# If the term free from

Question:

If the term free from $\mathrm{x}$ in the expansion of $\sqrt{x}-\frac{k}{x^{2}}$ is $\mathbf{4 0 5}$, find the value of $k$.

Solution:

Given $\sqrt{x}-\frac{k}{x^{2}}$

From the standard formula of Tro1 we can write given expression as

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

$={ }^{10} C_{r}(x)^{5-\frac{r}{2}-2 r}(-k)^{r}={ }^{10} C_{r} x^{\frac{10-5 r}{2}}(-k)^{r}$

For the term free from $x$ we have $(10-5 r) / 2=0$

Which implies $r=2$

So, the term free from $x$ is

$T_{2+1}={ }^{10} C_{2}(-k)^{2}$

${ }^{10} C_{2}(-k)^{2}=405$

$\frac{10 \times 9 \times 8 !}{2 ! \times 8 !}(-k)^{2}=405$

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