NCERT Solutions for Class 11 Maths Chapter 7 - Binomial Theorem

Download PDF

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 a^{n – 1} b′ + ⁿC₂ a^{n – 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 a^{n – 1} b + ⁿC₂ a^{n – 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

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

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

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

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

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

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

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 a^{n – 1} b + ⁿC₂ a^{n – 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 a^{n – 1} b + ⁿC₂ a^{n – 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 a^{n – 1} b + ⁿC₂ a^{n – 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 a^{n – 1} b + ⁿC₂ a^{n – 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 a^{n – 1} b + ⁿC₂ a^{n – 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 a^{n – 1} b + ⁿC₂ a^{n – 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 b⁰ – ⁴C₁ 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 b⁴ – ⁴C0 a⁴ + ⁴C₁ a³b – ⁴C₂ a²b²+ ⁴C₃ a¹b³ – ⁴C₄ 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 + ⁶C₄x²1⁴ + 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 x⁶ – ⁶C₁ x⁵ + ⁶C₂ x⁴ – ⁶C₃ x³ + ⁶C₄ x² – ⁶C₅ x + ⁶C₆]

= 2[⁶C₀x⁶ + ⁶C₂ x⁴+ ⁶C₄x²+ 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 9^{n + 1} – 8n – 9 is divisible by 64, whenever n is a positive integer.

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

In order to show that 9^{n + 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 a^{n – 2}b² + …

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

Therefore,

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

(9)^{n + 1} = 1 + (n + 1)8 + 8²[^{n + 1}C₂ (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

9^{n + 1} – 8n – 9 is divisible by 64.

So, we can say

9^{n + 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,

T_{r + 1} = ⁿC_r a^{n – r}b^r …(i)

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

a = x

b = 3 and n = 8

From equation (i),

T_{r + 1} = ⁸C_r x^{8 – r} 3^r
We need coefficient of x⁵

Therefore,

x^{8 – r} = x⁵

8 – r = 5 (∵ compairing powers)

r = 3

T_{3 + 1} = ⁸C₃ x^{8 – 3} (3)³

T_{3 + 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

T_{r + 1} = ⁿC_r a^{n – r}b^r …(i)

T_{r + 1} = ¹²C_r a^{12 – r} (– 2b)^r

We need coefficient of a5b7

a^{12 – r} = a⁵

12 – r = 5 (∵ comparing powers)

r = 12 – 5

equation r = 7

From (i)

T_{7 + 1} = ¹²C₇ a^{12 – 7} (– 2b)⁷

= ¹²C₇ a⁵b⁷ (– 2)⁷

Coefficient of a⁵b⁷is 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

T_{r + 1} = ⁿC_r a^{n – 1}b^r

T_{r + 1} = ⁶C_r (x²)^{6 – r}(– y)^r

= ⁶C_r x^{12 – 2r}(– 1 × y)^r

= ⁶C_r x^{12 – 2r} (– 1)^r y^r.

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

T_{r + 1} = ⁿC_r a^{n – r}b^r

T_{r + 1} = ¹²C_r (x²)^{12 – r} (– yx)^r

T_{r + 1} = ¹²C_r x^{24 – 2r} (– 1 × y × x)^r

= ¹²C_r x^{24 – 2r} (– 1)^r (y)^r (x)^r

= (– 1)^r ¹²C_r x^{24 – 2r} y^r x^r.

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

T_{r + 1} = ⁿC_r a^{n – r}b^r

We need 4th term

T₄ = T_{3 + 1} = ¹²C₃ x^{12 – 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

T_{r + 1} = ⁿC_r a^{n – r}b^r

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

T_{r + 1} = ⁿC_r a^{n – r}b^r

T₄ = T_{3 + 1} = ⁷C₃ a^{7 – 3}b³ (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 = T_{4 + 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

T^{r + 1} = ⁿC_r a^{n – r}b^r

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

= 6 × 42 × x⁵ × 243 × y⁵

= 61236 x⁵y⁵ (6^th term).

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

Sol. We need to find coefficient of a^m and aⁿ

As we know the general term is given by

T_{r + 1} = ⁿC_r a^{n – r}b^r

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

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

So, T_{r + 1} = ^{m + n}C_r (1)^{m + n – r} (a)^r …(i)

We need coefficient am and an

a^r = a^m

r = m [comparing powers]

and a^r = aⁿ

r = n [comparing powers]

When r = m from equation (i)

T_{m + 1} = ^{m + n}C_m (1)^{m + n – m} (a)^m

= ^{m + n}C_m· (a)^m [1n = 1]

Coefficient of a^m = ^{m + n}C_m …(ii)

Similarly when r = n from equation (i)

T_{n + 1} = ^{m + n}C_n (1)^{m + n – n} (aⁿ)

= ^{m + n}C_n aⁿ [(1)^m = 1]

Coefficient of aⁿ = ^{m + n}C_n …(iii)

[from equation (ii) and (iii)]

According to question

^{m + n}C_m = ^{m + n}C_n

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

So, coefficient of a^m and aⁿ are equal.

Hence Proved.

10. The coefficient of the (r – 1)^th, r^th and (r + 1)^thterms 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

T_{r + 1} = ⁿC_r a^{n – r}b^r

For (r – 1)^th term

T_{r – 1} = ⁿC_{r – 2} (x)^{n – (r – 2)} (1)^{r – 2} [r = r – 2]

= ⁿC_{r – 2}x^{n – r + 2} …(i)

For (r^thterm)

T_r = ⁿC_{r – 1} x^{n – (r – 1)} (1)^{r – 1} [r = r – 1]

= ⁿC_{r – 1} x^{n – r + 1} …(ii)

For (r + 1)^th term

T_{r + 1} = ⁿC_r x^{n – r} (1)^r[r = r]

= ⁿC_r x^{n – 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

T_{r + 1} = ⁿC_r a^{n – r} b^r

T_{r + 1} = ²ⁿC_r (1)^{2n – r} x^r

We need coefficient of xn

x^r = xⁿ

r = n [comparing powers]

T_{n + 1} = ²ⁿC_n (1)^{2n – n} xⁿ

= ²ⁿC_n 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

T_{r + 1} = ⁿC_r a^{n – 1} b^r

T_{r + 1} = 2^{n – 1}C_r (1)^{2n – 1 – r}(x^r)

We need xn coefficient

xⁿ = x^r

r = n (comparing power)

T_{n + 1} = ^{2n – 1}C_n (1)^{2n – 1 – n} (x)ⁿ

T_{n + 1} = ^{2n – 1}C_n 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

T_{r + 1} = ⁿC_r a^{n – r} b^r

T_{r + 1} = ^mC_r (1)^{m – r} (x)^r …(i)

We need coefficient of x²

x^r = x²

r = 2 [comparing powers]

Therefore from (i)

T_{2 + 1} = ^mC₂ (1)^{m – 2} x²

= ^mC₂ x² …(ii) [1^{n – 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

m + 3 = 0

m = 4

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

T_{r + 1} = ⁿC_r a^{n – r} b^r

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]

= na^{n – 1} b

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

na^{n– 1} b = 7290 …(ii)

T₃ = ⁿC₂ a^{n – 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

T_{r + 1} = ⁿC_r a^{n – 1} b^r

T_{r + 1} = ⁹C_r (3)^{n – r} (ax)^r

We need coefficient of x²

x^r = x²

r = 2 (comparing powers)

T_{2 + 1} = ⁹C₂ (3)^{9 – 2} (ax)²

T₃ = ⁹C₂ (3)⁷ a²x²

We need coefficient of x³

x^r = x³

r = 3

T_{3 + 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 a^{n – 1} b + ⁿC₂ a^{n – 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 a^{n – 1} b + ⁿC₂ a^{n – 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+ … + ⁿC_n (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 a^{n – 1} b + ⁿC₂ a^{n – 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 a⁶ – ⁶C₁ a⁵ (b) + ⁶C₂ a⁴ (b)² – ⁶C₃ a³ (b)³ + ⁶C₄ a(b)⁴ – ⁶C₅ 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 b⁶ – ⁶C₀a⁶ + ⁶C₁ a⁵b – ⁶C₂ a⁴b²
+ ⁶C₃ b³ – ⁶C₄ 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 a^{n – 1} b + ⁿC₂ a^{n – 2} b² + …

(x + y)⁴ is compare with (a + b)ⁿthen

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 x⁴ – ⁴C₁ x³y + ⁴C₂ x²y² – ⁴C₃ 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 a^{n – 1} b + ⁿC₂ a^{n – 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 a^{n – 1} b¹ + ⁿC₂ a^{n – 2} b² + … + ⁿC_{n – 1} a¹ b^{n – 1} + ⁿC_na⁰bⁿ

Fifth term from begining = ⁿC₄ a^{n – 4} b⁴

Fifth term from end = ⁿC_{n – 4} a⁴ b^{n – 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 a^{n – 1} b¹ + ⁿC₂ a^{n – 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+ a^{n – 1} b¹ + ⁿC₂ a^{n – 2} b² + …