Find the 9th term in the expansion of

To find: $9^{\text {th }}$ term in the expansion of $\left(\frac{a}{b}-\frac{b}{2 a^{2}}\right)^{12}$

Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$

(ii) $T_{r+1}={ }^{n} C_{r} a^{n-r} b^{r}$

For $9^{\text {th }}$ term, $r+1=9$

$\Rightarrow r=8$

$\ln ,\left(\frac{a}{b}-\frac{b}{2 a^{2}}\right)^{12}$

$9^{\text {th }}$ term $=T_{8+1}$

$\Rightarrow 12 \mathrm{C}_{8}\left(\frac{a}{b}\right)^{12-8}\left(\frac{-b}{2 a^{2}}\right)^{8}$

$\Rightarrow \frac{12 !}{8 !(12-8) !}\left(\frac{a}{b}\right)^{4}\left(\frac{-b}{2 a^{2}}\right)^{8}$

$\Rightarrow 495^{\left(\frac{a^{4}}{b^{4}}\right)}\left(\frac{b^{8}}{256 a^{16}}\right)$

$\Rightarrow\left(\frac{495 b^{4}}{256 a^{12}}\right)$

Ans) $\left(\frac{495 b^{4}}{256 a^{12}}\right)$