## Binomial Theorem Examples

Algebra > Binomial Theorem Examples in Middle term

4. Write the middle terms in the following binomial expansions:

1. (2x2 – 3/x )10
2. ( 4/y + y²/2x)13

Solution:

1. (2x2 – 3/x )10

In (2x2 – 3/x )10 , the binomial index n is 10, an even integer.

We know that when the binomial index n is an even integer, then there is only one middle term. And the middle term is:

T[(n/2) +1)] = T [(10/2 + 1) ] = T6

So, sixth term is the middle term in the given binomial expansion. Now write this sixth term T6 with the help of the general term.

Tr + 1 = ncr .xn – r . y r

{ here, x = 2x2 and y = - 3/x and n = 10}

T6 = T5+1 = 10c5 . (2x2)10 – 5 ( -3/x )5

= 10c5 .(2)5 .(-3)5 .(x2)5.( 1/x )5                   { .(x2)5. ( 1/x )5 = x10-5 = x5}

= - 10c5 . (2)5 .(3)5.(x)5

2. ( 4/y + y² /2x)13

In this binomial term, index n is 13, an odd number. Therefore, there are two middle terms. The two middle terms are found by using the formulas:

T( n + 1)/2 and T( n + 3 ) /2.

Put 13 in n to find the two middle terms required in the given binomial.

T( 13 + 1 )/2 and T( 13 + 3 )/2 .

Therefore, the two middle terms are T7 and T8.

Now, use the general term to write the two middle terms

Tr + 1 = ncr .x n – r .y r

T7 = T6 + 1 = 13c6 . (4/y )13 - 6 . (y²/2x)6

= 13c6 . (4)7. (1/y )7 . (y2)6 . (1/2 )6 . (1/x )6

= 13c6 . (2)8 .[(y5) / (x6)]

T8 = 13c7 . (4/y )13 - 7 . (y²/2x)7 13c7 . (4 )6 (1/y )7 (y2)7. (1/2)7 .(1/x )7

= 13c7 . (4 )6 . (1/2)7 . (y)7 . (1/x)7

= 13c7 . (2)5 .(y/x)7