The sum of the rational terms in the expansion of $(\sqrt{2}+\sqrt[5]{3}{)}^{10}$, is

Let $T_{r+1}$ be $(r+1{)}^{\text{th }}$term in the expansion of

$(\sqrt{2}+\sqrt[5]{3}{)}^{10}$. Then

$T_{r+1}{=}^{10}{C}_r\left(\sqrt{2}{)}^{10-r}\right(\sqrt[5]{3}{)}^r{=}^{10}{C}_r\times 2^{5-r/2}\times 3^{r/5}$

Here, $0\le r\le 10$

For $T_{r+1}$ to be rational

$\frac{r}{2}$and $\frac{r}{5}$must be integers. 

$\Rightarrow$ $r$must be a multiple of $10$and $0\le r\le 10$

$\Rightarrow r=0, 10$

Thus, $T_1$and $T_{11}$ are rational terms. 

Now,

$T_1+T_{11}{=}^{10}{C}_0\times 2^5\times 3^0+{ }^{10}{C}_{10}\times 2^0\times 3^2= 2^5+ 9=41$