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)

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