The sum of the coefficient of x^ 2 / 3 and x^ -2 / 5 in the...

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

Options

A $21 / 4$
B $63 / 16$
C $19 / 4$
D $69 / 16$

Correct answer

A. $21 / 4$

Step-by-step 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}$

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