Question

Find the third term in the expansion of
$\rm \bigg(2x^{2}+\frac{3}{2x}\bigg)^{8}$

Answer 1

Solution :

$\rm \bigg(2x^{2}+\frac{3}{2x}\bigg)^{8}$

Comparing with ( X + a )ⁿ , we get

X = 2x² , a = 3/2x , n = 8

In the expansion ,we have ( n + 1 ) = 8 + 1

= 9 terms

We know that ,

t( r + 1 ) = ⁿCr x^n-r × a^r

t3 = t ( 2 + 1 )

= 8C2 ( 2x² )^8-2 ( 3/2x )²

= 8C2 ( 2x² )^6 ( 9/4x² )

= 28 × 2^6 × 9/4 × x^12/x²

= 7 × 64 × 9 × x^10

= 4032 × x^10

Therefore ,

3rd term in the expansion = 4032 × x^10

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