Question:
In the expansion of $\left(x^{2}-\frac{1}{3 x}\right)^{9}$, the term without $x$ is equal to
(a) $\frac{28}{81}$
(b) $\frac{-28}{243}$
(c) $\frac{28}{243}$
(d) none of these
Solution:
(c) $\frac{28}{243}$
Suppose the (r + 1)th term in the given expansion is independent of x.
Then , we have:
$T_{r+1}={ }^{9} C_{r}\left(x^{2}\right)^{9-r}\left(\frac{-1}{3 x}\right)^{r}$
$=(-1)^{r}{ }^{9} C_{r} \frac{1}{3^{r}} x^{18-2 r-r}$
For this term to be independent of $x$, we must have:
$18-3 r=0$
$\Rightarrow r=6$
$\therefore$ Required term $=(-1)^{6}{ }^{9} C_{6} \frac{1}{3^{6}}=\frac{9 \times 8 \times 7}{3 \times 2} \times \frac{1}{3^{6}}=\frac{28}{243}$