If the coefficients of $x^2$ and $x^4$ in the expansion of ${\left(x^{1/3}+\frac{1}{2x^{1/3}}\right)}^{18},(x>0)$

are $m$ and $n$ respectively then , $\frac{m}{n}$is equal to 

Let $T_{r+1}$ be the $(r+1{)}^{\text{th }}$ term in the expansion of 

${\left(x^{1/3}+\frac{1}{2x^{1/3}}\right)}^{18}$.The 

$T_{r+1}{=}^{18}{C}_r{\left(x^{1/3}\right)}^{18-r}{\left(\frac{1}{2x^{1/3}}\right)}^r{=}^{18}{C}_r{x}^{6-\frac{2r}{3}}{2}^{-r}$

This will contain $x^{-2},$if $6-\frac{2r}{3}=-2\Rightarrow r=12$

$\therefore m=$Coefficient of $x^{-2}{=}_{C_{12}}^{18}{2}^{-12}$ 

For the coefficient of $x^4$ , we must have 

$6-\frac{2r}{3}=-4\Rightarrow r=15$

$\therefore n=$Coefficient of $x^{-4}{=}^{18}{\mathrm{C}}_{15}{2}^{-15}$

Hence, 

$\frac{m}{n}=\frac{{ }^{18}{C}_{12}{2}^{-12}}{{ }^{18}{C}_{15}{2}^{-15}}=\frac{18!}{6!12!}\times \frac{3!15!}{18!}\times 2^3=\frac{15\times 14\times 13\times 8}{6\times 5\times 4}=182$