Mathematics - Binomial Theorem Question with Solution | TestHub

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