HOW TO FIND THE INDICATED TERM OF THE BINOMIAL EXPANSION

The formula given below can be used to find a particular term of a binomial expansion.

General term :

T(r+1)ncr x(n-r)ar

Example 1 :

Find the coefficient of x5 in the expansion of (x + 1/x3)7

Solution :

T(r+1) =  ncr x(n-r) ar

Comparing the given expression with the form (x + a)nwe get x = x, a = 1/x3 and n = 7

T(r+1) =  17cr x(17-r) (1/x3)r

=  17cr x(17-r) (x-3)r

=  17cr x(17-r) (x-3r)

=  17cr x(17-r-3r) 

=  17cr x(17-4r) 

Let Tr + 1 be the term containing x5

17 − 4r = 5 ⇒ r = 3

Tr + 1 = T3 + 1

= 17C3 x17 − 4(3)

= 680x5

Hence the coefficient of x= 680.

Example 2 :

Find the coefficient of x5 in the expansion of (x - 1/x)11

Solution :

T(r+1) =  ncr x(n-r) ar

Comparing the given expression with the form (x + a)n, we get x = x, a = -1/x and n = 11

T(r+1) =  11cr x(11-r) (-1/x)r

=  11cr x(11-r) (-x-1)r

=  11cr x(11-r) (-x-r)

=  -11cr x(11-r-r) 

= - 11cr x(11-2r) 

Let Tr + 1 be the term containing x5

11 − 2r = 5 ⇒ r = 3

Tr + 1 = T3 + 1

= -11C3 x11 − 2(3)

11C =  (11⋅10⋅9)/(3⋅2⋅1)

= -165x5

Hence the coefficient of x= -165.

Example 3 :

Find the constant term in the expansion (√x - 2/x2)10

Solution :

T(r+1) =  ncr x(n-r) ar

Comparing the given expression with the form (x + a)n, we get x = √x, a = -2/x2 and n = 10

T(r+1) =  10cr x(10-r) (-2/x2)r

=  10cr x1/2(10-r) (-2x-2)r

=  10cr x(10-r)/2 (-2x-2)r

= (-2)r 10cr x(10-r)/2 x-2r

(-2)r 10cr x(10-r-4r)/2

(-2)r  10cr x(10-5r)/2

Let Tr + 1 be the constant term 

(10 - 5r)/2  =  0 ⇒ r = 2

Tr + 1 = T2 + 1

(-2)2  10c2 x(10-5(2))/2

10C =  (10⋅9)/(2⋅1)

= 4(45x0)

Hence the constant term is 180.

Example 4 :

Find the constant term in the expansion (2x2 + 1/x)12

Solution :

T(r+1) =  ncr x(n-r) ar

Comparing the given expression with the form (x + a)n, we get x = 2x2, a = 1/x and n = 12

T(r+1) =  12cr (2x2)(12-r) (1/x)r

 =  12cr (212-r)(x2)(12-r) (x-r)

 212-r [12cr x(24-3r)]

Let Tr + 1 be the constant term 

24 - 3r  =  0 ⇒ r = 8

Tr + 1 = T8 + 1

 212-8 [12c8 x24-3(8)]

 2(495) x ==> 7920

Hence the constant term is 7920.

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