6. If $(2 \dot{x}+4)^{2 n}=\sum_{i=0}^{2 n} a_{i} x^{i}\left(n \in \mathbf{Z}_{+}\right)$, then the remainder of $\sum_{i=1}^{n} a_{2 i}$ when divided by 3 is $\qquad$
6.1.

Let $x=0$, we get $a_{0}=4^{2 n}$.
By setting $x=1$ and $x=-1$ respectively, and adding the two resulting equations, we get
$$
\begin{array}{l}
\sum_{i=0}^{n} a_{2 i}=\frac{1}{2}\left(6^{2 n}+2^{2 n}\right) . \\
\text { Therefore, } \sum_{i=1}^{n} a_{2 i}=\frac{1}{2}\left(6^{2 n}+2^{2 n}\right)-4^{2 n} \\
\equiv(-1)^{2 n-1}-1^{2 n} \equiv 1(\bmod 3) .
\end{array}
$$