Question

If the second term in the expansion ${\left(\sqrt[13]{a}+\frac{a}{\sqrt{a^{-1}}}\right)}^n$ is $14a^{5/2}$, then $\frac{{}^n{C}_3}{{}^n{C}_2}=$

• A. $4$
• B. $3$
• C. $12$
• D. $6$

Answer 1

Answer: (A)

We have $T_2=14a^{\frac{5}{2}}$
${\Rightarrow }^n{C}_1{\left(a^{\frac{1}{13}}\right)}^{n-1}\left(a^{\frac{3}{2}}\right)=14a^{\frac{5}{2}}$
$n{a}^{\frac{n-1}{13}+\frac{3}{2}}=14a^{\frac{5}{2}}\Rightarrow n=14$
$\Rightarrow \frac{{}^n{C}_3}{{}^n{C}_2}=\frac{{}^{14}{C}_3}{{}^{14}{C}_2}=\frac{12}{3}=4$