Answer 1

Here, a = x², b = `(-2)/x`, n = 12

Now, n is even

∴ `("n" + 2)/2 = (8 + 2)/2` = 5

∴ Middle term is t₅, for which r = 4.

We have, t_r+1 = ⁿC_r a^n–r .b^r

∴ t₅ = `""^8"C"_4 (x^2)^4 ((-2)/x)^4`

= `(8!)/(4!4!) (x^8)(16/x^4)`

= `(8 xx 7 xx 6 xx 5)/(4 xx 3 xx 2 xx 1) xx x^8/x^4 xx 16`

= 1120x⁴

∴ Middle term is 1120x⁴.