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Question
Answer the following:
Find the middle term (s) in the expansion of `((2"a")/3 - 3/(2"a"))^6`
Solution
Here a = `(2"a")/3`, b = `(-3)/(2"a")`, n = 6
Since n is even
∴ `("n" + 2)/2 = (6 + 2)/2` = 4
∴ Middle term is t4, for which r = 3
We have tr+1 = nCr an–r.br
∴ t4 = `""^6"C"_3 ((2"a")/3)^3 ((-3)/(2"a"))^3`
= `(6!)/(3!3!) xx (-1)`
= `(-6 xx 5 xx 4 xx 3!)/(3 xx 2 xx 1 xx 3!)`
= – 20
∴ Middle term is – 20.
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