For \(\Big(x^4- \frac{1}{x^3}\Big)^{11}\)
a = x4, b = \(\frac{-1}{x^4}\) and n = 11
As n is odd, there are two middle terms i.e
II. \(\Big(\frac{n+1}{2}\Big)^{th}\) 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 \(\Big(\frac{n+1}{2}\Big)^{th}\) = \(\Big(\frac{11+1}{2}\Big)^{th}\) = \(\Big(\frac{12}{2}\Big)^{th}\) = (6)th
Therefore, for the 6th middle term, r = 5
Therefore, the first middle term is
t6 = t5+1
= -462.x9
II. The second middle term is \(\Big(\frac{n+3}{2}\Big)^{th}\) = \(\Big(\frac{11+3}{2}\Big)^{th}\) = \(\Big(\frac{14}{2}\Big)^{th}\) = (7)th
Therefore, for the 7th middle term, r = 6
Therefore, the second middle term is
t7 = t6+1
= 462.x2