Class 11 NCERT Solutions- Chapter 8 Binomial Theorem – Miscellaneous Exercise on Chapter 8

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Question 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.

Solutions:

As, we know that (r+1)^th term of (a+b)ⁿ is denoted by,

T_r+1 = ⁿC_r a^n-r b^r

Here, it is given that first three terms of the expansion are 729, 7290 and 30375.

When, T₁= 729, T₂ = 7290 and T₃ = 30375

T₀₊₁ (r=0) = ⁿC₀ a^n-0 b⁰ = aⁿ= 729 …………………(1)

T₁₊₁(r=0) = ⁿC₁ a^n-1 b¹ = na^n-1 b = 7290 …………………(2)

T₂₊₁ (r=0) = ⁿC₂ a^n-2 b² = $\frac{n!}{2!(n-2)!}$ a^n-2b²= $\frac{n(n-1)a^{n-2}b^2}{2}$ = 30375 …………………(3)

Dividing (1) and (2), we get

$\frac{na^{n-1} b}{a^n} = \frac{7290}{729}$

na^n-1-n b = 10

$\frac{nb}{a}$ = 10 ……………………….(I)

Now, dividing (3) and (2), we get

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

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

$\frac{nb-b}{a} = \frac{25}{3}$

$\frac{nb}{a} - \frac{b}{a} = \frac{25}{3}$

From (I), we can substitute

$10 - \frac{b}{a} = \frac{25}{3}$

$\frac{b}{a}$ = 10 – $\frac{25}{3}$

$\frac{b}{a} = \frac{30-25}{3} = \frac{5}{3}$ ………………….(II)

Substituting (II) in (I), we get

$\frac{nb}{a}$ = 10

$n(\frac{5}{3})$ = 10

n = $\frac{10 \times3}{5}$

n = 6

Substituting n = 6 in (1), we get

aⁿ = 729

a⁶ = 729

a = 3

Substituting a = 3 in (II), we get

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

b = $\frac{5\times3}{3}$

b = 5

Hence, a = 3, b = 5 and n = 6

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

Solutions:

As, we know that (r+1)^th term of (a+b)ⁿ is denoted by,

T_r+1 = ⁿC_r a^n-r b^r

Here, a = 3 and b = ax and n = 9.

T_r+1 = ⁹C_r 3^9-r (ax)^r

T_r+1 = ⁹C_r $(\frac{3^9}{3^r})$ a^rx^r

T_r+1 = ⁹C_r $(\frac{3^9 \times a^r}{3^r}) x^r$

So, here if you want the power of x² and x³. Then r=2 and r=3.

r = 2, T_{2+1 =} $[^9C_2 (\frac{3^9 \times a^2}{3^2})] x^2$

r = 3, T₃₊₁ = $[^9C_3 (\frac{3^9 \times a^3}{3^3})] x^3$

Coefficient of x² = Coefficient of x³

$^9C_2 (\frac{3^9 \times a^2}{3^2}) = ^9C_3 (\frac{3^9 \times a^3}{3^3})$

$\frac{9!}{2! (9-2)!} \times \frac{3^9 \times a^2}{3^2} = \frac{9!}{3! (9-3)!} \times \frac{3^9 \times a^3}{3^3}$

$\frac{1}{2! (9-2)!} \times \frac{a^2}{3^2} = \frac{1}{3! (9-3)!} \times \frac{a^3}{3^3}$

$\frac{3! (9-3)!}{2! (9-2)!} = \frac{a^3 \times 3^2}{a^2 \times 3^3}$

$\frac{a}{3} = \frac{3! 6!}{2! 7!}$

$\frac{a}{3} = \frac{3}{7}$

a = $\frac{3\times 3}{7} = \frac{9}{7}$

Hence, a = $\frac{9}{7}$

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

Solutions:

For getting the coefficient of x⁵, lets expand both the binomials for more clear understanding.

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

= 1 + 6 (2x) + 15 (2x)² + 20 (2x)³ + 15 (2x)⁴ + 6 (2x)⁵ + (2x)⁶

= 1 + 12 x + 60x² + 160 x³ + 240 x⁴ + 192 x⁵+ 64x⁶

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

= 1 – 7x + 21x² – 35x³ + 35x⁴ – 21x⁵ + 7x⁶ – x⁷

Now, the product will be seen as follows

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

Here, what we can see that x⁵ will be obtained when two terms will be multiplied of having sum of power of x as 5. Those two terms will be as follows:

First binomial Second binomial

1^st term 6^th term

2^nd term 5^th term

3^rd term 4^th term

4^th term 3^rd term

5^th term 2^nd term

6^th term 1^st term

So,

Coefficient of x⁵ = (1)(-21) + (12)(35) + (60)(-35) + (160)(21) + (240)(-7) + (192)(1)

Coefficient of x^{5 =} -21 + 420 – 2100 + 3360 – 1680 + 192

Coefficient of x⁵ = -21 + 420 – 2100 + 3360 – 1680 + 192

Coefficient of x⁵ = 171

Hence, the coefficient of x⁵ in the expression (1+2x)⁶ (1-x)⁷ is 171.

First binomial	Second binomial
1^st term	6^th term
2^nd term	5^th term
3^rd term	4^th term
4^th term	3^rd term
5^th term	2^nd term
6^th term	1^st term

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

[Hint write an = (a – b + b)ⁿ and expand]

Solutions:

To prove that (a – b) is a factor of (aⁿ – bⁿ),

aⁿ – bⁿ = k (a – b) where k is some natural number or constant.

a can be written as = a – b + b

aⁿ = (a – b + b)ⁿ = [(a – b) + b]ⁿ

[(a – b) + b]ⁿ = ⁿC₀ (a – b)ⁿ + ⁿC₁ (a – b)^n-1b + ……….ⁿC_n-1(a – b)b^n-1 + ⁿC_n bⁿ

aⁿ = (a – b)ⁿ + n (a – b)^n-1 b + ……….ⁿC_n-1(a – b)b^n-1+ bⁿ

Now, aⁿ – bⁿ will be

aⁿ – bⁿ = [(a – b)ⁿ + n (a – b)^n-1 b + ……….ⁿC_n-1(a – b)b^n-1 + bⁿ] – bⁿ

aⁿ – bⁿ = (a – b)ⁿ+ n (a – b)^n-1 b + ……….ⁿC_n-1(a – b)b^n-1

Taking (a-b) common, we have

aⁿ – bⁿ = (a – b) [(a –b)^n-1 + n (a – b)^n-2 b + …… + ⁿC_n-1 b^n-1]

aⁿ – bⁿ = (a – b) k

Where k = [(a –b)^n-1 + n (a – b)^n-2 b + …… + ⁿC_n-1 b^n-1] is a natural number

Hence, it is proved a – b is a factor of aⁿ – bⁿ, where n is a positive integer

Question 5. Evaluate (√3+√2)⁶−(√3−√2)⁶.

Solutions:

Using binomial theorem the expression (a + b)⁶ and (a – b)⁶, can be expanded as follows:

(a + b)⁶ = ⁶C0 a⁶ + ⁶C₁ a⁵ b + ⁶C₂ a⁴ b² + ⁶C₃ a³ b³ + ⁶C₄ a² b⁴ + ⁶C₅ a b⁵ + ⁶C₆ b⁶

(a – b)⁶ = ⁶C₀ a⁶ – ⁶C₁ a⁵ b + ⁶C₂a⁴ b² – ⁶C₃ a³ b³ + ⁶C₄ a² b⁴ – ⁶C₅ a b⁵+ ⁶C₆ b⁶

Now adding them,

(a + b)⁶ – (a – b)⁶ = ⁶C₀ a⁶ + ⁶C₁ a⁵ b + ⁶C₂ a⁴ b² + ⁶C₃ a³ b³ + ⁶C₄ a² b⁴ + ⁶C₅ a b⁵ + ⁶C₆ b⁶ – [⁶C₀ a⁶ – ⁶C₁ a⁵ b + ⁶C₂ a⁴ b² – ⁶C₃ a³ b³ + ⁶C₄ a² b⁴ – ⁶C₅ a b⁵ + ⁶C₆ b⁶]

(a + b)⁶ – (a – b)⁶ = 2[⁶C₁ a⁵ b + ⁶C₃a³ b³ + ⁶C₅ a b⁵]

Substituting a = √3 and b = √2, we get

(√3 + √2)⁶ – (√3 – √2)⁶ = 2 [6 (√3)⁵ (√2) + 20 (√3)³ (√2)³ + 6 (√3) (√2)⁵]

= 2 [54(√6) + 120 (√6) + 24 √6]

= 2 (√6) (198)

= 396 √6

Question 6. Find the value of $(a^2 + \sqrt{a^2-1})^4 + (a^2 - \sqrt{a^2-1})^4$ .

Solutions:

Using binomial theorem the expression (x+y)⁴ and (x – y)⁴, can be expanded as follows:

(x + y)⁴ = ⁴C₀ x⁴ + ⁴C₁ x³ y + ⁴C₂ x² y²+ ⁴C₃ x y³ + ⁴C₄ y⁴

(x – y)⁴ = ⁴C₀ x⁴ – ⁴C₁ x³ y + ⁴C₂ x² y² – ⁴C₃ x y³ + ⁴C₄ y⁴

Now adding them,

(x + y)⁴ + (x – y)⁴ = ⁴C₀ x⁴ + ⁴C₁ x³ y + ⁴C₂ x² y² + ⁴C₃ x y³ + ⁴C₄ y⁴ + [⁴C₀ x⁴ – ⁴C₁ x³ y + ⁴C₂ x² y² – ⁴C₃ x y³ + ⁴C₄ y⁴]

(x + y)⁴ + (x – y)⁴ = 2[⁴C₀ x⁴ + ⁴C₂ x² y² + ⁴C₄ y⁴]

Substituting x = a² and y = $\sqrt{a^2-1}$ , we get

$(a^2 + \sqrt{a^2-1})^4 + (a^2 - \sqrt{a^2-1})^4 = 2[(a^2)^4 +6 (a^2)^2 (\sqrt{a^2-1})^2 + (\sqrt{a^2-1})^4]$

= 2[a⁸ + 6a⁴ (a²-1) + (a²-1)²]

= 2[a⁸ + 6a⁶ – 6a⁴ + (a⁴ + 1 – 2(a²)(1))]

= 2[a⁸ + 6a⁶ – 6a⁴ + a⁴ + 1 – 2a²]

= 2a⁸ + 12a⁶ – 10a⁴ – 4a² + 2

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

Solutions:

To make 0.99 in binomial form,

0.99 = 1 – 0.01

Now by applying binomial theorem, we get

(0. 99)⁵ = (1 – 0.01)⁵

Taking first three terms of its expansion, we have

= ⁵C₀ (1)⁵ – ⁵C₁ (1)⁴ (0.01) + ⁵C₂ (1)³ (0.01)²

= 1 – 5 (0.01) + 10 (0.01)²

= 1 – 0.05 + 0.001

= 0.951

Approximation of (0.99)⁵ = 0.951.

Question 8. Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $(\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}})^n$ is √6:1.

Solutions:

As, here it is said we have to calculate

(Fifth term from the beginning : Fifth term from the end) of the Binomial = $(\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}})^n$

Instead of taking fifth term from the end, lets reverse the binomial term and take it from beginning.

Fifth term from the end of binomial $(\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}})^n$ = Fifth term from the beginning $(\frac{1}{\sqrt[4]{3}}+ \sqrt[4]{2})^n$

Lets move further with this

As, we know that (r+1)^th term of (a+b)ⁿis denoted by,

T_r+1 = ⁿCr a^n-r b^r

Fifth term from the beginning of $(\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}})^n$ ,

T₅ = T₄₊₁ = ⁿC₄ $(\sqrt[4]{2})^{n-4} (\frac{1}{\sqrt[4]{3}})^4$

T₅ = ⁿC₄ $(\sqrt[4]{2})^n(\frac{1}{\sqrt[4]{2} \times \sqrt[4]{3}})^4$

T₅ = ⁿC⁴ $(\sqrt[4]{2})^n (\frac{1}{2 \times 3})$

T₅ = ⁿC₄ $(\frac{(\sqrt[4]{2})^n}{6})$ …………………………….(1)

Now, Fifth term from the beginning of $(\frac{1}{\sqrt[4]{3}}+ \sqrt[4]{2})^n$ ,

T₅ = T₄₊₁ = ⁿC₄ $(\frac{1}{\sqrt[4]{3}})^{n-4} (\sqrt[4]{2})^4$

T₅ = ⁿC₄ $\times \frac{(\frac{1}{\sqrt[4]{3}})^n}{(\frac{1}{\sqrt[4]{3}})^4} \times (2)$

T₅ = ⁿC₄ $\frac{(\frac{1}{\sqrt[4]{3}})^n}{(\frac{1}{3})} (2)$

T₅ = ⁿC₄ $(\frac{3}{(\sqrt[4]{3})^n}) (2)$

T₅ = ⁿC₄ $(\frac{6}{(\sqrt[4]{3})^n})$ ……………………….(2)

Now taking ratio of (1) and (2), which is equal to √6:1

$\frac{^nC_4 (\frac{(\sqrt[4]{2})^n}{6})}{^nC_4 (\frac{6}{(\sqrt[4]{3})^n})} = \sqrt{6}$

$\frac{(\sqrt[4]{2})^n \times (\sqrt[4]{3})^n}{(6\times 6)} = \sqrt{6}$

$(\sqrt[4]{6})^n = \sqrt{6} \times 36$

$6^{\frac{n}{4}} = 6^{\frac{5}{2}}$

$\frac{n}{4} = \frac{5}{2}$

n = $\frac{5\times 4}{2}$

n = 10

Question 9. Expand using Binomial Theorem $(1+\frac{x}{2} - \frac{2}{x})^4$ , x≠0.

Solutions:

Grouping $(1+\frac{x}{2} - \frac{2}{x})^4$ in binomial form, we have

$[(1+\frac{x}{2}) - \frac{2}{x}]^4$

Comparing it with (a+b)ⁿ,

a = $(1+\frac{x}{2})$ , b = $\frac{2}{x}$ and n = 4

$[(1+\frac{x}{2}) - \frac{2}{x}]^4$ = ⁴C₀ $(1+\frac{x}{2})^4$ – ⁴C₁ $(1+\frac{x}{2})^3 (\frac{2}{x})$ + ⁴C₂ $(1+\frac{x}{2})^2 (\frac{2}{x})^2$ – ⁴C₃ $(1+\frac{x}{2}) (\frac{2}{x})^3$ + ⁴C₄ $(\frac{2}{x})^4$

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

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

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

Now, lets get the value of $(1+\frac{x}{2})^4, (1+\frac{x}{2})^3$ and $(1+\frac{x}{2})^2$

$(1+\frac{x}{2})^2 = 12 + (\frac{x}{2})^2 + 2(1)(\frac{x}{2})$

$(1+\frac{x}{2})^2 = 1 + \frac{x^2}{4} + x$

$(1+\frac{x}{2})^3$ = ³C₀ (1)³ + ³C₁ (1)² $(\frac{x}{2})$ + ³C₂ (1) $(\frac{x}{2})^2$ + ³C₃ $(\frac{x}{2})^3$

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

$(1+\frac{x}{2})^4$ = ⁴C₀ (1)⁴ + ⁴C₁ (1)³ $(\frac{x}{2})$ + ⁴C₂(1)² $(\frac{x}{2})^2$ + ⁴C₃ (1) $(\frac{x}{2})^3$ + ⁴C₄ $(\frac{x}{2})^4$

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

Now, substituting these values in the main equation, we get

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

= $1 + 2x + \frac{3x^2}{2} + \frac{x^3}{2} + \frac{x^4}{16} - (\frac{8}{x} + (\frac{8}{x})(\frac{3x}{2}) + (\frac{8}{x})(\frac{3x^2}{4}) + (\frac{8}{x})(\frac{x^3}{8})) + (\frac{24}{x^2} + (\frac{24}{x^2})(\frac{x^2}{4}) + (\frac{24}{x^2}) (x)) - (\frac{32}{x^3} + \frac{16}{x^2}) + (\frac{16}{x^4})$

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

= $\frac{16}{x} + \frac{8}{x^2} - \frac{32}{x^3} + \frac{16}{x^4} - 4x + \frac{x^2}{2} + \frac{x^3}{2} + \frac{x^4}{16} - 5$

Question 10. Find the expansion of (3x²– 2ax + 3a²)³ using binomial theorem.

Solutions:

Grouping (3x²– 2ax + 3a²)³ in binomial form, we have

[3x²+ (- 2ax + 3a²)]³

Comparing it with (a+b)ⁿ,

a = 3x², b = -a (2x-3a) and n = 3

[3x² + (-a (2x-3a))]³

= ³C₀ (3x²)³ + ³C₁(3x²)²(-a (2x-3a)) + ³C₂ (3x²) (-a (2x-3a))²+ ³C₃(-a (2x-3a))³

= 27x⁶ + 3 (9x⁴) (-a) (2x-3a) + 3 (3x²) (-a)²(2x-3a)² + (-a)³ (2x-3a)³

= 27x⁶ + (-54ax⁵ + 81a²x⁴) + 9a²x² (2x-3a)² – a³ (2x-3a)³

Now, lets get the value of (2x-3a)² and (2x-3a)³.

(2x-3a)² = (2x)² + (3a)² – 2(2x)(3a)

(2x-3a)²= 4x² + 9a² -12xa

(2x-3a)³ = (2x)³ – (3a)³ – 3(2x)(3a)(2x-3a)

(2x-3a)³ = 8x³ – 27a³ – 36x²a +54xa²

Now, substituting these values in the main equation, we get

= 27x⁶ – 54ax⁵ + 81a²x⁴ + 9a²x² (4x² + 9a² -12xa) – a³ (8x³– 27a³ – 36x²a + 54xa²)

= 27x⁶ – 54ax⁵ + 81a²x⁴ + 36a²x⁴ + 81a⁴x² -108x³a³ – (8a³x³ – 27a⁶ – 36x²a⁴+ 54xa⁵⁾

= 27x⁶– 54ax⁵ + 117a²x⁴ + 81a⁴x² -108x³a³ – 8a³x³ + 27a⁶ + 36x²a⁴ – 54xa⁵

= 27x⁶ – 54ax⁵+ 117a²x⁴ – 116a³x³ + 117a⁴x² – 54a⁵x + 27a⁶

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Last Updated : 07 Apr, 2021

Class 11 NCERT Solutions - Chapter 8 Binomial Theorem - Exercise 8.2

Class 11 NCERT Solutions- Chapter 9 Sequences And Series - Exercise 9.1

Chapter 1: Sets

Chapter 2: Relations and Functions

Chapter 3: Trigonometric Functions

Chapter 4: Principle of Mathematical Induction

Chapter 5: Complex Numbers and Quadratic Equations

Chapter 6: Linear Inequalities

Chapter 7: Permutations and Combinations

Chapter 8: Binomial Theorem

Chapter 9: Sequences and Series

Chapter 10: Straight Lines

Chapter 11: Conic Sections

Chapter 12: Introduction to Three-dimensional Geometry

Chapter 13: Limits and Derivatives

Chapter 14: Mathematical Reasoning

Chapter 15: Statistics

Chapter 16: Probability

Chapter 1: Sets

Chapter 2: Relations and Functions

Chapter 3: Trigonometric Functions

Chapter 4: Principle of Mathematical Induction

Chapter 5: Complex Numbers and Quadratic Equations

Chapter 6: Linear Inequalities

Chapter 7: Permutations and Combinations

Chapter 8: Binomial Theorem

Chapter 9: Sequences and Series

Chapter 10: Straight Lines

Chapter 11: Conic Sections

Chapter 12: Introduction to Three-dimensional Geometry

Chapter 13: Limits and Derivatives

Chapter 14: Mathematical Reasoning

Chapter 15: Statistics

Chapter 16: Probability

Class 11 NCERT Solutions- Chapter 8 Binomial Theorem – Miscellaneous Exercise on Chapter 8

Question 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.

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

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

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

[Hint write an = (a – b + b)n and expand]

Question 5. Evaluate (√3​+√2​)6−(√3​−√2​)6.

Question 6. Find the value of .

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

Question 8. Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of is √6:1.

Question 9. Expand using Binomial Theorem , x≠0.

Question 10. Find the expansion of (3x2– 2ax + 3a2)3 using binomial theorem.

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Question 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.

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

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

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

[Hint write an = (a – b + b)ⁿ and expand]

Question 5. Evaluate (√3+√2)⁶−(√3−√2)⁶.

Question 6. Find the value of $(a^2 + \sqrt{a^2-1})^4 + (a^2 - \sqrt{a^2-1})^4$ .

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

Question 8. Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $(\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}})^n$ is √6:1.

Question 9. Expand using Binomial Theorem $(1+\frac{x}{2} - \frac{2}{x})^4$ , x≠0.

Question 10. Find the expansion of (3x²– 2ax + 3a²)³ using binomial theorem.