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Question
Find the middle terms in the expansion of
`(2x^2 - 3/x^3)^10`
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Solution
`(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 tr+1 = nCr xn-r ar
t6 = t5+1 = tr+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 × x10-15 = 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`
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