Question

Find the real value of $k$ if the term independent of $x$ in the expansion of ${(\sqrt{x}+\frac{k}{x^2})}^{10}$ is $405$.

Answer 1

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

$={}^{10}{C}_r{k}^r{x}^{5-\frac{x}{2}-2r}$

$={}^{10}{C}_r{k}^r{x}^{5-\frac{5x}{2}}$

For term independent of $x$

$5-\frac{5x}{2}=0$

Or, $\frac{5x}{2}=5\Rightarrow r=2$

$\therefore$ The term is ,$T_3={}^{10}{C}_2{k}^2=405$

Or, $45k^2=405$

Or,$k^2=9$

$\therefore k=\pm 3$ Ans