# FIND THE COEFFICIENT OF SPECIFIC TERM IN BINOMIAL EXPANSION

The number of terms in a binomial expansion with an exponent of n is equal to n + 1. To find a particular term in the expansion of (a + b)n we make use of the general term formula.

The general term of the binomial expansion is

Tr+1 = nCr an-r br

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

Tr+1 = nCr an-r br

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

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

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

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

 x6-r ∙ x = x5x7-r = x57 - r = 5r = 2 x6 -r = x56 - r = 5r = 6 - 5r = 1

= [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

Tr+1 = nCr an-r br

Here a = x², b = 1, n = 8.

= (x + 2) [8Cr (x²)8-r (1)r]

= (x + 2) [8Cr (x)16-2r (1)r]

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

 x16-2r ∙ x = x5x17-2r = x517 - 2r = 512 = 2rr = 6 x16-2r  = x516 - 2r = 52r = 11r = 11/2

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

Tr+1 = nCr an-r br

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

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

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

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

 x9-r = x69 - r = 6r = 9 - 6r = 3 x9 -r ∙ x1 = x6x10 -r = x610 - r = 6r = 10 - 6r = 4

= [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

Tr+1 = nCr an-r br

Putting a = 3x², b = 1/x, n = 8

Tr+1 = 8Cr (3x²)8-r (1/x)r

= 8Cr (3)8-r (x)16-2r (1)r (x)-r

= 8Cr (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

Tr+1 = nCr an-r br

Here a = 2x, b = -3/x², n = 12.

= 12Cr (2x)12-r (-3/x²)r

= 12Cr (2)12-r (x)12-r (-3)r (x)-2r

= 12Cr (2)12-r (x)12-3r (-3)r

x12-3r = 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

Tr+1 = nCr an-r br

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

= (2x + 3) [6Cr (x)6-r (-2)r]

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

 x6-r ∙ x = x5x7-r = x57 - r = 5r = 7 - 5r = 2 x6 -r = x56 - r = 5r = 6 - 5r = 1

= [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

Tr+1 = nCr an-r br

Here a = 2x, b = 1/ax², n = 9.

= 9Cr (2x)9-r (1/ax²)r

= 9Cr (2x)9-r (1/a)r (x)-2r

= 9Cr (2)9-r (x)9-3r (1/a)r

x9-3r = 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

Tr+1 = nCr an-r br

Here a = 3x, b = -2/x², n = 6.

= 6Cr (3x)6-r (-2/x²)r

= 6Cr (3)6-r (x)6-r (-2)r (x)-2r

Tr+1 = 6Cr (3)6-r (x)6-3r (-2)r

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

x6-3r = 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

3rd term = 4860.

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