The number of integral terms in the expansion of $\left(3^{1/2}+5^{1/4}\right)^{680}$ is equal to
Step-by-Step Solution
Key Concept: $T_{r+1}=\binom{680}{r}3^{(680-r)/2}\cdot 5^{r/4}$. For both powers to be integers, $(680-r)/2\in\mathbb{Z}$ (always) and $r/4\in\mathbb{Z}$, so $r$ must be a multiple of 4.
$r\in\{0,4,8,\ldots,680\}$: $171$ values.
Correct Answer: 171