Question

Find the middle terms in the expansion of ${(3x-\frac{x^3}{6})}^7$

Answer 1

The general term in the given expansion is given by
$T_{r+1}=(-1{)}^r{\times }^7{C}_r\times (3x{)}^{7-r}\times {\left(\frac{x^3}{6}\right)}^r\Rightarrow T_{r+1}=(-1{)}^r{\times }^7{C}_r\times 3^{(7-2r)}\times 2^{-r}\times x^{(7+2r)}$
Clearly, the given expansion has 8 terms.
So, it has two middle terms namely $\left(\frac{8}{2}\right)$th, i.e. 4th and 5th.
Now, $T_4=T_{(3+1)}=(-1{)}^3{\times }^7{C}_3\times 3^{(7-6)}\times 2^{-3}\times x^{13}=-(\frac{7\times 6\times 5}{3\times 2\times 1}\times 3^1\times \frac{1}{2^3}\times x^{13})=-\left(\frac{35\times 3\times x^{13}}{8}\right)=\frac{-105x^{13}}{8}\text{And,}{T}_5=T_{(4+1)}=(-1{)}^4{\times }^7{C}_4\times 3^{(7-8)}\times 2^{-4}\times x^{15}{=}^7{C}_3\times 3^{-1}\times 2^{-4}\times x^{15}=(\frac{7\times 6\times 5}{3\times 2\times 1}\times \frac{1}{3}\times \frac{1}{2^4}\times x^{15})=\frac{35x^{15}}{48}$
Hence, the required middle terms are $\frac{-105x^{13}}{8}\text{and}\frac{35x^{15}}{48}$