# GENERAL AND MIDDLE TERMS IN BINOMIAL THEOREM

## About "General and middle terms in binomial theorem"

General and middle terms in binomial theorem :

Here we are going to see how to find the middle terms in binomial theorem.

General term :

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

The number of terms in the expansion of (x + a)depends upon the index n. The index is either even (or) odd.

Let us find the middle terms.

Case (i) : n is even

The number of terms in the expansion is (n + 1), which is odd. Hence, there is only one middle term and it is given by T(n/2) + 1

Case (ii) : n is odd

The number of terms in the expansion is (n + 1), which is even. Hence, there are two middle terms and they are given by T(n + 1)/2 and T(n + 3)/2

## General and middle terms in binomial theorem - Examples

Example 1 :

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)nwe 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

Let us see the next example on "General and middle terms in binomial theorem".

Example 2 :

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)nwe 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.

Let us see the next example on "General and middle terms in binomial theorem".

Example 3 :

Find the middle term in the expansion of (3x - 2x2/3)8

Solution :

Here n = 8, that is even

So, the middle term  = T(n/2) + 1

=  T (8/2) + 1

=  T (4 + 1)  ==>  T 5

General term :

T(r+1) =  ncr x(n-r) ar

x = 3x, a = 2x2/3,  r = 4 and n = 8

T (4 + 1)  =  8c4 (3x)(8-4) (2x2/3)4

=  (8  7  6  5)/ (4  3  2  1)(3x)4 (2x2/3)4

=  (8  7  6 ⋅ 5)/ (4  3  2  1)(3x)4 (2x2/3)4

=  70(81x4)(16x8/81)

=  70(16)x12

=  1120 x12

Let us see the next example on "General and middle terms in binomial theorem".

Example 4 :

Find the middle term in the expansion of (b/x  - x/b)16

Solution :

Here n = 16, that is even

So, the middle term  = T(n/2) + 1

=  T (16/2) + 1

=  T (8 + 1)  ==>  T 9

General term :

T(r+1) =  ncr x(n-r) ar

x = b/x, a = x/b,  r = 8 and n = 16

(8 + 1)  =  16c8 (b/x)(16-8) (x/b)8

=  16c8 (b/x)8 (x/b)8

= 16c8

Let us see the next example on "General and middle terms in binomial theorem".

Example 5 :

Find the middle term in the expansion of (a/x  - x)16

Solution :

Here n = 16, that is even

So, the middle term  = T(n/2) + 1

=  T (16/2) + 1

=  T (8 + 1)  ==>  T 9

General term :

T(r+1) =  ncr x(n-r) ar

x = a/x, a = x,  r = 8 and n = 16

(8 + 1)  =  16c8 (a/x)(16-8) (x)8

=  16c8 a8 x-8

=  16c8 a8 x-4

=  16c8 a8/x4

Let us see the next example on "How to find middle term of an expansion".

Example 6 :

Find the middle term in the expansion of (x  - 2y)13

Solution :

Here n = 13, that is even

So, the middle term  =  T(n + 1)/2 and T(n + 3)/2

T(n + 1)/2   =   T (13+1)/2  ==>  T 7

General term :

T(r+1) =  ncr x(n-r) ar

x = x, a = -2y,  r = 6 and n = 13

(6 + 1)  =  13c6 (x)(13-6) (-2y)6

=  13c6 x7 (-2)y6

=  13c6 x7 2y6

T(n + 3)/2   =   T (13+3)/2  ==>  8

x = x, a = -2y,  r = 7 and n = 13

(7 + 1)  =  13c6 (x)(13-7) (-2y)7

=  13c6 x6 (-2)y7

= - 13c6 x6 2y7

Example 7 :

Find the middle term in the expansion of (x  + 2/x2)17

Solution :

Here n = 17, that is even

So, the middle term  =  T(n + 1)/2 and T(n + 3)/2

T(n + 1)/2   =   T (17+1)/2  ==>  9

General term :

T(r+1) =  ncr x(n-r) ar

x = x, a = 2/x2,  r = 8 and n = 17

(8 + 1)  =  17c8 (x)(17-8) (2/x2)8

=  17c8 x9 (2)x-16

=  17c8 x9-16 28

=  17c8 x-7 28

=  17c8 (28/x7)

T(n + 3)/2   =   T (17+3)/2  ==>  10

x = x, a = 2/x2,  r = 9 and n = 17

(9 + 1)  =  17c9 (x)(17-9) (2/x2)9

=  17c9 (x)(29/x18)

=  17c9 (x)(29x-18)

=  17c9 (x)8-18 29

=  17c9 x-10 29

=  17c9 (29/x10)

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