The middle term in the expansion of (x^2-1/x)^8 is

TS ICET 2015 Question Paper with Solutions for online practice

Question : 137 of 200

Marks: +1, -0

The middle term in the expansion of

(x^2-1/x)^8

Solution:

Given :

(x^2-1/x)^8

n = 8 is even [Middle Term

=(n/2+1)^{\th}

term]
∴ Middle term is fifth term.

t_{5}=^{8}{C}_{4}(x^{2})^{4}(-\;\;{1}/{x})^{4}=70x^{8}(\;\;{1}/{x^{4}})=70x^{4}

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