The sum of all rational terms in the expansion of $\left(2^{\frac{1}{5}}+5^{\frac{1}{3}}\right)^{15}$ is equal to :

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{T}_{\mathrm{r}+1}={ }^{15} \mathrm{C}_{\mathrm{r}}\left(5^{\frac{1}{3}}ight)^{\mathrm{r}}\left(\frac{1}{2^5}ight)^{15-\mathrm{r}} \ & ={ }^{15} \mathrm{C}_{\mathrm{r}} 5^{\frac{\mathrm{r}}{3}} \cdot 2^{\frac{15-\mathrm{r}}{5}} \ & \mathrm{R}=3 \lambda, 15 \mu \ & \Rightarrow \mathrm{r}=0,15 \end{aligned}$ 2 rational terms $\begin{aligned} & \Rightarrow{ }^{15} \mathrm{C}_0 2^3+{ }^{15} \mathrm{C}_{15}(5)^5 \ & =8+3125=3133 \end{aligned}$

The sum of all rational terms in the expansion of $\left(2^{\frac{1}{5}}+5^{\frac{1}{3}}\right)^{15}$ is…

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