# FIND A SPECIFIC TERM IN A 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:

Without simplifying, find:

The 6th term of (2x + 5)15

Solution :

General term of expansion (a + b)n

Tr+1 = nCr an-r br

To find 6th term,

T6 = T5+1 of expansion (2x + 5)15

Putting r = 5, a = 2x, b = 5, n = 15

T6 = 15C5 (2x)15-5 (5)5

T6 = 15C5 (2x)10 (5)5

Problem 2 :

The 4th term of (x² + 5/x)9

Solution :

General term of expansion (a + b)n

Tr+1 = nCr an-r br

To find 4th term,

T4 = T3+1 of expansion (x² + 5/x)9

Putting r = 3, a = x², b = 5/x, n = 9

T4 = 9C3 (x²)9-3 (5/x)3

T4 = 9C3 (x²)6 (5/x)3

Problem 3 :

The 10th term of (x - 2/x)17

Solution :

General term of expansion (a + b)n

Tr+1 = nCr an-r br

To find 10th term,

T10 = T9+1 of expansion (x - 2/x)17

Putting r = 9, a = x, b = -2/x, n = 17

T10 = 17C9 (x)17-9 (-2/x)9

T10 = 17C9 (x)8 (-2/x)9

Problem 4 :

The 9th term of (2x² - 1/x)21

Solution :

General term of expansion (a + b)n

Tr+1 = nCr an-r br

To find 9th term,

T9 = T8+1 of expansion (2x² - 1/x)21

Putting r = 8, a = 2x², b = -1/x, n = 21

T9 = 21C8 (2x²)21-8 (-1/x)8

T9 = 21C8 (2x²)13 (-1/x)8

Problem 5 :

Find the coefficient of:

x10 in the expansion of (3 + 2x²)10

Solution :

General term of expansion (a + b)n

Tr+1 = nCr an-r br

Putting a = 3, b = 2x², n = 10

Tr+1 = 10Cr (3)10-r (2x²)r

= 10Cr (3)10-r 2r x²r

To find the coefficient of x10, take

x2r = x10

2r = 10

r = 5

T6 = 10C5 (3)10-5 25 (5)

T6 = 10C5 35 25 x10

So, coefficient of x10 is

10C5 35 25

Problem 6 :

x3 in the expansion of (2x² - 3/x)6

Solution :

General term of expansion (a + b)n

Tr+1 = nCr an-r br

Putting a = 2x², b = -3/x, n = 6

Tr+1 = 6Cr (2x²)6-r (-3/x)r

= 6Cr (2)6-r (x²)6-r (-3/x)r

= 6Cr (2)6-r (x)12-2r (-3/x)r

= 6Cr (2)6-r (x)12-2r (-3)r (x)-r

= 6Cr (2)6-r (-3)r (x)12-3r

To find the coefficient of x3, take

x12-3r = x3

12 - 3r = 3

-3r = 3 - 12

-3r = -9

r = 3

T4 = 6C3 (2)6-3 (-3)3 x1²-9

T4 = 6C3 23 (-3)3 x3

So, coefficient of x3 is

6C3 23 (-3)3

Problem 7 :

x12 in the expansion of (2x² - 1/x)12

Solution :

General term of expansion (a + b)n

Tr+1 = nCr an-r br

Putting a = 2x², b = -1/x, n = 12

Tr+1 = 12Cr (2x²)12-r (-1/x)r

= 12Cr (2)12-r (x²)12-r (-1/x)r

= 12Cr (2)12-r (x)24-2r (-1/x)r

= 12Cr (2)12-r (x)24-2r (-1)r (x)-r

= 12Cr (2)12-r (-1)r (x)24-3r

To find the coefficient of x12, take

x24-3r = x12

24 - 3r = 12

-3r = 12 - 24

-3r = -12

r = 4

T5 = 12C4 (2)12-4 (-1)4 4-12

T5 = 12C4 28 (-1)4 x12

So, coefficient of x12 is

12C4 28 (-1)4

Problem 8 :

Find the constant term in:

The expansion of (x + 2/x²)15

Solution:

General term of expansion (a + b)n

Tr+1 = nCr an-r br

Putting a = x, b = 2/x², n = 15

Tr+1 = 15Cr (x)15-r (2/x²)r

= 15Cr (x)15-r 2r (x)-2r

= 15Cr 2r (x)15-3r

Constant term:

15 - 3r = 0

-3r = -15

r = 5

T6 = 15C5 25 (x)15-15

T6 = 15C5 25

So, the constant term is

15C5 25

Problem 9 :

The expansion of (x - 3/x²)9

Solution :

General term of expansion (a + b)n

Tr+1 = nCr an-r br

Putting a = x, b = -3/x², n = 9

Tr+1 = 9Cr (x)9-r (-3/x²)r

= 9Cr (x)9-r (-3)r (x)-2r

= 9Cr (-3)r (x)9-3r

Constant term:

9 - 3r = 0

-3r = -9

r = 3

T4 = 9C3 (-3)3 (x)9-9

T4 = 9C3 (-3)3

So, the constant term is

9C3 (-3)3

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