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

<p>$$ \begin{aligned} & T_{r+1}={ }^9 C_r\left(\frac{x^{-2 / 5}}{2}\right)^r\left(x^{2 / 3}\right)^{9-r} \\ & ={ }^9 C_r \frac{1}{2^r} x^{\frac{2}{3}(9-r)+\left(\frac{-2 r}{5}\right)} \\ & ={ }^9 C_r \cdot \frac{1}{2^r} \cdot x^{6-\frac{16 r}{15}} \end{aligned} $$</p> <p>For coefficient of $x^{2 / 3}$</p> <p>$$\begin{aligned} & \Rightarrow 6-\frac{16 r}{15}=\frac{2}{3} \\ & \Rightarrow 90-16 r=10 \\ & \Rightarrow r=5 \end{aligned}$$</p> <p>For coefficient of $x^{-2 / 5}$</p> <p>$$\begin{aligned} & \Rightarrow 6-\frac{16 r}{15}=\frac{-2}{5}\\ & \Rightarrow 90-16 r=-6 \\ & \Rightarrow r=6 \end{aligned}$$</p> <p>Sum of coefficient of $x^{2 / 3}$ & $x^{-2 / 5}$</p> <p>$$\begin{aligned} & ={ }^9 C_5 \cdot \frac{1}{2^5}+{ }^9 C_6 \cdot \frac{1}{2^6} \\ & =\frac{9!}{5!4!}\left(\frac{1}{2^5}\right)+\frac{9!}{6!3!}\left(\frac{1}{2^6}\right)=\frac{21}{4} \end{aligned}$$</p>