7. In the expansion of $(2+\sqrt{x})^{2 n+1}$, the sum of the coefficients of the terms where the power of $x$ is an integer is $\qquad$
Let $(2+t)^{2 n+1}=a_{0}+a_{1} t+a_{2} t^{2}+\cdots+a_{2 n+1} t^{2 n+1}$, i.e., find $a_{0}+a_{2}+\cdots+a_{2 n}$. Thus, by taking $t=1,-1$, we get $a_{0}+a_{2}+\cdots+a_{2 n}=\frac{3^{2 n+1}+1}{2}$.