+^5\text{C}_4\bigg(\frac{x}{3}\bigg)^1\bigg(\frac{1}{x}\bigg)^4$$

$$^5\text{C}_5\bigg(\frac{x}{3}\bigg)^0\bigg(\frac{1}{x}\bigg)^5\space\text{...(i)}\\\text{We know}\space^n\text{C}_r=\frac{n!}{r!(n-r)!}\\\text{So, from equation (i)}\\\bigg(\frac{x}{3}+\frac{1}{x}\bigg)^5=\frac{5!}{0!(5-0)!}\bigg(\frac{x}{3}\bigg)^{5}\bigg(\frac{1}{x}\bigg)^{0}\\+\frac{5!}{1!(5-1)!}\bigg(\frac{x}{3}\bigg)^4\bigg(\frac{1}{x}\bigg)^{1}\\+\frac{5!}{2!(5-2)!}\bigg(\frac{x}{3}\bigg)^{3}\bigg(\frac{1}{x}\bigg)^{2}\\+\frac{5!}{3!(5-3)!}\bigg(\frac{x}{3}\bigg)\bigg(\frac{1}{x}\bigg)^{3}$$

$$+\frac{5!}{4!(5-4)!}\bigg(\frac{x}{3}\bigg)^{1}\bigg(\frac{1}{x}\bigg)^{4}\\+\frac{5!}{5!(5-5)!}\bigg(\frac{x}{3}\bigg)^{0}\bigg(\frac{1}{x}\bigg)^{5}\\=\frac{x^{5}}{243}+5\bigg(\frac{x^4}{81}\bigg)\bigg(\frac{1}{x}\bigg)\\+\frac{10x^{3}}{27}\bigg(\frac{1}{x^{2}}\bigg)+10\frac{x^{2}}{9}\bigg(\frac{1}{x^{3}}\bigg)\\+5\bigg(\frac{x}{3}\bigg)\bigg(\frac{1}{x^{4}}\bigg)+\frac{1}{x^{5}}$$

$$=\frac{x^{5}}{243}+\frac{5}{81}x^{3}+\frac{10}{27}x+\frac{10}{9x}+\frac{5}{3x^{3}}+\frac{1}{x^{5}}.$$

$$\textbf{5.}\space\bigg(\textbf{x}+\frac{\textbf{1}}{\textbf{x}}\bigg)^{\textbf{6}}\textbf{.}$$

Sol. We know that

$$(a+b)^{n}=\displaystyle\sum_{k=0}^n\space^n\text{C}_k\space a^{n-k}b^{k}$$

= nC0 an + nC1 an – 1 b + nC2 an – 2 b2 +…

$$\text{Here a = x, b =}\frac{1}{x},\text{n=6}$$

$$\text{Expansion of}\space\bigg(x+\frac{1}{x}\bigg)^{6}\\\text{can be written as}$$

$$\text{Expansion of}\space\bigg(x+\frac{1}{x}\bigg)^{6}\\\text{can be written as}\\\bigg(x+\frac{1}{x}\bigg)^{6}=\space^6\text{C}_0(x)^{6}\bigg(\frac{1}{x}\bigg)^{0}\\+^{6}\text{C}_1x^{5}\bigg(\frac{1}{x}\bigg)^{1}+\space^6\text{C}_2x^{4}\bigg(\frac{1}{x}\bigg)^{2}\\+^{6}\text{C}_3x^{3}\bigg(\frac{1}{x}\bigg)^{3}+\space^{6}\text{C}_4(x)^{2}\bigg(\frac{1}{x}\bigg)^{4}\\+^{6}\text{C}_5(x)^{1}\bigg(\frac{1}{x}\bigg)^{5}+\space^{6}\text{C}_6(x)^0\bigg(\frac{1}{x}\bigg)^{6}\space\text{...(i)}\\\text{We know}\space ^n\text{C}_r=\frac{n!}{r!(n-r)!}$$

So, from equation (i)

$$\bigg(x+\frac{1}{x}\bigg)^{6}=\frac{6!}{0!(6-0)!}x^6\\+\frac{6!}{1!(6-1)!}\bigg(\frac{1}{x}\bigg)+\frac{6!}{2!(6-2)!}x^4\bigg(\frac{1}{x}\bigg)^2\\+\frac{6!}{3!(6-3)!}(x)^3\bigg(\frac{1}{3}\bigg)^3\\+\frac{6!}{4!(6-4)!}x^2\bigg(\frac{1}{x}\bigg)^4+\frac{6!}{5!(6-5)!}x^1\bigg(\frac{1}{x}\bigg)^5$$

$$+\frac{6!}{6!(6-6)!}(x)^0\bigg(\frac{1}{x}\bigg)^6\\=x^6+6(x)^5\bigg(\frac{1}{x}\bigg)+15x^{4}\bigg(\frac{1}{x^2}\bigg)\\+20x^3\bigg(\frac{1}{x^3}\bigg)+15x^2\bigg(\frac{1}{x^4}\bigg)\\+6x\bigg(\frac{1}{x^5}\bigg)+\frac{1}{x^6}\\=x^6+6x^4+15x^2+20\\+\frac{15}{x^2}+\frac{6}{x^4}+\frac{1}{x^6}.$$

Direction (Q. Nos. 6 to 9): Using binomial theorem, evaluate each of the following:

6. (96)3.

Sol. (96)3 = (100 – 4)3

Since, 96 can be expressed as the sum or difference of two numbers and then binomial theorem can be applied.

We know that

$$(a+b)^n=\displaystyle\sum_{k=0}^n\space ^n\text{C}_k a^{n-k}b^k$$

= nC0 an + nC1 an – 1 b + nC2 an – 2 b2

Here a = 100, b = – 4, n = 3

Therefore, expansion of (100 – 4)3 can be written as (100 – 4)3 = 3C0 (100)3 (– 4)0 + 3C1 (100)2 (– 4)1 + 3C2 (100) (– 4)2 + 3C3 (100)0 (– 4)3 …(i)

We know that

$$^n\text{C}_r=\frac{n!}{r!(n-r)!}$$

So, from equation (i)

$$(100-4)^{3}=\frac{3!}{0!(3-0)!}(100)^{3}(-4)^0\\+\frac{3!}{1!(3-1)!}(100)^2(-4)^1+\\\frac{3!}{2!(3-2)!}(100)(-4)^2\\+\frac{3!}{3!(3-3)!}(100)^0(-4)^3$$

= 1000000 + 3 × 10000 (– 4) + 3 × 100 × 16 + (– 64)

= 1000000 – 120000 + 4800 – 64

= 1000000 + 4800 – 120064

= 884736.

7. (102)5.

Sol. (102)5 = (100 + 2)5

Since, 102 can be expressed as the sum or different of two numbers and then binomial theorem can be applied.

We know that

$$(a+b)n=\displaystyle\sum_{k=0}^n\space ^n\text{C}_k\space a^{n-k}\space b^{k}$$

= nC0 an + nC1 an – 1 b + nC2 an – 2 b2 + …

Here

a = 100, b = 2, n = 5

Therefore, expansion of (100 + 2)5 can be written as

(100 + 2)5 = 5C0 (100)5 + (2)0 + 5C1 (100)4 (2)1 + 5C2 (100)3 (2)2 + 5C3 (100)2 (2)3 + 5C4 (100) (2)4 + 5C5 (2)5 …(i)

$$\text{We know}\space^n\text{C}_r=\frac{n!}{r!(n-r)!}\\\text{So, from equation}\\\text{(100 × 2)}^{5}=\frac{5!}{0!(5-0)!}(100)^{5}(2)^0\\+\space\frac{5!}{1!(5-1)!}(100)^4(2)^1\\+\frac{5!}{2!(5-2)!}(100)^3(2)^2\\+\frac{5!}{3!(5-3)!}(100)^2(2)^3\\+\frac{5!}{4!(5-4)!}(100)(2)^4\\+\frac{5!}{5!(5-5)!}(2)^5$$

= 10000000000 + 5 × 100000000 × 2 + 10 × 1000000 × 4 + 10 × 10000 × 8 + 5 × 100 × 16 + 32

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

= 11040808032.

8. (101)4.

Sol. (101)4 = (100 + 1)4

Since, 101, can be expressed as the sum or different of two numbers can then binomial theorem can be applied.

We know that

$$(a+b)^n=\displaystyle\sum_{k=0}^n\space^n\text{C}_k\space a^{n-k}\space b^k$$

= nC0 an + nC1 an – 1 b + nC2 an – 2 b2 + …

Here  a = 100, b = 1, n = 4

Expansion of (100 + 1)4 can be written as

(100 + 1)4 = 4C0 (100)4 (1)0 + 4C1 (100)3 × 4 + 4C2 (100)2 (4)2 + 4C3 (100)1 (4)3 + 4C4 (4)4 …(i)

We know that

$$^n\text{C}_r=\frac{n!}{r!(n-r)!}\\\text{From equation}\\(100+1)^4=\frac{4!}{0!(4-0)!}(100)^4-(1)^0\\+\frac{4}{1!(4-1)!}(100)^3(1)\\+\frac{4!}{2!(4-2)!}(100)^{2}(1)^2\\+\frac{4!}{3!(4-3)!}(100)^1(1)^3\\+\frac{4!}{4!(4-4)!}(1)^4$$

= 100000000 + 4 × 1000000 × 1 + 6 × 10000 + 4 × 100 × 1 + 1

= 100000000 + 4000000 + 60000 + 400 + 1

= 104060401.

9. (99)5.

Sol. (99)5 = (100 – 1)5

Since, 99 can be expressed as the sum or difference of two numbers then binomial theorem can be applied.

We know that

$$(a+b)^n=\displaystyle\sum_{k=0}^n\space^n\text{C}_k\space a^{n-k}\space b^k$$

= nC0 an + nC1 an – 1 b + nC2 an – 2 b2 + …

Here a = 100, b = – 1, n = 5

Expansion of (100 – 1)5 can be written as

(100 – 1)5 = 5C0 (100)5 (– 1)0 + 5C1 (100)4 (– 1)1
+ 5C2 (100)3 (– 1)2 + 5C3 (100)2 (– 1)3
+ 5C4 (100)1 (– 1)4 + 5C5 (100)0 (– 1)5 …(i)

We know that

$$^n\text{C}_r=\frac{n!}{r!(n-r)!}\\\text{From equation}\\(100-1)^5=\frac{5!}{0!(5-0)!}(100)^5\\+\frac{5!}{1!(5-1)!}(100)^4(-1)^1\\+\frac{5!}{2!(5-2)!}(100)^3(-1)^2\\+\frac{5!}{3!(5-3)!}(100)^2(-1)^3\\+\frac{5!}{4!(5-4)!}(100)^1(-1)^4\\\frac{5!}{5!(5-5)!}(100)^0(-1)^5$$

= 1 × 10000000000 + 5 × 100000000 × (– 1) + 10 × 1000000 + 10 × 10000(– 1) + 5 × 100 × (1) + (– 1)

= 10000000000 – 500000000 + 1000000 – 100000 + 500 – 1

= 9509900499.

10. Using binomial theorem, indicate which number is larger (1·1)10000 or 1000.

Sol. (1·1)10000

By splitting the given 1·1 and then applying binomial theorem, the first few terms of (1·1)10000 can be obtained as

(1 + 0·1)10000

We know

$$(a+b)^n=\displaystyle\sum_{k=0}^n\space^n\text{C}_k\space a^{n-k}\space b^k$$

= nC0 an + nC1 an – 1 b + nC2 an – 2 b2 + …

Here a = 1, b = 0·1, n = 10000

Expansion of (1 + 0·1)10000 can be written as

(1 + 0·1)10000 = 10000C0 (1)10000 (0·1)0 + 10000C1 (1)9999 (0·1) + 10000C2 (1)9998 (0·1)2 + …

$$=\frac{10000!}{0!(10000-0)!}(1)^{10000}+\\\frac{10000}{1!(10000-1)!}(1)^{9999}(0.1)+...\\=1+10000×\frac{1}{10}+...$$

= 1 + 1000 + …

= 1001 + …

So, (1·1)10000 is greater than 1000.

11. Find (a + b)4 – (a – b)4. Hence, evaluate

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

Sol. (a + b)4 – (a – b)4

$$\text{So,}\space(a+b)^4=\displaystyle\sum_{k=0}^n\space^n\text{C}_k\space a^{n-k}b^k\\=\space^n\text{C}_0a^n+\space^n\text{C}_1a^{n-1}b+\space^n\text{C}_2a^{n-2}b^2+...$$

Here  n = 4

$$(a+b)^4=\space^4\text{C}_0a^4b^0+^4\text{C}_1a^3b^1+\space^4\text{C}_2a^2b^2+\\\space^4\text{C}_3a^1b^3+\space^4\text{C}_4a^0b^4$$

(a – b)4 = 4C0 a4 b04C1 a3 b1 + 4C2 a2 b2 4C3 a1 b3 + 4C1 b4

(a + b)4 – (a – b)4

= [4C0 a4 + 4C1 a3b + 4C2 a2b2 + 4C3 a1b3 + 4C4 b44C0 a4 + 4C1 a3b – 4C2 a2b2 + 4C3 a1b34C4 b

= [4C1 a3b + 4C3 a1b3 + 4C1 a3b + 4C3 a1b3]

= 2[4C1 a3 b + 4C3 a1 b3]

$$=2\bigg[\frac{4!}{1!(4-1)!}a^3b+\frac{4!}{3!(4-3)!}a^1b^3\bigg]$$

= 2[4a3b + 4a1b3]

= 8ab[a2 + b2] …(i)

Here given,

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

$$\text{So,}\space a=\sqrt{3},b=\sqrt{2}$$

Therefore,

$$8ab[a^2+b^2]=8(\sqrt{3})(\sqrt{2})[(\sqrt{3})^2+(\sqrt{2})^2]$$

[from equation (i)]

$$=(8\sqrt{6})×(5)\\=40\sqrt{6}.$$

12. Find (x + 1)6 + (x – 1)6. Hence or otherwise evaluate

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

Sol. (x + 1)6 + (x – 1)6

Here n = 6

We know that

$$(a+b)^n=\displaystyle\sum_{k=0}^n\space^n\text{C}_k\space a^{n-k}b^k\\=\space^n\text{C}_0a^n+\space^n\text{C}_1a^{n-1}b+\space^n\text{C}_2a^{n-2}b^2+...$$

So, (ax + 1)6 = 6C0 x6 (1)0 + 6C1 x5 (1)1 + 6C2 x412 + 6C3 x313 + 6C4 x214 + 6C5 x15 + 6C6 16

…(i)

(x – 1)6 = 6C0 x6 + 6C1 x5 (– 1) + 6C2 x4 (– 1)2 + 6C3 x3 (– 1)3 + 6C2 x2 (– 1)4 + 6C5 x (– 1)5 + 6C6 (– 1)6 …(ii)

Equation (i) + equation (ii)

(x + 1)6 + (x – 1)6 = [6C0 x6 + 6C1 x5 + 6C2 x4 + 6C3 x3 + 6C4 x2 + 6C5 x + 6C6 + 6C0 x66C1 x5 + 6C2 x46C3 x3 + 6C4 x26C5 x + 6C6]

= 2[6C0 x6 + 6C2 x4 + 6C4 x2 + 6C6]

$$=2\bigg[\frac{6!}{0!(6-0)!}x^6+\frac{6!}{2!(6-2)!}x^4\\\frac{6!}{4!(6-4)!}x^2+\frac{6!}{6!(6-6)!}\bigg]$$

= 2[x6 + 15x4 + 15x2 + 1]

$$\text{Here, given}\\(\sqrt{2}+1)^6+(\sqrt{2}-1)^6\\\text{So,}\space x=\sqrt{2}\\\text{Therefore,} 2[x^6 + 15x^4 + 15x^2 + 1]\\=2[(\sqrt{2})^6+15(\sqrt{2})^4+15(\sqrt{2})^2+1]$$

= 2[8 + 15 × 4 + 15 × 2 + 1]

= [8 + 60 + 30 + 1]

= 2[99]

= 198.

13. Show that 9n + 1 – 8n – 9 is divisible by 64, whenever n is a positive integer.

Sol. 9n + 1 = (1 + 8)n + 1

In order to show that 9n + 1 – 8n – 9 = 64k, where k is some natural number.

We know that

$$(a+b)^n=\displaystyle\sum_{k=0}^n\space^n\text{C}_ra^{n-k}b^k$$

= nC0 an + nC1 acn – 1b + nC2 an – 2b2 + …

Here a = 1, b = 8, n = n + 1

Therefore,

(1 + 8)n + 1 = n + 1C0 (1)n + 1 + n + 1C1 (1)n (8) + n + 1C2 (1)n – 1 (8)2 + …

(9)n + 1 = 1 + (n + 1)8 + 82[n + 1C2 (1)n – 1 + …]

(9)n + 1 = 1 + 8n + 8 + 64(k)

(9)n + 1 = 9 + 8n + 64k

9n + 1 – 8n – 9 = 64(k)

So, we can say

9n + 1 – 8n – 9 is divisible by 64.

So, we can say

9n + 1 – 8n – 9 is divisible by 64.

$$\textbf{14. Prove that}\space\displaystyle\sum_{\textbf{r=0}}^\textbf{n}\textbf{3r}\space^\textbf{n}\textbf{C}_\textbf{r}=4^\textbf{n}.$$

Sol. By Binomial theorem,

$$\displaystyle\sum_{r=0}^n\space^n\text{C}_r\space a^{n-r}b^{r}=(a+b)^n\space\text{...(i)}\\\displaystyle\sum_\text{r=0}^n\space^{3r}\text{C}_{r}=(4)^{n}\space\text{...(ii)}$$

On right side we need 4n, so we will put the values as putting b = 3 and a = 1 in the above equation.

So, (a + b)n = 4n

a + b = 4

Put a = 1 and b = 3

$$\displaystyle\sum_{r=0}^n\space3^r\space^n\text{C}_r=4n\\\displaystyle\sum_{r=0}^n\space^n\text{C}_r(1)^{n-r}3^r=(7+3)^n\\\displaystyle\sum_{r-0}^n\space ^n\text{C}_r3^r=4^n.$$

Exercise 8.2

Find the coefficient of:

1. x5 in (x + 3)8.

Sol. We know general term,

Tr + 1 = nCr an – rbr …(i)

By comparing (x + 3)8 with (a + b)n, we get

a = x

b = 3 and n = 8

From equation (i),

Tr + 1 = 8Cr x8 – r 3r
We need coefficient of x5

Therefore,

x8 – r = x5

8 – r = 5 (∵ compairing powers)

r = 3

T3 + 1 = 8C3 x8 – 3 (3)3

T3 + 1 = 8C3 (3)3 x5

Coefficient of x2 is 8C3 (3)3

$$=\frac{8!}{3!5!}×27\\=\frac{8×7×6×5!}{3×2×5!}×27$$

= 1512.

2. a5b7 in (a – 2b)12.

Sol. a5b7 in (a – 2b)12

Compare (a – 2b)12 with (a + b)n, we get

a = a, b = – 2b, n = 12

As we know, general term

Tr + 1 = nCr an – rbr …(i)

Tr + 1 = 12Cr a12 – r (– 2b)r

We need coefficient of a5b7

a12 – r = a5

12 – r = 5 (∵ comparing powers)

r = 12 – 5

equation r = 7

From (i)

T7 + 1 = 12C7 a12 – 7 (– 2b)7

= 12C7 a5b7 (– 2)7

Coefficient of a5b7 is 12C7 (– 2)7

$$=\frac{12!}{7!(12-7)!}(-2)^{2}\\=\frac{12×11×10×9×8×7!}{7!\space5×4×3×2×1}(-128)$$

= 12 × 11 × 6 × (– 128)

= – 101376.

Direction (Q. Nos. 3 and 4): Write the general term in the given expansions.

3. (x2 – y)6.

Sol. Compare (x2 – y)6 with (a + b)n, we get

a = x2, b = – y, n = 6

As we know the general term

Tr + 1 = nCr an – 1br

Tr + 1 = 6Cr (x2)6 – r (– y)r

= 6Cr x12 – 2r (– 1 × y)r

= 6Cr x12 – 2r (– 1)r yr.

4. (x2 – yx)12, x ≠ 0.

Sol. Compare (x2 – yx) with (a + b)n, we get

a = x2, b = – yx, n = 12

As we know the general term

Tr + 1 = nCr an – rbr

Tr + 1 = 12Cr (x2)12 – r (– yx)r

Tr + 1 = 12Cr x24 – 2r (– 1 × y × x)r

= 12Cr x24 – 2r (– 1)r (y)r (x)r

= (– 1)r 12Cr x24 – 2r yr xr.

5. Find the 4th term in the expansion of (x – 2y)12.

Sol. Compare (x – 2y)12 with (a + b)n, we get

a = x, b = – 2y, n = 12

As we know the general term

Tr + 1 = nCr an – rbr

We need 4th term

T4 = T3 + 1 = 12C3 x12 – 3 (– 2y)3 [r = 3]

= 12C3 x9 (– 2)3 y3

$$=\frac{12!}{3!(12-3)!}x^9(-8)y^3\\=\frac{12×11×10×9!}{3!9!}x^9(-8)y^3\\=\frac{12×11×10×(-8)}{3×2}x^9y^3$$

= 20 × 11 × (– 8) x9y3

= – 1760 x9y3.

6. Find the 13th term in the expansion of

$$\bigg(\textbf{9x}-\frac{\textbf{1}}{\textbf{3}\sqrt{\textbf{x}}}\bigg)^{\textbf{18}},\textbf{x}\neq\textbf{0.}$$

$$\textbf{Sol.}\space\text{Compare}\bigg(9x-\frac{1}{3\sqrt{x}}\bigg)\text{with (a + b)}^{n},\\\text{we get}\\\text{a = 9x,}\space b=\frac{-1}{3\sqrt{x}},n=18$$

As we know the general term

Tr + 1 = nCr an – rbr

We need 13th term

$$\text{T}_{13} = \text{T}_{12 + 1} =\space^{18}\text{C}_{12}(9x)^{18-12}\bigg(\frac{-1}{3\sqrt{x}}\bigg)^{12}\\=\frac{18!}{12!6!}×(9x)^6×(-1)^{12}\bigg(\frac{1}{3}\bigg)^{12}\bigg(\frac{1}{\sqrt{x}}\bigg)^{12}\\=\frac{18!}{12!6!}×9^6x^6×\frac{1}{3^{12}}×\frac{1}{[(x)^{1/2}]^{1/2}}\\=\frac{18×17×16×15×14×13×12!}{12!×6×5×4×3×2×1}\\×9^6x^6\frac{1}{3^{2×6}}×\frac{1}{x^{\bigg(\frac{1}{2}×12\bigg)}}$$

$$= 17 × 2 × 3 × 14 × 13 × 9^6x^6\frac{1}{9^6}×\frac{1}{x^6}$$

17 × 2 × 3 × 14 × 13

= 18564.

Direction (Q. Nos. 7 to 8): Find the middle term in the given expressions.

$$\textbf{7.}\space\bigg(\textbf{3}-\frac{\textbf{x}^\textbf{3}}{\textbf{6}}\bigg)^\textbf{7}\textbf{.}$$

Sol. In this expansion n is odd so there are two middle

$$\text{term}\bigg(\frac{n+1}{2}\bigg)^{\text{th}}\text{and}\bigg(\frac{n+1}{2}+1\bigg)^{\text{th}}\\\text{Compare}\bigg(3-\frac{x^{3}}{6}\bigg)^7\space\text{with (a + b)}^n\\\text{Here}\space a=3,b=\frac{x^{3}}{6},n=7\\\text{Middle term =}\bigg(\frac{7+1}{2}\bigg)^{\text{th}}=\bigg(\frac{8}{2}\bigg)^{\text{th}}\\=\text{4}^{\text{th}}\space\text{term}\\\text{and}\space\bigg(\frac{7+1}{2}\bigg)^{\text{th}}=5^{\text{th}}\space\text{term}$$

As we know the general term

Tr + 1 = nCr an – r br

T4 = T3 + 1 = 7C3 a7 – 3b3   (r = 3)

$$=\space^7\text{C}_3(3)^4\bigg(-\frac{x^3}{6}\bigg)^3\\=\frac{7!}{3!(7-3)!}(3)^4\bigg(\frac{-x^3}{6}\bigg)^3\\=\frac{-7×6×5×4!}{3×2×4!}×81×\frac{x^9}{6^3}\\=\frac{-35×81}{216}x^9\\=\frac{-105}{8}x^9(4^{\text{th}}\space\text{term})$$

T5 = T4 + 1 = 7C4 (a)7 – 4 (b4)  [r = 4]

$$=\space^7\text{C}_4\space(3)^3\bigg(\frac{-x^3}{6}\bigg)^4\\=\frac{7!}{4!(7-4)!}(27)\frac{(x^3)^4}{6^4}\\=\frac{7×6×5×4!}{3!}×27×\frac{x^{12}}{6^4}\\=\frac{7×6×5}{3×2}×27×\frac{x^{12}}{6^{4}}\\=\frac{35×27}{36×36}x^{12}\\=\frac{35}{48}x^{12}(5^{\text{th}\space }\text{term}).$$

$$\textbf{8.}\space\bigg(\frac{\textbf{x}}{\textbf{3}}\textbf{+9y}\bigg)^{\textbf{10}}.$$

Sol. In this expansion n is even, so single middle term

$$\text{exist and given by}\bigg(\frac{n}{2}+1\bigg)^{\text{th}}\space\text{term}\\\text{Middle term =}\bigg(\frac{10}{2}+1\bigg)^{\text{th}}=\text{6}^{\text{th}}\space\text{term}\\\text{Compare}\bigg(\frac{x}{3}+9y\bigg)^{10}\text{with (a + b)}^n\\\text{We get a =}\frac{x}{3},\text{b = 9y, n = 10}$$

As we know the general term

Tr + 1 = nCr an – rbr

$$\text{T}_6=\text{T}_{5+1}=\space^{10}\text{C}_5\bigg(\frac{x}{3}\bigg)^{10-5}\space(9y)^5\space[r=5]\\=\frac{10!}{5!(10-5)!}\bigg(\frac{x}{3}\bigg)^{5}(9^5)×(y^5)\\=\frac{10×9×8×7×6×5!}{5!×5×4×3×2×1}×\\\frac{x^5}{3^5}×9^5×y^5\\=\frac{3×2×7×6}{35}×x^5×(3^2)^5×y^5\\=\frac{6×42×x^5×3^{10}}{3^5}×y^5$$

= 6 × 42 × x5 × 243 × y5

= 61236 x5y5 (6th term).

9. In the expansion of (1 + a)m + n, prove that the coefficient of am and an are equal.

Sol. We need to find coefficient of am and an

As we know the general term is given by

Tr + 1 = nCr an – rbr

Compare (1 + a)m + n with (a + b)n, we get

a = 1, b = a, n = m + n

So, Tr + 1 = m + nCr (1)m + n – r (a)r …(i)

We need coefficient am and an

ar = am

r = m [comparing powers]

and ar = an

r = n [comparing powers]

When r = m from equation (i)

Tm + 1 = m + nCm (1)m + n – m (a)m

= m + nCm · (a)m [1n = 1]

Coefficient of am = m + nCm …(ii)

Similarly when r = n from equation (i)

Tn + 1 = m + nCn (1)m + n – n (an)

= m + nCn an [(1)m = 1]

Coefficient of an = m + nCn …(iii)

[from equation (ii) and (iii)]

According to question

m + nCm = m + nCn

$$\frac{(m+n)!}{m!(m+n-m)!}=\frac{(m+n)!}{n!(m+n-n)!}\\=\frac{m+n!}{m!n!}=\frac{m+n!}{n!(m)!}$$

So, coefficient of am and an are equal.

Hence Proved.

10. The coefficient of the (r – 1)th, rth and (r + 1)th terms in the expansion of (x + 1)n are in the ratio 1 : 3 : 5. Find n and r.

Sol. (x + 1)n

Compare (x + 1)n with (a + b)n

Here

a = x, b = 1, n = n

As we know the general term

Tr + 1 = nCr an – rbr

For (r – 1)th term

Tr – 1 = nCr – 2 (x)n – (r – 2) (1)r – 2 [r = r – 2]

= nCr – 2 xn – r + 2 …(i)

For (rth term)

Tr = nCr – 1 xn – (r – 1) (1)r – 1 [r = r – 1]

= nCr – 1 xn – r + 1 …(ii)

For (r + 1)th term

Tr + 1 = nCr xn – r (1)r     [r = r]

= nCr xn – r …(iii)

[from (i), (ii) and (iii)]

According to question

$$\frac{^n\text{C}_{r-2}}{^n\text{C}_{r-1}}=\frac{1}{3}\space\text{and}\space\frac{^n\text{C}_{r-1}}{^n\text{C}_{r}}=\frac{3}{5}\\\frac{\frac{n!}{(r-2)!(n-r+2)!}}{\frac{n!}{(r-1)!(n-r+1)!}}=\frac{1}{3}\\\text{and}\space\frac{\frac{n!}{(r-1)(n-r+1)!}}{\frac{n!}{r!(n-r)!}}=\frac{3}{5}\\\frac{n!×(r-1)!×(n-r+1)!}{(r-2)!(n-r+2)!×n!}=\frac{1}{3}\\\text{and}\space\frac{n!(r)!(n-r)!}{n!(r-1)!(n-r+1)!}=\frac{3}{5}\\\Rarr\space\frac{(r-1)(r-2)!(n-r+1)!}{(r-2)!(n-r+2)(n-r+1)!}=\frac{1}{3}$$

$$\text{and}\space\frac{r(r-1)!(n-r)!}{(r-1)!(n-r+1)(n-r)!}=\frac{3}{5}\\\Rarr\space\frac{(r-1)}{(n-r+2)}=\frac{1}{3}\\\text{and}\space\frac{r}{n-r+1}=\frac{3}{5}$$

3(r – 1) = n – r + 2 and 5r = 3n – 3r + 3

3r – 3 = n – r + 2

– n + 4r = 5 …(iv)

and – 3n + 8r = 3 …(v)

From (iv) and (v)

$$– 3n + 12r = 15\space\space\lbrack\text{equation (v) × 3}\rbrack\\– 3n + 8r = 3\\\frac{+\qquad-}{4r=12}\\r=3$$

Put r = 3 in (iv)

– n + 4 × 3 = 5

– n = 5 – 12

– n = – 7

n = 7.

11. Prove that the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n – 1.

sol. (1+x)2n

Compare (1 + x)2n with (a + b)n

Here, a = 1, b = x, n = 2n

As we know that general term

Tr + 1 = nCr an – r br

Tr + 1 = 2nCr (1)2n – r xr

We need coefficient of xn

xr = xn

r = n [comparing powers]

Tn + 1 = 2nCn (1)2n – n xn

= 2nCn xn …(i)

and (1 + x)2n – 1

Compare (1 + x)2n – 1 with (a + b)n

Here, a = 1, b = x, n = 2n – 1

Lower term

Tr + 1 = nCr an – 1 br

Tr + 1 = 2n – 1Cr (1)2n – 1 – r (xr)

We need xn coefficient

xn = xr

r = n (comparing power)

Tn + 1 = 2n – 1Cn (1)2n – 1 – n (x)n

Tn + 1 = 2n – 1Cn xn …(ii)

From (i)

$$^{2n}\text{C}_{n}=\frac{\phase{2n}}{\phase{n}\phase{2n-n}}\\=\frac{\phase{2n}}{\phase{n}\phase{n}}=\frac{\phase{2n}}{(\phase{n})^{2}}$$

$$^{2n}\text{C}=\frac{\phase{2n}}{(\phase{n})^{2}}\space\text{...(iii)}$$

From equation (ii)

$$^{2n-1}\text{C}_n=\frac{\phase{2n-1}}{\phase{n}.\phase{2n-1-n}}\\=\frac{\phase{2n-1}}{\phase{n}\phase{n-1}}×\frac{2n}{2n}\\=\frac{2n\phase{2n-1}}{2n\phase{n}\phase{n-1}}\\=\frac{\phase{2n}}{2\phase{n}\phase{n}}\\\begin{bmatrix}\because\space2n\phase{2n-1}=\phase{2n} \\n\phase{n-1}=\phase{n}\end{bmatrix}$$

$$^{2n-1}\text{C}_n=\frac{\phase{2n}}{2(\phase{n})^{2}}\space\text{...(iv)}$$

Now from equation (iii) and (iv), we get

$$^{2n-1}\text{C}_n=\frac{1}{2}×^{2n}\text{C}_n$$

So, coefficient of (1 + x)2n is twice of (1 + x)2n – 1.

12. Find a positive value of m for which the coefficient of x2 in the expansion (1 + x)m is 6.

Sol. (1 + x)m

Compare (1 + x)n with (a + b)n

Here a = 1, b = x, n = m

General term is given by

Tr + 1 = nCr an – r br

Tr + 1 = mCr (1)m – r (x)r …(i)

We need coefficient of x2

xr = x2

r = 2 [comparing powers]

Therefore from (i)

T2 + 1 = mC2 (1)m – 2 x2

= mC2 x2 …(ii) [1n – 2 = 1]

According to question, from equation (ii)

mC2 = 6

$$\frac{m!}{2!(m-2)!}=6\\\frac{m×(m-1)(m-2)!}{2×1(m-2)!}=6$$

m(m – 1) = 12

m2 – m – 12 = 0

m2 – 4m + 3m – 12 = 0

m(m – 4) + 3(m – 4) = 0

(m – 4) (m + 3) = 0

(m – 4) = 0

or

m + 3 = 0

m = 4

or

m = – 3 (–ve)

So, for m = 4 coefficient of x2 is 6.

Miscellaneous Exercise

1. Find a, b and n in the expansion of (a + b)n, if the first three terms of the expansion are 729, 7290 and 30375 respectively.

Sol. We know that

Tr + 1 = nCr an – r br

The first three terms of the expansion are given as 729, 7290 and 30375 respectively. Then,

T1 = nC0 (a)n b0 [r = 0]

= an [nC0 = 1, b0 = 1]

an = 729 …(i)

T2 = nC1 (a)n – 1 b1 [r = 1]

= nan – 1 b

$$\bigg[\space^{n}\text{C}_1=\frac{n!}{1!(n-1)!}=n\bigg]$$

nan– 1 b = 7290 …(ii)

T3 = nC2 an – 2 b2       [r = 2]

$$=\frac{n(n-1)}{2}a^{n-2}b^2\\\bigg[\frac{n!}{2!(n-2)!}=\space^n\text{C}_2=\frac{n(n-1)}{2}\bigg]\\\frac{n(n-1)}{2}a^{n-2}b^{2}=30375\space\text{...(iii)}$$

Divide equation (ii) by (i)

$$\frac{na^{n-1}b}{a^{n}}=\frac{7290}{729}\\\frac{(n).(b)a^n(a)^{\normalsize-1}}{a^n}=10\\\frac{(n)b}{a}=10\space\text{...(iv)}$$

Divide equation (iii) by (ii)

$$\frac{n(n-1)a^{n-2}b^{2}}{2(n)^2a^{n-1}b}=\frac{30375}{7290}\\=\frac{(n-1)b}{2a}=\frac{30375}{7290}$$

$$\begin{bmatrix}\frac{a^{n-2}}{a^{n-1}}=a^{n-2(n-1)}\\=a^{(n-2-n+1)}\\=a^{\normalsize-1}=\frac{1}{a}\end{bmatrix}$$

$$\frac{(n-1)b}{a}=\frac{30375×2}{7290}\\\frac{(n-1)b}{a}=\frac{25}{3}\space\text{...(v)}\\\frac{nb}{a}-\frac{b}{a}=\frac{25}{3}\\10-\frac{b}{a}=\frac{25}{3}\\\bigg(\text{from equation (iv)}\space\frac{nb}{a}=10\bigg)\\10-\frac{25}{3}=\frac{b}{a}\\\frac{30-25}{3}=\frac{b}{a}\\\frac{b}{a}=\frac{5}{3}$$

From equation (iv)

$$\frac{nb}{a}=10\\n×\frac{5}{3}=10\space\bigg[\text{Put}\frac{b}{a}=10\bigg]\\n=\frac{30}{5}$$

n = 6

From equation (i),

9n = 729

a6 = 729

a6 = 36

a = 3

$$\frac{b}{a}=\frac{5}{3}\\b=\frac{5×3}{3}$$

b = 5.

2. Find a if the coefficients of x2 and x3 in the expansion of (3 + ax)9 are equal.

Sol. Compare (3 + ax)9 with (a + b)n

We get a = 3, b = ax, n = 9

As we know that general term

Tr + 1 = nCr an – 1 br

Tr + 1 = 9Cr (3)n – r (ax)r

We need coefficient of x2

xr = x2

r = 2 (comparing powers)

T2 + 1 = 9C2 (3)9 – 2 (ax)2

T3 = 9C2 (3)7 a2x2

We need coefficient of x3

xr = x3

r = 3

T3 + 1 = 9C3 (3)9 – 3 (9x)3

T4 = 9C3 (3)6 a3 x3

According to question,

9C3 36 a3 = 9C2 37 a2

$$a=\frac{^9\text{C}_23^7}{^9\text{C}_3 3^6}\\a=\frac{\frac{9!}{2!(9-2)!}.3}{\frac{9!}{3!(9-3)!}}\\=\frac{9!(3)!6!}{2!7!9!}.3\\a=\frac{3×2!×6!}{2!×7.6!}×3\\a=\frac{9}{7}.$$

3. Find the coefficient of x5 in the product (1 + 2x)6 (1 – x)7 by using binomial theorem.

Sol. Binomial expansion is given by

$$(a+b)^n=\displaystyle\sum_{k=0}^n\space^n\text{C}_k\space a^{n-k}b^k$$

= nC0 an + nC1 an – 1 b + nC2 an – 2 b2 + …

a = 1, b = 2x, n = 6

[compare (1 + 2x)6 with (a + b)n]

(1 + 2x)6 = 6C0 16 (2x)0 + 6C1 (1)5 (2x)1 + 6C2 (1)4 (2x)2 + 6C3 (1)3 (2x)4 + 6C4 (1)2 (2x)4 + 6C5 (1)1 (2x)5 + 6C6 (1)0 (2x)6

$$=1+\frac{6!}{1!(6-1)!}(2x)^1+\frac{6!}{2!(6-2)!}(2x)^2\\+\frac{6!}{3!(6-3)!}(2x)^3+\frac{6!}{4!(6-4)!}(2x)^4\\+\frac{6!}{5!(6-5)!}(2x)^5+\frac{6!}{6!(6-6)!}(2x)^6$$

= 1 + 6 × 2x + 15 × 4x2 + 20 × 8x3 + 15 × 16x4
+ 6 × 32 × x5 + 64x6
= 1 + 12x + 60x2 + 160x3 + 240x4 + 192x5 + 64x6
…(i)
Similarly, compare (1 – x)7 with (a + b)n

Here a = 1, b = – x, n = 7

So, (1 – x)7 = 7C0 (x)0 + 7C1 (– x)1 + 7C2 (– x)2 + 7C3 (– x)3 + 7C4 (– x)4 + 7C5 (– x)5 + 7C6 (– x)6 + 7C7 (– x)7

$$=1+\frac{7!}{1!(7-1)!}(-x)+\frac{7!}{2!(7-2)!}(-x)^2\\+\frac{7!}{3!(7-3)!}(-x)^3+\frac{7!}{4!(7-4)!}(-x)^4\\+\frac{7!}{5!(7-5)!}(-x)^5+\frac{7!}{6!(7-6)!}(-x)^6\\+\frac{7!}{7!(7-7)!}(-x)^7$$

= 1 – 7x + 21x2 – 35x3 + 35x4 – 21x5 + 7x b – x7 …(ii)

From equation (i) and (ii)

(1 + 2x)6 (1 – x)7 = (1 + 12x + 60x2 + 160x3 + 240x4+ 192x5 + 64x6) × (– 1 – 7x + 21x2 – 35x3
+ 35x4 – 21x5 + 7x6 – x7)

The terms containing x5 are

(1 + 2x)6 (1 – x)7

= – 21x5 + 420x5 – 2100x5 + 3360x5 – 1680x5 + 192x5

= – 3801x5 + 3972x5

= 171x5

Coefficient of x5 in 171.

4. If a and b are distinct integers, prove that a – b is a factor of an – bn, whenever n is a positive integer.

Sol. (a – b) is factor of an – bn

i.e., an – bn = (a – b) (k)

[k is some natural value]

So, an = (a – b + b)n

= [(a – b) + b]n

By using Binomial theorem

(a + b)n = nC0 an + nC1 an – 1 b + nC2 an – 2 b2 + …

Compare (a + b)n with [(a – b) + b]n we get

a = a – b, b = b, n = n

So, [(a – b) + b]n = nC0 (a – b)n + nC1 (a – b)n – 1.b+ … + nCn (a – b)0 bn

(a – b + b)n = (a – b)n + (a – b)n – 1 nC1 b + … + bn

$$\begin{bmatrix}\space^n\text{C}_0=1\\^n\text{C}_n=1\end{bmatrix}$$

an – bn = (a – b)

[(a – b)n – 1 + (a – b)n – 2 nC1 b + …]

an – bn = (a – b) [natural value]

So, (a – b) is factor of an – bn.

$$\textbf{5. Evaluate}\space(\sqrt{\textbf{3}}+\sqrt{\textbf{2}})^\textbf{6}-(\sqrt{\textbf{3}}-\sqrt{\textbf{2}})^\textbf{6.}$$

Sol. The given expression in the form of (a + b)6 – (a – b)6

$$\text{So, here}\space a=\sqrt{3},a=\sqrt{2}$$

Expansion can be done as

(a + b)n = nC0 an + nC1 an – 1 b + nC2 an – 2 b2 + …

So,  n = 6

(a + b)6 = 6C0 a6 (b)0 + 6C1 a5 (b) + 6C2 a4 (b)2 + 6C3 a3 (b)3 + 6C4 a2 (b)4 + 6C5 a1 (b)5 + 6C6 a0 (b)6

(a – b)6 = 6C0 a6 + 6C1 a5 (– b) + 6C2 a4 (– b)2 + 6C3 a3 (– b)3 + 6C4 a2 (– b)4 + 6C5 a1 (– b)5 + 6C6 a0 (– b)6

= 6C0 a66C1 a5 (b) + 6C2 a4 (b)26C3 a3 (b)3 + 6C4 a(b)46C5 a(b)5 + 6C6 (a)0 (b)6

(a + b)6 – (a – b)6 = 6C0 a6 + 6C1 a5b1 + 6C2 a4b2 + 6C3 a3b3 + 6C4 a2b4 + 6C5 ab5 + 6C6 b66C0 a6 + 6C1 a5b – 6C2 a4b2
+ 6C3 b36C4 a2b4 + 6C5 ab5 6C6 b6

= 2[6C1 a5 + 6C3 a3b3 + 6C5 ab5]

$$=2\bigg[\frac{6!}{1!(6-1)!}a^5+\frac{6!}{3!(6-3)!}a^3b^3+\\\frac{6!}{5!(6-5)!}ab^5\bigg]$$

= 2[6a5b + 20a3b3 + 6ab5]

$$\text{Put}\space a=\sqrt{3}\text{and}\space b=\sqrt{2}\text{we get}\\=2[6(\sqrt{3})^5(\sqrt{2})+20(\sqrt{3})^3(\sqrt{2})^3+6(\sqrt{3})(\sqrt{2})^5]\\=2[6×9\sqrt{3}×\sqrt{2}+20×3\sqrt{3}×2\sqrt{2}\\+6×\sqrt{3}×4×\sqrt{2}]$$

$$=2(54\sqrt{6}+120\sqrt{6}+24\sqrt{6})\\=2(198\sqrt{6})\\=396\sqrt{6}.$$

6. Find the value of

$$\textbf{(a}^\textbf{2}\textbf{+}\sqrt{\textbf{a}^\textbf{2}\textbf{-1}})^\textbf{4}\textbf{+}(\textbf{a}^\textbf{2}\textbf{-}\sqrt{\textbf{a}^\textbf{2}\textbf{-1}})^\textbf{4}\textbf{.}$$

Sol. The given expression is in the form of (x + y)4 and (x – y)4.

$$\text{So,}\space\text{x=a}^2\text{and}\space y=\sqrt{a^2-1}$$

Binomial expansion is given by

(a + b)n = nC0 an + nC1 an – 1 b + nC2 an – 2 b2 + …

(x + y)4 is compare with (a + b)n then

a = x, b = y = n = 4

(x + y)4 = 4C0 x4y0 + 4C1 x3y1 + 4C2 x2y2 + 4C3 x y3 + 4C4 x0y4

In (x – y)9, a = x, b = – y, n = 4

(x – y)4 = 4C0 x4 (– y)0 + 4C1 x3(– y)1 + 4C2 x2 (– y)2 + 4C3 x(– y)3 + 4C4 x0 (– y)0

= 4C0 x44C1 x3y + 4C2 x2y24C3 xy3 + 4C4 y4

$$=2\bigg[x^4+\frac{4!}{(4-2)!2!}x^2y^2+\frac{4!}{4!(4-4)}y^4\bigg]$$

= 2[x4 + 6x2y2 + y4]

$$\text{Put x = a2 and y =}\sqrt{a^2-1}\space\\\text{we get}\\=2[(a^2)^4+6(a^2)^2(\sqrt{a^2-1})^2+(\sqrt{a^2-1})^4]$$

= 2[a8 + 6a4 (a2 – 1) + (a2 – 1)2]

= 2[a8 + 6a6 – 6a4 + a4 – 1 – 2a2]

= 2[a8 + 6a6 – 5a4 – 2a2 + 1]

= 2a8 + 12a6 – 10a4 – 4a2 + 2.

7. Find an approximation of (0.99)5 using the first three terms of its expansion.

Sol. (0·99)5     [given]

It can be written as (1 – 0·01)5

Compare it with (a + b)n

Here

a = 1, b = – 0·01, n = 5

$$(a+b)^n=\displaystyle\sum_{k=0}^n\space^n\text{C}_k\space a^{n-k}\space b^k$$

= nC0 an + nC1 an – 1 b + nC2 an – 2 b2 + …

Expansion is given by (1 – 0·01)5.

So, (1 – 0·01)5 = 5C0 (1)5 (– 0·01)0 + 5C1 (1)4 (– 0·01)1 (– 0·01) + 5C2 (1)3 (– 0·01)2

[upto three terms]

$$(0.99)^5=1-\frac{5!}{1!(5-1)!}(0.01)\\+\frac{5!}{2!(5-2)!}(0.01)^2$$

= 1 – 5 × 0·01 + 10 × 0·01 × 0·01

$$1-0.05+10×\frac{1}{100}×0.01$$

= 1 – 0·05 + 0·001

= 1·001 – 0·05

= 0·951.

Ans.

8. Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of

$$\bigg(\sqrt[\textbf{4}]{\textbf{2}}\textbf{+}\frac{\textbf{1}}{\sqrt[\textbf{4}]{\textbf{3}}}\bigg)^\textbf{n}\textbf{is}\space\sqrt{\textbf{6}}\space\textbf{:\space1.}$$

Sol. Binomial expansion

(a + b)n = nC0 an + nC1 an – 1 b1 + nC2 an – 2 b2 + … + nCn – 1 a1 bn – 1 + nCn a0bn

Fifth term from begining = nC4 an – 4 b4

Fifth term from end = nCn – 4 a4 bn – 4

$$\Rarr\space\bigg(4\sqrt{2}+\frac{1}{4\sqrt{3}}\bigg)^n\space\text{compare with (9 + b)}^n$$

$$\text{Here a=4}\sqrt{2},b=\frac{1}{4\sqrt{3}},n=n$$

So, fifth term from beginning

$$=\space^n\text{C}_4(\sqrt[4]{2})^{n-4}\bigg(\frac{1}{\sqrt[4]{3}}\bigg)^4$$

Fifth term from end

$$=\space^n\text{C}_{n-4}(\sqrt[4]{2})^4\bigg(\frac{1}{\sqrt[4]{3}}\bigg)^{n-4}$$

According to question

$$\frac{^n\text{C}_4(\sqrt[4]{2})^n-4\bigg(\frac{1}{4\sqrt{3}}\bigg)^4}{^n\text{C}_{n-4}(\sqrt[4]{2})^4\bigg(\frac{1}{\sqrt[4]{3}}\bigg)^{n-4}}=\frac{\sqrt{6}}{1}$$

$$\frac{\frac{n!}{4!(n-4)}(2)^\frac{n-4}{4}\space\bigg(\frac{1}{(3)^{1/4}}\bigg)^4}{\frac{n!}{(n-4)!}(2^{1/4})^4\bigg(\frac{1}{3^{1/4}}\bigg)^{n-4}}=\frac{\sqrt{6}}{1}\\\frac{(2)^\frac{n-4}{4}.\bigg(\frac{1}{3}\bigg)}{2.\frac{1}{(3)^{\frac{n-4}{4}}}}=\frac{\sqrt{6}}{1}\\\Rarr\space(2)^{\frac{n-4}{4}}.(3)^{\frac{n-4}{4}}×\frac{1}{6}=\frac{\sqrt{6}}{1}\\(6)^{\frac{n-4}{4}}=6\sqrt{6}\\(6)^{\frac{n-4}{4}}=6(6)^{1/2}\\(6)^\frac{n-4}{4}=(6)^{3/2}$$

Comparing powers

$$\frac{n-4}{4}=\frac{3}{2}$$

2(n – 4) = 3 × 4

2n – 8 = 12

2n = 20

$$n=\frac{20}{2}$$

n = 10.

9. Expand using binomial theorem

$$\bigg(\textbf{1}+\frac{\textbf{x}}{\textbf{2}}-\frac{\textbf{2}}{\textbf{x}}\bigg)^\textbf{4}\textbf{,}\space\textbf{x}\neq\textbf{0}.$$

$$\textbf{Sol.}\space\bigg(1+\frac{x}{2}-\frac{2}{x}\bigg)^4=\bigg[\bigg(1+\frac{x}{2}\bigg)-\frac{2}{x}\bigg]^4$$

The given expression is in the form of (a + b)4

By comparing this with (a + b)n

$$\text{Here},a=\bigg(1+\frac{x}{2}\bigg),b=\frac{-2}{x},n=4$$

Binomial expansion is given by

(a + b)n = nC0 an + nC1 an – 1 b1 + nC2 an – 2 b2 + …

$$\bigg[\bigg(1+\frac{x}{2}\bigg)-\frac{2}{x}\bigg]^4=\space^{4}\text{C}_0\bigg(1+\frac{x}{2}\bigg)^4\bigg(\frac{-2}{x}\bigg)^0+$$

$$^4\text{C}_1\bigg(1+\frac{x}{2}\bigg)^3\bigg(\frac{-2}{x}\bigg)^1+\space^4\text{C}_2\bigg(1+\frac{x}{2}\bigg)\bigg(\frac{-2}{x}\bigg)^2\\+^4\text{C}_3\bigg(1+\frac{x}{2}\bigg)^1\bigg(\frac{-2}{x}\bigg)^3+\space^4\text{C}_4\space\bigg(1+\frac{x}{2}\bigg)^0\bigg(\frac{-2}{x}\bigg)^4$$

$$=1\bigg(1+\frac{x}{2}\bigg)^4+\frac{4!}{1!(4-1)!}\bigg(1+\frac{x}{2}\bigg)^3\bigg(\frac{2}{x}\bigg)\\+\frac{4!\bigg(1+\frac{x}{2}\bigg)^2}{2!(4-2)!}\bigg(\frac{-2}{x}\bigg)^2+\\\frac{4!}{3!(4-3)!}\bigg(1+\frac{x}{2}\bigg)^1\bigg(\frac{-2}{x}\bigg)^3\\+^4\text{C}_4\bigg(1+\frac{x}{2}\bigg)^0\bigg(\frac{-2}{2}\bigg)^0$$

$$=\bigg(1+\frac{x}{2}\bigg)^4-\frac{8}{x}\bigg(1+\frac{x}{2}\bigg)^3+\\6\bigg(1+\frac{x^2}{4}+x\bigg)\frac{4}{x^2}-\\4\bigg(1+\frac{x}{2}\bigg)\bigg(\frac{8}{x^3}\bigg)+\frac{16}{x^4}\space\text{...(i)}$$

$$\text{Therefore}\space\bigg(1+\frac{x}{2}\bigg)^4\text{can be expanded binomially.}\\\text{Here}\space a=1,b=\frac{x}{2},n=4\\\bigg(1+\frac{x}{2}\bigg)^4=\space^4\text{C}_0(1)^4\bigg(\frac{x}{2}\bigg)^0+\\\space^4\text{C}_1(1)^3\bigg(\frac{x}{2}\bigg)^1+\space^4\text{C}_2(1)^2\bigg(\frac{x}{2}\bigg)^2+\\\space^4\text{C}_3(1)^1\bigg(\frac{x}{2}\bigg)^3+\space^4\text{C}_4(1)^{0}\bigg(\frac{x}{2}\bigg)^4$$

$$=1+\frac{4!}{1!(4-1)!}\bigg(\frac{x}{2}\bigg)+\frac{4!}{2!(4-2)!}(1)\bigg(\frac{x}{2}\bigg)^2\\\frac{4!}{3!(4-3)!}\bigg(\frac{x}{2}\bigg)^3+\frac{4!}{4!(4-4)!}\bigg(\frac{x}{2}\bigg)^4\\\text{...(ii)}$$

$$1+\frac{4x}{2}+\frac{6x^2}{4}+\frac{4x^3}{8}+\frac{x^4}{16}\\=1+2x+\frac{3x^2}{2}+\frac{x^3}{2}+\frac{x^4}{16}$$

and

$$\bigg(1+\frac{x}{2}\bigg)^3=\space^3\text{C}_0(1)^3+^3\text{C}_1(1)^2\frac{x}{2}+\\\space^3\text{C}_2(1)\bigg(\frac{x}{2}\bigg)^2+\space^3\text{C}_3(1)^0\bigg(\frac{x}{2}\bigg)^3\\=1+\frac{3!}{(3-1)!(1)!}\bigg(\frac{x}{2}\bigg)^1+\\\frac{3!}{2!(3-2)!}\bigg(\frac{x}{2}\bigg)^2+\\\frac{3!}{3!(3-3)!}\bigg(\frac{x}{2}\bigg)^3\\=1+\frac{3x}{2}+\frac{3x^2}{4}+\frac{x^3}{8}\space\text{...(iii)}$$

Put the value of equation (ii) and equation (iii) in we get from equation (ii).

$$= 1+2x+3x^2+\frac{x^3}{2}+\frac{x^4}{16}-\frac{8}{x}\\\bigg(1+\frac{3x}{4}+\frac{3}{4}x^2+x^3\bigg)$$

$$+\frac{24}{x^2}+\frac{24}{x^2}+\frac{24x}{x^2}+\frac{24x^2}{4x^2}-\\4\bigg(\frac{8}{x^3}+\frac{8x}{2x^3}\bigg)+\frac{16}{x^4}$$

$$= 1+2x+\frac{3}{2}x^2+\frac{x^3}{2}+\frac{x^4}{16}-\frac{8}{x}-\frac{24x}{2x}-\frac{8}{x}$$

$$×\frac{3x^2}{4}-\frac{8}{x}.\frac{x^3}{8}+\frac{24}{x^2}+\frac{24}{x}+6\\-\frac{32}{x^3}-\frac{4×8x}{2x^3}+\frac{16}{x^4}$$

$$= 1+2x+\frac{3}{2}x^2+\frac{x}{2}^3+\frac{x^4}{16}-\frac{8}{x}-12-6x-x^2\\+\frac{24}{x}+\frac{24}{x}+6-\frac{32}{x^3}-\frac{16}{x^2}+\frac{16}{x^4}\\=\frac{16}{x}+\frac{8}{x^2}-\frac{32}{x^3}+\frac{16}{x^4}-4x+\\\frac{x^2}{2}+\frac{x^3}{2}+\frac{x^4}{16}-5.$$

10. Find the expansion of (3x2 – 2ax + 3a)3 using binomial theorem.

Sol. (3x2 – 2ax + 3a2)3 = [(3x2 – 2ax) + 3a2]3

The expression can be written as (a + b)3

where a = 3x2 – 2ax and b = 3a2 and n – 3

Expansion can be written as

(a + b)n = nC0 an + nC1 + an – 1 b1 + nC2 an – 2 b2 + …

So [(3x2 – 2ax) + 3a2]3 = 3C0 (3x2 – 2ax)3 (3a2)0+ 3C1 (3x2 – 2ax)2 (3a2)1 + 3C2 (3x2 – 2ax)1 (3a2)2+ 3C3 (3x2 – 2ax)0 (3a2)3

$$=1(3x^2-2ax)^3+\frac{3!}{1!(3-1)!}(3x^2-2ax)(3a^2)\\+\frac{3!}{2!(3-2)!}(3x^2-2ax)^1(3a^2)^2\\+\frac{3!}{3!(3-3)!}(3x^2-2ax^0)(3a^2)^3$$

= (3x2 – 2ax)3 + 3a2 (3x2)2 + (2ax)2 – (2 × 3x2 × 2ax) + 3(3x2 – 29x) (4a4) + 27a6

= (3x2 – 2ax)3 + 9a2(9x4 + 4a2x2 – 12ax2) + 81a4x2 – 54a5x + 27a6 …(i)

So, (3x2 – 2ax)3 = 3C0 (3x2)33C1 (3x2)2 (2ax) + 3C2 (3x2)1 (2ax)23C3 (2ax)3

$$=(3x^2)^3-\frac{3!}{1!(3-1)!}(3x)^2(2ax)\\+\frac{3!}{2!(3-2)!}(3x)^2(2ax)^2-\frac{3!}{3!(3-3)!}(2ax)^3$$

= 27x6 – 3ax4 · 2ax + 3·3x2 · 4a2x2 – 8a3x3

= 27x6 – 54ax5 + 36x4a2 – 8a3x3

Put this value in equation (i), we get,

= 27x6 – 54ax5 + 36x4a2 – 8a3x3 + 81a2x4 + 36a4x2 – 108a3x3 + 81a4x2 – 54a5x + 27a6

= 27x6 – 54ax5 + 117a2x4 – 11a3x3 + 117a4x2 – 54a5x + 27a6.