Question

If the second term of the expansion ${[a^{\frac{1}{13}}+\frac{a}{\sqrt{a^{-1}}}]}^n$ is $14a^{\frac{5}{2}}$, then the value of $\frac{{}^n{C}_3}{{}^n{C}_2}$ is

Answer 1

$T_2={}^n{C}_1{\left(a^{\frac{1}{13}}\right)}^{n-1}\cdot a\sqrt{a}=14a^{\frac{5}{2}}\Rightarrow n(a{)}^{\frac{n-1}{13}}=14a\Rightarrow n(a{)}^{\frac{n-14}{13}}=14$
Comparing L.H.S. and R.H.S.
$\Rightarrow a^{\frac{n-14}{13}}=a^0\Rightarrow n=14\Rightarrow \frac{{}^{14}{C}_3}{{}^{14}{C}_2}=\frac{14!}{3!\cdot 11!}\times \frac{2!\cdot 12!}{14!}=4$