Question

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

• A. 2
• B. 1
• C. 3
• D. 9

Answer 1

Answer: (C)

${\left(\sqrt{x}-\frac{k}{x^2}\right)}^{10}$
$T_{r+1}={}^{10}{C}_r(\sqrt{x}{)}^{10-r}{\left(\frac{-k}{x^2}\right)}^r$
$T_{r+1}={}^{10}{C}_r\cdot x^{\frac{10-t}{2}}\cdot (-k{)}^r\cdot x^{-2r}$
$T_{r+1}={}^{10}{C}_r{x}^{\frac{10-5r}{2}}(-k{)}^r$
Constant term $:\frac{10-5r}{2}=0\Rightarrow r=2$
$T_3={}^{10}{C}_2\cdot (-k{)}^2=405$
$k^2=\frac{405}{45}=9$
$k=\pm 3\Rightarrow |k|=3$