If the coefficients of x^{-2} and x^{-4} in t | vedclass

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Question

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

Vedclass · Accepted Answer

$182$

Vedclass · Answer

$T_{r+1}=18 C_{r}\left(x^{\frac{1}{3}}ight)^{18-r}\left(\frac{1}{2 x^{\frac{1}{3}}}ight)^{r}$

$=^{18} C_{r} x^{6-\frac{2 r}{3}} \frac{1}{2^{r}}$

$\left\{ \begin{gathered}
  6 - \frac{{2r}}{3} =  - 2 \Rightarrow r = 12 \hfill \
  \& \,6 - \frac{{2r}}{3} =  - 4 \Rightarrow r = 15 \hfill \ 
\end{gathered}  ight\}$

$\Rightarrow \quad \frac{	ext { coefficient of } x^{-2}}{	ext { coefficient of } x^{-4}}=\frac{^{18} C_{12} \frac{1}{2^{12}}}{^{18} C_{15} \frac{1}{2^{15}}}=182$

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

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