Question

Answer the following:

Find the middle term (s) in the expansion of (x²+ 2y²)⁷ 

Answer 1

Here a = x², b = 2y², n = 7

Since n is odd

∴ `("n" + 1)/2 = (7 + 1)/2` = 4, `("n" + 3)/2 = (7 + 3)/2` = 5

∴ Middle terms are t₄ and t₅, for which r = 3 and r = 4 respectively.

We have, t_r+1 = ⁿC_r a^n–r. b^r 

∴ t₄ = ⁷C₃ (x²)⁴ . (2y²)³

= `(7!)/(3!4!) (x^8) (8y^6)`

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

= 280x⁸y⁶

and t₅ = ⁷C₄ (x²)³ . (2y²)⁴

= `(7!)/(4!3!)(x^6).(16y^8)`

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

= 560x⁶y⁸

∴ Middle terms are 280x⁸y⁶ and 560x⁶y⁸.