Question

If the constant term in the binomal expansion of ${(\sqrt{x}-\frac{k}{x^2})}^{10}$ is $405$, then $\left|k\right|$ equals:

• A. $1$
• B. $3$
• C. $2$
• D. $9$

Answer 1

The correct option is B $3$

General term
$T_{r+1}{=}^{10}{C}_r{\left(\frac{-k}{x^2}\right)}^r\left(\sqrt{x}{)}^{10-r}{=}^{10}{C}_r\right(-k{)}^r(x{)}^{5-\frac{5r}{2}}$
For constant term
$5-\frac{5r}{2}=0\Rightarrow r=2$
$T_3{=}^{10}{C}_2{k}^2=405$
$k^2=\frac{405}{45}=\frac{81}{9}=9$
$\left|k\right|=3$