Example 7. Given $a_{n}=C_{1994}^{3 n-1}$. Find $\sum_{n=1}^{655} a_{n}$.
Consider the binomial expansion
$$
(1+x)^{1994}=\sum_{k=0}^{1994} C_{1994}^{k} x^{k} \text {. }
$$

Substitute \(1, \omega, \omega^{2}\left(\omega=\frac{-1+\sqrt{3} i}{2}\right)\) into (1), we get
$$
\begin{array}{l}
2^{1994}=\sum_{k=0}^{1994} C_{1994}^{k}, \\
(1+\omega)^{1994}=\sum_{k=0}^{1994} C_{1994}^{k} \cdot \omega^{k}, \\
\left(1+\omega^{2}\right)^{1994}=\sum_{k=0}^{1994} C_{1994}^{k} \omega^{2 k} .
\end{array}
$$
(2) + (3) \(\times \omega\) + (4) \(\times \omega^{2}\), we obtain
$$
\begin{aligned}
3 & \sum_{n=1}^{665} a_{n} \\
= & 2^{1994}+(1+\omega)^{1994} \cdot \omega \\
& +\left(1+\omega^{2}\right)^{1994} \cdot \omega^{2} \\
= & 2^{1994}+\left(-\omega^{2}\right)^{1994} \cdot \omega+(-\omega)^{1994} \cdot \omega^{2} \\
= & 2^{1994}+\omega^{2}+\omega \\
= & 2^{1994}-1 . \\
\therefore & \sum_{n=1}^{665} a_{n}=\frac{2^{1994}-1}{3} .
\end{aligned}
$$