Question
What are the values of $$k$$ if the term independent of $$x$$ in the expansion of $${\left( {\sqrt x + \frac{k}{{{x^2}}}} \right)^{10}}\,$$ is 405 ?
A.
$$ \pm 3$$
B.
$$ \pm 6$$
C.
$$ \pm 5$$
D.
$$ \pm 4$$
Answer :
$$ \pm 3$$
Solution :
Given expansion is $${\left( {\sqrt x + \frac{k}{{{x^2}}}} \right)^{10}}$$
$$\eqalign{
& {\left( {r + 1} \right)_{th}}\,{\text{term}},\,{T_{r + 1}} = {\,^{10}}{C_r}{\left( {\sqrt x } \right)^{10 - r}}{\left( {\frac{k}{{{x^2}}}} \right)^r} \cr
& \Rightarrow \,{T_{r + 1}} = {\,^{10}}{C_r}{x^{\frac{{5 - r}}{2}}} \cdot {\left( k \right)^r} \cdot {x^{ - 2r}} \cr
& \therefore \,{T_{r + 1}} = {\,^{10}}{C_r}{x^{\frac{{\left( {10 - 5r} \right)}}{2}}}{\left( k \right)^r} \cr} $$
Since, $${T_{r + 1}}$$ is independent of $$x$$
$$\eqalign{
& \therefore \,\frac{{10 - 5r}}{2} = 0 \cr
& \Rightarrow \,r = 2 \cr
& \therefore 405 = {\,^{10}}{C_2}{\left( k \right)^2} \cr
& 405 = 45 \times {k^2} \cr
& \Rightarrow \,{k^2} = 9 \cr
& \Rightarrow \,k = \pm \,3 \cr} $$