Question:
If the term independent of $x$ in the expansion of $\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{9}$ is $k$, then $18 k$ is equal to :
Correct Option: , 3
Solution:
General term $=T_{r+1}={ }^{9} C_{r}\left(\frac{3 x^{2}}{2}\right)^{9-r}\left(-\frac{1}{3 x}\right)^{r}$
$={ }^{9} C_{r}\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^{r} x^{18-3 r}$
The term is independent of $x$, then
$18-3 r=0 \Rightarrow r=6$
$\therefore T_{7}={ }^{9} C_{6}\left(\frac{3}{2}\right)^{3}\left(-\frac{1}{3}\right)^{6}={ }^{9} C_{3}\left(\frac{1}{6}\right)^{3}$
$=\frac{9 \times 8 \times 7}{3 \times 2 \times 1}\left(\frac{1}{6}\right)^{3}=\left(\frac{7}{18}\right)$
$\therefore 18 k=18 \times \frac{7}{18}=7$.