# In the expansion of

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 $=(-1)^{6}{ }^{9} C_{6} \frac{1}{3^{6}}=\frac{9 \times 8 \times 7}{3 \times 2} \times \frac{1}{3^{6}}=\frac{28}{243}$