The sum of the coefficient of x^{2 / 3} and x | vedclass

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Question

$ \mathrm{T}_{\mathrm{r}+1}={ }^9 \mathrm{C}_{\mathrm{r}}\left(\mathrm{x}^{2 / 3}\right)^{9-\mathrm{r}}\left(\frac{\mathrm{x}^{-2 / 3}}{2}\right)^{\ma

Vedclass · Accepted Answer

$21 / 4$

Vedclass · Answer

$ \mathrm{T}_{\mathrm{r}+1}={ }^9 \mathrm{C}_{\mathrm{r}}\left(\mathrm{x}^{2 / 3}ight)^{9-\mathrm{r}}\left(\frac{\mathrm{x}^{-2 / 3}}{2}ight)^{\mathrm{r}} $

$ ={ }^9 \mathrm{C}_{\mathrm{r}}\left(\frac{1}{2}ight)^{\mathrm{r}}(\mathrm{r})^{\left(6-\frac{2 \mathrm{r}-2 \mathrm{r}}{3}ight)}$

for coefficient of $x^{2 / 3}$, put $6-\frac{2 r}{3}-\frac{2 r}{5}=\frac{2}{3}$

$\Rightarrow \mathrm{r}=5$

$	herefore$ Coefficient of $\mathrm{x}^{2 / 3}$ is $={ }^9 \mathrm{C}_3\left(\frac{1}{5}ight)^5$

For coefficient of $x^{-2 / 5}$, put $6-\frac{2 r}{3}-\frac{2 r}{5}=-\frac{2}{5}$

$\Rightarrow \mathrm{r}=6$

Coefficient of $\mathrm{x}^{-2 / 5}$ is ${ }^9 \mathrm{C}_6\left(\frac{1}{2}ight)^6$

$	ext { Sum }={ }^9 \mathrm{C}_5\left(\frac{1}{2}ight)^5+{ }^9 \mathrm{C}_6\left(\frac{1}{2}ight)^6=\frac{21}{4}$

The sum of the coefficient of $x^{2 / 3}$ and $x^{-2 / 5}$ in the binomial expansion of $\left(x^{2 / 3}+\frac{1}{2} x^{-2 / 5}\right)^9$ is :

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