Question

Find the two middle terms in the expansion of : (x² + a²)⁵

Answer 1

For (x² + a²)⁵ , 

a= x² , b= a² and n=5 

As n is odd, there are two middle terms i.e.

1. \(\Big(\frac{n+1}{2}\Big)^{th}\) and II. \(\Big(\frac{n+3}{2}\Big)^{th}\) 

General term t_r+1 is given by,

t_r+1 = (_rⁿ) a^n-rb^r

I. The first, middle term is   \(\Big(\frac{n+1}{2}\Big)^{th}\)  =   \(\Big(\frac{5+1}{2}\Big)^{th}\)  =   \(\Big(\frac{6}{2}\Big)^{th}\)  = (3)^rd

Therefore, for the 3^rd middle term, r=2 

Therefore, the first middle term is

t₃ = t₂₊₁

= (₂⁵)(x²)^5-2(a²)²

= (₂⁵)(x²)³(a)⁴

= (₂⁵)(x)⁶(a)⁴

= \(\frac{5\times4}{2\times1}\) .(x)⁶(a)⁴

= 10^.a^4.x⁶

II. The second middle term is   \(\Big(\frac{n+1}{2}\Big)^{th}\)  =   \(\Big(\frac{5+3}{2}\Big)^{th}\)  =   \(\Big(\frac{8}{2}\Big)^{th}\)  = (4)^th

Therefore, for the 4^th middle term, r = 3 

Therefore, the second middle term is

t₄ = t₃₊₁

= 10. a⁶ . x⁴