$$
\begin{array}{l}
\text { 7. Let }\left(3+x+2 x^{2}\right)^{n} \\
=a_{0}+a_{1} x+\cdots+a_{2 n} x^{2 n}\left(n \in \mathbf{N}_{+}\right)
\end{array}
$$

hold for $x \in \mathbf{R}$. Then
$$
a_{1}+a_{2}+\cdots+a_{2 n-1}=
$$
$\qquad$
Let $x=1$. Then $a_{0}+a_{1}+\cdots+a_{2 n}=6^{n}$.
It is easy to see that $a_{0}=3^{n}, a_{2 n}=2^{n}$.
Therefore, $a_{1}+a_{2}+\cdots+a_{2 n-1}=6^{n}-3^{n}-2^{n}$.