Example 3 Given $a_{n}=C_{1994}^{3 n-1}$, find $\sum_{n=1}^{665} a_{n}$.
Solution: Construct the binomial $(1+x)^{1994}=\sum_{k=0}^{1994} C_{1994}^{k} x^{k}$.
Substitute $x=1, \omega, \omega^{2}\left(\omega=\frac{-1+\sqrt{3} i}{2}\right)$ into the above expression to get
$$
2^{1994}=\sum_{k=0}^{1994} C_{1994}^{k},(1+\omega)^{1994}=\sum_{k=0}^{1994} C_{1994}^{k} \omega^{k},\left(1+\omega^{2}\right)=\sum_{k=0}^{1994} C_{1994}^{k} \omega^{2 k} .
$$

Thus,
$$
\begin{aligned}
3 \sum_{n=1}^{665} a_{n} & =2^{1994}+(1+\omega)^{1994} \cdot \omega+\left(1+\omega^{2}\right)^{1994} \cdot \omega^{2} \\
& =2^{1994}+\left(-\omega^{2}\right)^{1994} \cdot \omega+(-\omega)^{1994} \cdot \omega^{2} \\
& =2^{1994}-1,
\end{aligned}
$$

Here, the property $1+\omega+\omega^{2}=0$ is fully utilized. Therefore,
$$
\sum_{n=1}^{665} a_{n}=\frac{2^{1994}-1}{3}
$$