$\left(2^{1/2} + 3^{1/5} \right)^{10} = ^{10}C_{0} \left(2^{1/2}\right)^{10} + ^{10}C_{1} \left(2^{1/2}\right)^{9} \left(3^{1/5}\right) + ..... + ^{10}C_{10} \left(3^{1/5}\right)^{10}$
There are only two rational terms - first term and last term.
Now sum of two rational terms
$ = \left(2\right)^{5} + \left(3\right)^{2}$
= 32 + 9 = 41