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 tr+1 is given by,
tr+1 = (rn) an-rbr
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 3rd 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.a4.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 4th middle term, r = 3
Therefore, the second middle term is
t4 = t3+1
= 10. a6 . x4