FIND THE COEFFICIENT OF SPECIFIC TERM IN BINOMIAL EXPANSION

Problem 1:

Find the coefficient of x⁵ in the expansion of

(x + 3) (2x - 1)⁶.

Solution :

= (x + 3) (2x - 1)⁶

General term of expansion (a + b)ⁿ

T_r+1 = ⁿC_r a^n-r b^r

Here a = 2x, b = -1, n = 6.

= (x + 3) [⁶C_r (2x)^6-r (-1)^r]

= (x + 3) [⁶C_r (2)^6-r (x)^6-r (-1)^r]

= [⁶C_r (2)^6-r (x)^6-r (-1)^r ∙ x] + [⁶C_r (2)^6-r (x)^6-r (-1)^r ∙ 3]

= [⁶C₂ (2)^6-2 (x)^6-2 (-1)² ∙ x] + [⁶C₁ (2)^6-1 (x)^6-1 (-1)¹ ∙ 3]

= 15(2)⁴ (x)⁵ + 6(2)⁵ (x)⁵ (-1) ∙ 3

= 240x⁵ - 576x⁵

= -336x⁵

So, coefficient of x⁵ is -336.

Problem 2 :

Find the coefficient of x⁵ in the expansion of

(x + 2) (x² + 1)⁸.

Solution:

= (x + 2) (x² + 1)⁸

General term of expansion (a + b)ⁿ

T_r+1 = ⁿC_r a^n-r b^r

Here a = x², b = 1, n = 8.

= (x + 2) [⁸C_r (x²)^8-r (1)^r]

= (x + 2) [⁸C_r (x)^16-2r (1)^r]

= [⁸C_r (x)^16-2r (1)^r ∙ x] + [⁸C_r (x)^16-2r (1)^r ∙ 2] --(1)

By applying the value of r in (1), we get

= [⁸C₆ (x)^16-2(6) (1)⁶ ∙ x] + [⁸C₆ (x)^16-2(6) (1)⁶ ∙ 2]

= [⁸C₆ (x)⁵ + [2∙⁸C₆ (x)⁴]

= (8 x 7)/(2 x 1)

= 28

Problem 3 :

Find the coefficient of x⁶ in the expansion of

(2 - x) (3x + 1)⁹

Solution :

= (2 - x) (3x + 1)⁹

General term of expansion (a + b)ⁿ

T_r+1 = ⁿC_r a^n-r b^r

Here a = 3x, b = 1, n = 9.

= (2 - x) [⁹C_r (3x)^9-r (1)^r]

= (2 - x) [⁹C_r (3)^9-r (x)^9-r (1)^r]

= [⁹C_r (3)^9-r (x)^9-r (1)^r ∙ 2] + [⁹C_r (3)^9-r (x)^9-r (1)^r ∙ (-x)]

= [⁹C₃ (3)^9-3 (x)^9-3 (1)³ ∙ 2] - [⁹C₄ (3)^9-4 (x)^9-4 (1)⁴ ∙ x]

= 168(3)⁶ (x)⁶ - 126 (3)⁵ (x)⁶

= 122472x⁶ - 30618x⁶

= 91854x⁶

So, coefficient of x⁶ is 91854.

Problem 4 :

Find the constant term in the expansion of (3x² + 1/x)⁸.

Solution :

General term of expansion (a + b)ⁿ

T_r+1 = ⁿC_r a^n-r b^r

Putting a = 3x², b = 1/x, n = 8

T_r+1 = ⁸C_r (3x²)^8-r (1/x)^r

= ⁸C_r (3)^8-r (x)^16-2r (1)^r (x)^-r

= ⁸C_r (3)^8-r (x)^16-3r

Constant term:

16 - 3r = 0

-3r = -16

r = 16/5

It does not have one.

Problem 5 :

Find the coefficient of x^-6 in the expansion of (2x - 3/x²)¹².

Solution :

= (2x - 3/x²)¹²

General term of expansion (a + b)ⁿ

T_r+1 = ⁿC_r a^n-r b^r

Here a = 2x, b = -3/x², n = 12.

= ¹²C_r (2x)^12-r (-3/x²)^r

= ¹²C_r (2)^12-r (x)^12-r (-3)^r (x)^-2r

= ¹²C_r (2)^12-r (x)^12-3r (-3)^r

x^12-3r = x^-6

12 - 3r = -6

-3r = -18

r = 6

= ¹²C₆ (2)^12-6 (x)^12-18 (-3)⁶

= ¹²C₆ (2)⁶ (-3)⁶ (x)^-6

So, coefficient of x^-6 is ¹²C₆ (2)⁶ (-3)⁶.

Problem 6 :

Find the coefficient of x⁵ in the expansion of

(2x + 3) (x - 2)⁶.

Solution:

= (2x + 3) (x - 2)⁶

General term of expansion (a + b)ⁿ

T_r+1 = ⁿC_r a^n-r b^r

Here a = x, b = -2, n = 6.

= (2x + 3) [⁶C_r (x)^6-r (-2)^r]

= [⁶C_r (x)^6-r (-2)^r ∙ 2x] + [⁶C_r (x)^6-r (-2)^r ∙ 3]

= [⁶C₂ (x)^6-2 (-2)² ∙ 2x] + [⁶C₁ (x)^6-1 (-2)¹ ∙ 3]

= 120x⁵ - 36x⁵

= 84x⁵

So, coefficient of x⁵ is 84.

Problem 7 :

Find the possible values of a if the coefficient of x³ in

(2x + 1/ax²)⁹ is 288.

Solution :

= (2x + 1/ax²)⁹

General term of expansion (a + b)ⁿ

T_r+1 = ⁿC_r a^n-r b^r

Here a = 2x, b = 1/ax², n = 9.

= ⁹C_r (2x)^9-r (1/ax²)^r

= ⁹C_r (2x)^9-r (1/a)^r (x)^-2r

= ⁹C_r (2)^9-r (x)^9-3r (1/a)^r

x^9-3r = x³

9 - 3r = 3

-3r = -6

r = 2

= ⁹C₂ (2)^9-2 (x)^9-6 (1/a)²

= ⁹C₂ (2)⁷ (x)³ (1/a)²

288 = 4608 x³ (1/a)²

(1/a)² = 4608/288

(1/a)² = 16

a = ± 4

So, value of a is ± 4.

Problem 8 :

Find the term independent of x in the expansion of

(3x - 2/x²)⁶.

Solution :

= (3x - 2/x²)⁶

General term of expansion (a + b)ⁿ

T_r+1 = ⁿC_r a^n-r b^r

Here a = 3x, b = -2/x², n = 6.

= ⁶C_r (3x)^6-r (-2/x²)^r

= ⁶C_r (3)^6-r (x)^6-r (-2)^r (x)^-2r

T_r+1 = ⁶C_r (3)^6-r (x)^6-3r (-2)^r

To find the term independent of x. So, power of x is 0.

x^6-3r = 0

6 - 3r = 0

-3r = -6

r = 2

T₂₊₁ = ⁶C₂ (3)^6-2 (x)^6-6 (-2)²

T₃ = ⁶C₂ (3)⁴ (x)⁰ (-2)²

= 4860

Hence, the term which is independent of x is

3^rd term = 4860.