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Q.
If the term independent of $x$ in $\left(\sqrt{x}-\frac{k}{x^{2}}\right)^{10}$ is $405$ , then find value of $k$.
Binomial Theorem
Solution:
$t _{ r +1}={ }^{10} C _{ r }(\sqrt{ x })^{10- r }\left(-\frac{ k }{ x ^{2}}\right)^{ r }$
$={ }^{10} C _{ r } x ^{5-\frac{5 r }{2}}(- k )^{ r }$
For the independent of $x , r$ must be $2$ so that
${ }^{10} C _{2} k ^{2}=405$
$\therefore k =3$