1. General term in the expansion

For $\left(2^{\frac15}+5^{\frac13}\right)^{15},$ the general term is $T_{r+1}=\binom{15}{r}\left(2^{\frac15}\right)^{15-r}\left(5^{\frac13}\right)^r,\qquad r=0,1,2,\dots,15.$

So, $T_{r+1}=\binom{15}{r}\,2^{\frac{15-r}{5}}\,5^{\frac r3}.$

2. Condition for a rational term

A term will be rational only if both exponents are integers:

• $\dfrac{15-r}{5}$ must be an integer,
• $\dfrac r3$ must be an integer.

That means:

• $15-r \equiv 0 \pmod 5 \Rightarrow r \equiv 0 \pmod 5$,
• $r \equiv 0 \pmod 3$.

Hence $r$ must be divisible by both $5$ and $3$, i.e. $r\equiv 0 \pmod{15}.$

Since $0\le r\le 15$, the possible values are $r=0,\;15.$

So there are only two rational terms.

3. Compute the rational terms

• For $r=0$: $T_1=\binom{15}{0}\left(2^{\frac15}\right)^{15}=2^3=8.$

• For $r=15$: $T_{16}=\binom{15}{15}\left(5^{\frac13}\right)^{15}=5^5=3125.$

4. Sum of all rational terms

Therefore, $8+3125=3133.$

5. Compare with stored answer

Our derived answer is 3133, which corresponds to Option C. The stored correct answer is also C.

So the stored answer is correct.