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

∴ ${\displaystyle \frac{\text{n}+1}{2}=\frac{7+1}{2}}$ = 4, ${\displaystyle \frac{\text{n}+3}{2}=\frac{7+3}{2}}$ = 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

= ${\displaystyle \frac{7!}{3!4!}\left(x^8\right)\left(8y^6\right)}$

= ${\displaystyle \frac{7\times 6\times 5\times 4!}{3\times 2\times 1\times 4!}\times 8x^8{y}^6}$

= 280x⁸y⁶

and t₅ = ⁷C₄ (x²)³ . (2y²)⁴

= ${\displaystyle \frac{7!}{4!3!}\left(x^6\right).\left(16y^8\right)}$

= ${\displaystyle \frac{7\times 6\times 5\times 4!}{4!\times 3\times 2\times 1}\times 16x^6{y}^8}$

= 560x⁶y⁸

∴ Middle terms are 280x⁸y⁶ and 560x⁶y⁸.