Find the two middle terms in the expansion of :

Solution:

For $\left(3 x-\frac{x^{3}}{6}\right)^{9}$

$\mathrm{a}=3 \mathrm{x}, \quad \mathrm{b}=\frac{-\mathrm{x}^{3}}{6}$ and $\mathrm{n}=9$

As n is odd, there are two middle terms i.e.

I. $\left(\frac{n+1}{2}\right)^{\text {th }}$ and II. $\left(\frac{n+3}{2}\right)^{\text {th }}$

General term $\mathrm{t}_{r+1}$ is given by,

$\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$

I. The first middle term is $\left(\frac{n+1}{2}\right)^{\text {th }}=\left(\frac{9+1}{2}\right)^{\text {th }}=\left(\frac{10}{2}\right)^{\text {th }}=(5)^{\text {th }}$

Therefore, for $5^{\text {th }}$ middle term, $r=4$

Therefore, the first middle term is

$t_{5}=t_{4+1}$

$=\left(\begin{array}{l}9 \\ 4\end{array}\right)(3 x)^{9-4}\left(\frac{-x^{3}}{6}\right)^{4}$

$=\left(\begin{array}{l}9 \\ 4\end{array}\right)(3 x)^{5}\left(x^{3}\right)^{4}\left(\frac{1}{6}\right)^{4}$

$=\left(\begin{array}{l}9 \\ 4\end{array}\right)(3)^{5}(\mathrm{x})^{5}(\mathrm{x})^{12}\left(\frac{1}{6}\right)^{4}$

$=\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \cdot \frac{243}{1296}(\mathrm{x})^{17}$

$=\frac{189}{8}(\mathrm{x})^{17}$

II. The second middle term is $\left(\frac{\mathrm{n}+3}{2}\right)^{\text {th }}=\left(\frac{9+3}{2}\right)^{\text {th }}=\left(\frac{12}{2}\right)^{\text {th }}=(6)^{\text {th }}$

Therefore, for the $6^{\text {th }}$ middle term, $r=5$

Therefore, the second middle term is

$t_{6}=t_{5+1}$

$=\left(\begin{array}{l}9 \\ 5\end{array}\right)(3 x)^{9-5}\left(\frac{-x^{3}}{6}\right)^{5}$

$=\left(\begin{array}{l}9 \\ 4\end{array}\right)(3 x)^{4}\left(-x^{3}\right)^{5}\left(\frac{1}{6}\right)^{5} \ldots\left[\because\left(\begin{array}{l}n \\ r\end{array}\right)=\left(\begin{array}{c}n \\ n-r\end{array}\right)\right]$

$=\left(\begin{array}{l}9 \\ 4\end{array}\right)(3)^{4}(\mathrm{x})^{4}(-\mathrm{x})^{15}\left(\frac{1}{6}\right)^{5}$

$=\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \cdot \frac{81}{7776}(-\mathrm{x})^{19}$

$=-\frac{21}{16}(\mathrm{x})^{19}$