If the coefficients of $x^{-2}$ and $x^{-4}$, in the expansion of ${\left(x^{\frac{1}{3}}+\frac{1}{2x^{\frac{1}{3}}}\right)}^{18},\left(x>0\right)$, are $m$ and $n$ respectively, then $\frac{m}{n}$ is equal to

We have, ${\left(x^{\frac{1}{3}}+\frac{1}{2x^{\frac{1}{3}}}\right)}^{18},\left(x>0\right)$

General term, $T_{r+1}={}^{18}{C}_r{\left(x^{\frac{1}{3}}\right)}^{18-r}{\left(\frac{1}{2x^{\frac{1}{3}}}\right)}^r={}^{18}{C}_r x^{6-\frac{2r}{3}}\frac{1}{2^r}$
Now,$\begin{array}{ccc}6-\frac{2r}{3}=-2& \Rightarrow & r=12\end{array}$ and, $\begin{array}{ccc}6-\frac{2r}{3}=-4& \Rightarrow & r=15\end{array}$
Hence, $\frac{m}{n}=\frac{\text{Coefficient of} x^{-2}}{\text{Coefficient of} x^{-4}}=\frac{\frac{{}^{18}C_{12}}{2^{12}}}{\frac{{}^{18}C_{15}}{2^{15}}}$

$=\frac{{\displaystyle \frac{18!}{6!.12!}}}{\frac{18!}{3!.15!}}\times 2^3$

$=\frac{15\times 14\times 13}{6\times 5\times 4}\times 2^3$

$=182$