Question

Find the middle terms in the expansion of

`(2x^2 - 3/x^3)^10`

Answer 1

`(2x^2 - 3/x^3)^10 = (2x^2 + (-3)/x^3)^10`  compare with the (x + a)ⁿ

Here x is 2x², a is `(-3)/x^3`, n = 10, which is even.

So the only middle term is `("t"_(n+1))/2 = "t"_10/2 + 1 = "t"_6`

General term t_r+1 = nC_r x^n-r a^r

t₆ = t₅₊₁ = t_r+1

`= 10"C"_5 (2x^2)^(10-5) ((-3)/x^3)^5`

= \[\ce{10C5 (2x^2)^5 \frac{(-3)^5}{(x^3)^5} = 10C5 2^5 x^10 \frac{(-3)^5}{x^15}}\]

= 10C₅ 2⁵ × (-3)⁵ × x^10-15 = 10C₅ 2⁵ × (-3)⁵ × x^-5

`= 10"C"_5 2^5 xx (-1)^5 xx 3^5 xx1/x^5 = - 10"C"_5 xx (2 xx 3)^5 xx 1/x^5 = - 10"C"_5 (6^5) 1/x^5`