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.

Answer 1

Let (r + 1)^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,
$T_{r+1}{=}^{10}{C}_r{\left(\sqrt{x}\right)}^{10-r}{\left(\frac{-k}{x^2}\right)}^r{=}^{10}{C}_r{x}^{5-\frac{5r}{2}}{\left(-k\right)}^r....\left(1\right)$

If T_{r + 1} is independent of x, then
$5-\frac{5r}{2}=0\Rightarrow r=2$

Putting r = 2 in (1), we obtain

$T_3{=}^{10}{C}_2{\left(-k\right)}^2=45k^2$

But it is given that the value of the term free from x is 405.
$\therefore 45k^2=405\Rightarrow k^2=9\Rightarrow k=\pm 3$

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