## Binomial Theorem Examples

** Algebra > Binomial Theorem Examples in Middle term**
**4. Write the middle terms in the following binomial expansions:**

1. (2x^{2} – 3/_{x} )^{10}

2. ( 4/_{y} + y²/_{2x})^{13}

*Solution:*

**1. (2x**^{2} – 3/_{x} )^{10}

In
(2x^{2} – 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) ]} = T_{6}

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.

T_{r + 1} = ^{n}c_{r} .x^{n – r} . y ^{r}

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

T_{6} = T_{5+1} = ^{10}c_{5} . (2x^{2})^{10 – 5} ( -3/_{x} )^{5}

= ^{10}c_{5} .(2)^{5} .(-3)^{5 }.(x^{2})^{5}.( 1/_{x} )^{5} { .(x^{2})^{5}.
( 1/_{x} )^{5} = x^{10-5} = x^{5}}

= - ^{10}c_{5} . (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 T_{7} and T_{8}.

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

T_{r + 1} = ^{n}c_{r} .x ^{n – r} .y ^{r}

T_{7} = T_{6 + 1} = ^{13}c_{6} . (4/_{y} )^{13 - 6} . (y²/_{2x})^{6}

= ^{13}c_{6} . (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}

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