The sum of the rational terms in the binomial expansion of $$\left(2^{\frac{1}{2}} + 3^{\frac{1}{5}}\right)^{10}$$ is :

The binomial expansion of $$\left(2^{\frac{1}{2}} + 3^{\frac{1}{5}}\right)^{10}$$ has a general term given by the formula for the $$(r+1)$$-th term: $$T_{r+1} = \binom{10}{r} \left(2^{\frac{1}{2}}\right)^{10-r} \left(3^{\frac{1}{5}}\right)^r$$.

Simplifying the exponents, we get:

$$T_{r+1} = \binom{10}{r} 2^{\frac{10-r}{2}} \cdot 3^{\frac{r}{5}}$$

For a term to be rational, both exponents must be integers. This is because integer powers of integers yield rational numbers (specifically integers when the base is an integer and the exponent is non-negative). Therefore, we require:

1. $$\frac{10-r}{2}$$ must be an integer.
2. $$\frac{r}{5}$$ must be an integer.

Set $$\frac{r}{5} = m$$, where $$m$$ is an integer. Then $$r = 5m$$. Since $$r$$ must be between 0 and 10 inclusive, we have:

$$0 \leq 5m \leq 10 \implies 0 \leq m \leq 2$$

So $$m$$ can be 0, 1, or 2, giving $$r = 0, 5,$$ or $$10$$.

Now check the condition $$\frac{10-r}{2}$$ is an integer for each $$r$$:

• For $$r = 0$$: $$\frac{10-0}{2} = \frac{10}{2} = 5$$, which is an integer.
• For $$r = 5$$: $$\frac{10-5}{2} = \frac{5}{2} = 2.5$$, which is not an integer.
• For $$r = 10$$: $$\frac{10-10}{2} = \frac{0}{2} = 0$$, which is an integer.

Thus, only $$r = 0$$ and $$r = 10$$ yield rational terms. The term for $$r = 5$$ is irrational because $$2^{\frac{5}{2}} = \sqrt{32}$$, which is irrational.

Now compute the rational terms:

• For $$r = 0$$: $$T_1 = \binom{10}{0} \cdot 2^{\frac{10-0}{2}} \cdot 3^{\frac{0}{5}} = 1 \cdot 2^5 \cdot 3^0 = 1 \cdot 32 \cdot 1 = 32$$.
• For $$r = 10$$: $$T_{11} = \binom{10}{10} \cdot 2^{\frac{10-10}{2}} \cdot 3^{\frac{10}{5}} = 1 \cdot 2^0 \cdot 3^2 = 1 \cdot 1 \cdot 9 = 9$$.

The sum of the rational terms is $$32 + 9 = 41$$.

Hence, the correct answer is Option D.