Find the second term of the binomial expansion of $\left(\sqrt[13]{a} + \dfrac{a}{\sqrt{a^{-1}}}\right)^m$ if $C^m_3 : C^m_2 = 4 : 1$.
Given: $\;\;$ $C^m_3 : C^m_2 = 4 : 1$
i.e. $\;$ $\dfrac{m!}{3! \left(m-3\right)!} : \dfrac{m!}{2! \left(m-2\right)!} = \dfrac{4}{1}$
i.e. $\;$ $\dfrac{m!}{3 \times 2! \left(m-3\right)!} \times \dfrac{2! \left(m-2\right) \left(m-3\right)!}{m!} = 4$
i.e. $\;$ $\dfrac{m-2}{3} = 4$
i.e. $\;$ $m - 2 = 12$ $\implies$ $m = 14$
$\therefore \;$ The problem is: $\;\;$ $\left(\sqrt[13]{a} + \dfrac{a}{\sqrt{a^{-1}}}\right)^m = \left(a^{\frac{1}{13}} + a^{1 + \frac{1}{2}}\right)^{14} = \left(a^{\frac{1}{13}} + a^{\frac{3}{2}}\right)^{14}$
$\therefore \;$ Second term in the expansion of $\left(a^{\frac{1}{13}} + a^{\frac{3}{2}}\right)^{14}$ is
$\begin{aligned}
T_2 = T_{1+1} & = C^{14}_{1} \; \left(a^{\frac{1}{13}}\right)^{14-1} \; \left(a^{\frac{3}{2}}\right)^1 \\\\
& = 14 \times \left(a^{\frac{1}{13}}\right)^{13} \times a^{\frac{3}{2}} \\\\
& = 14 \; a^{1+\frac{3}{2}} \\\\
& = 14a^{\frac{5}{2}}
\end{aligned}$