The sum of the coefficient of x^2/3 and x^-2/5 in the binomial expansion of (x^2/3+(1)/(2)…

Question

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

Accepted Answer

$\begin{gathered} {T}_{r+1}={}^{9}{C}_{r}{(\frac{{x}^{-2/5}}{2})}^{r}{({x}^{2/3})}^{9-r} \ ={}^{9}{C}_{r}\frac{1}{{2}^{r}}{x}^{\frac{2}{3}(9-r)+(\frac{-2r}{5})} \ ={}^{9}{C}_{r}\cdot \frac{1}{{2}^{r}}\cdot {x}^{6-\frac{16r}{15}} \end{gathered}$

For coefficient of ${x}^{2/3}$

$\begin{gathered} \Rightarrow 6-\frac{16r}{15}=\frac{2}{3} \ \Rightarrow 90-16r=10 \ \Rightarrow r=5 \end{gathered}$

For coefficient of ${x}^{-2/5}$

$\begin{gathered} \Rightarrow 6-\frac{16r}{15}=\frac{-2}{5} \ \Rightarrow 90-16r=-6 \ \Rightarrow r=6 \end{gathered}$

Sum of coefficient of ${x}^{2/3}$ \& ${x}^{-2/5}$

$\begin{gathered} ={}^{9}{C}_{5}\cdot \frac{1}{{2}^{5}}+{}^{9}{C}_{6}\cdot \frac{1}{{2}^{6}} \ =\frac{9!}{5!4!}(\frac{1}{{2}^{5}})+\frac{9!}{6!3!}(\frac{1}{{2}^{6}})=\frac{21}{4} \end{gathered}$

Solutions · 1