Question:
If the constant term in the binomial expansion of $\left(\sqrt{x} \frac{k}{x^{2}}\right)^{10}$ is 405, then $|k|$ equals:
Correct Option: , 3
Solution:
General term $=T_{r+1}={ }^{10} C_{r}(\sqrt{x})^{10-r} \cdot\left(-\frac{k}{x^{2}}\right)^{\eta}$
$={ }^{10} C_{r}(-k)^{r} \cdot x^{\frac{10-r}{2}-2 r}={ }^{10} C_{r}(-k)^{r} \cdot x^{\frac{10-5 r}{2}}$
Since, it is constant term, then
$\frac{10-5 r}{2}=0 \Rightarrow r=2$
$\therefore{ }^{10} C_{2}(-k)^{2}=405$
$\Rightarrow k^{2}=\frac{405 \times 2}{10 \times 9}=\frac{81}{9}=9$
$\therefore|k|=3$