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Find the two middle terms in the expansion of : \(\Big(3x - \frac{x^3}{6}\Big)^9\)

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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

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