# If the term free from x in the expansion of

Question:

If the term free from $x$ in the expansion of $\left(\sqrt{x}-\frac{k}{x^{2}}\right)^{10}$ is 405, find the value of $k$.

Solution:

Let $(r+1)^{\text {th }}$ term, in the expansion of $\left(\sqrt{x}-\frac{k}{x^{2}}\right)^{10}$, be free from $x$ and be equal to $T_{r+1}$. Then,

$T_{r+1}={ }^{10} C_{r}(\sqrt{x})^{10-r}\left(\frac{-k}{x^{2}}\right)^{r}={ }^{10} C_{r} x^{5-\frac{5 r}{2}}(-k)^{r} \quad \ldots(1)$

If $T_{r+1}$ is independent of $x$, then

$5-\frac{5 r}{2}=0 \Rightarrow r=2$

Putting r = 2 in (1), we obtain

$T_{3}={ }^{10} C_{2}(-k)^{2}=45 k^{2}$

But it is given that the value of the term free from x is 405.

$\therefore 45 k^{2}=405 \Rightarrow k^{2}=9 \Rightarrow k=\pm 3$

Hence, the value of $k$ is $\pm 3$.