7. In the expansion of $(\sqrt{x}+2)^{2 n+1}$, the sum of the coefficients of the terms where the power of $x$ is an integer is $\qquad$
7. $\frac{1}{2}\left(3^{2 n+1}+1\right)$ Detailed Explanation: The $(r+1)$-th term in the expansion is $T_{r+1}=C_{2 n+1} x^{\frac{2 n+1-1}{2}} \cdot 2^{r}$, since the exponent of $x$ is an integer, hence $r$ is odd, denoted as $S=C_{2 n+1}^{1} \cdot 2+C_{2 n+1}^{3} \cdot 2^{3}+C_{2 n+1}^{5} \cdot 2^{5}+\cdots+C_{2 n+1}^{2 n+1} \cdot 2^{2 n+1}$, then using the expansions of $(1+2)^{2 n+1}$ and $(1-2)^{2 n+1}$ we get $S=\frac{1}{2}\left(3^{2 n+1}+1\right)$.