If the constant term in the expansion of $\left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[3]{5}} ight)^{12}, x
eq 0$, is $\alpha imes 2^8 imes \sqrt[5]{3}$, then $25 \alpha$ is equal to :

Question

If the constant term in the expansion of $\left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[3]{5}}ight)^{12}, x 
eq 0$, is $\alpha 	imes 2^8 	imes \sqrt[5]{3}$, then $25 \alpha$ is equal to :

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{T}_{\mathrm{r}+1}={ }^{12} \mathrm{C}_{\mathrm{r}}\left(\frac{3^{1 / 5}}{\mathrm{x}}ight)^{12-\mathrm{r}}\left(\frac{2 \mathrm{x}}{5^{1 / 3}}ight)^{\mathrm{r}} \ & \mathrm{T}_{\mathrm{r}+1}=\frac{{ }^{12} \mathrm{C}_{\mathrm{r}}(3)^{\frac{12-\mathrm{r}}{5}}(2)^{\mathrm{r}}(\mathrm{x})^{2 \mathrm{r}-12}}{(5)^{\mathrm{r} / 3}} \ & \mathrm{r}=6 \ & \mathrm{~T}_7=\frac{{ }^{12} \mathrm{C}_6(3)^{6 / 5}(2)^6}{5^2}=\left(\frac{9 	imes 11 	imes 7}{25}ight) 2^8 \cdot 3^{1 / 5} \ & 25 \alpha=693\end{aligned}$

If the constant term in the expansion of $\left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[3]{5}}\right)^{12}, x…

Solution