The sum of the coefficient of $x^{2 / 3}$ and $x^{-2 / 5}$ in the binomial expansion of $\left(x^{2 /…

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
  1. $21 / 4$
  2. $63 / 16$
  3. $19 / 4$
  4. $69 / 16$

Solution

$\begin{aligned} & \mathrm{T}_{\mathrm{r}+1}={ }^9 \mathrm{C}_{\mathrm{r}}\left(\mathrm{x}^{2 / 3}\right)^{9-\mathrm{r}}\left(\frac{\mathrm{x}^{-2 / 5}}{2}\right)^{\mathrm{r}} \\ & ={ }^9 \mathrm{C}_{\mathrm{r}}\left(\frac{1}{2}\right)^{\mathrm{r}}(\mathrm{r})^{\left(6 \frac{2 \mathrm{r}}{3} \frac{2 \mathrm{r}}{5}\right)}\end{aligned}$
for coefficient of $x^{2 / 3}$, put $6-\frac{2 r}{3}-\frac{2 r}{5}=\frac{2}{3}$ $\Rightarrow \mathrm{r}=5$ $\therefore$ Coefficient of $\mathrm{x}^{2 / 3}$ is $={ }^9 \mathrm{C}_5\left(\frac{1}{5}\right)^5$ 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 ${ }^9 \mathrm{C}_6\left(\frac{1}{2}\right)^6$ $\text {Sum }={ }^9 \mathrm{C}_5\left(\frac{1}{2}\right)^5+{ }^9 \mathrm{C}_6\left(\frac{1}{2}\right)^6=\frac{21}{4}$

Asked in: JEE Main 2024 (09 Apr Shift 2)

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