Find the $13^{\text {th }}$ term in the expansion of $\left(9 x-\frac{1}{3 \sqrt{x}}\right)^{18}, x \neq 0$
It is known that $(r+1)^{\text {th }}$ term, $\left(T_{r+1}\right)$, in the binomial expansion of $(a+b)^{n}$ is given by $T_{r+1}={ }^{n} C_{r} a^{n-r} b^{r}$.
Thus, $13^{\text {th }}$ term in the expansion of $\left(9 x-\frac{1}{3 \sqrt{x}}\right)^{18}$ is
$\mathrm{T}_{13}=\mathrm{T}_{12+1}={ }^{18} \mathrm{C}_{12}(9 \mathrm{x})^{18-12}\left(-\frac{1}{3 \sqrt{\mathrm{x}}}\right)^{12}$
$=(-1)^{12} \frac{18 !}{12 ! 6 !}(9)^{6}(x)^{6}\left(\frac{1}{3}\right)^{12}\left(\frac{1}{\sqrt{x}}\right)^{12}$
$=\frac{18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13 \cdot 12 !}{12 ! \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2} \cdot x^{6} \cdot\left(\frac{1}{x^{6}}\right) \cdot 3^{12}\left(\frac{1}{3^{12}}\right) \quad\left[9^{6}=\left(3^{2}\right)^{6}=3^{12}\right]$
$=18564$