For \(\Big(3x - \frac{x^3}{6}\Big)^9\)
a = 3x, b = \(\frac{-x^3}{6}\) and n = 9
As n is odd, there are two middle terms i.e.
I. \(\Big(\frac{n+1}{2}\Big)^{th}\) 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 \(\Big(\frac{n+1}{2}\Big)^{th}\)= \(\Big(\frac{9+1}{2}\Big)^{th}\) = \(\Big(\frac{10}{2}\Big)^{th}\) = (5)th
Therefore, for 5th middle term, r=4
Therefore, the first middle term is
t5 = t4+1
II. The second middle term is \(\Big(\frac{n+1}{2}\Big)^{th}\)= \(\Big(\frac{9+3}{2}\Big)^{th}\) = \(\Big(\frac{12}{2}\Big)^{th}\) = (6)th
Therefore, for the 6th middle term, r=5
Therefore, the second middle term is
t6 = t5+1