For example, find $C_{n}^{1}+C_{n}^{4}+C_{n}^{7}+\cdots+C_{n}^{3 m+1}$, where $3 m+1$ is the largest integer not greater than $n$.

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Let $1, q, q^{2}$ be the 3rd roots of unity. By the binomial theorem, we have
$$
\begin{array}{l}
(1+1)^{n}=C_{n}^{0}+C_{n}^{1}+C_{n}^{2}+C_{n}^{3}+\cdots+C_{n}^{n} \\
(1+q)^{n}=C_{n}^{0}+C_{n}^{1} q+C_{n}^{2} q^{2}+C_{n}^{3}+C_{n}^{4} q+\cdots \\
\left(1+q^{2}\right)^{n}=C_{n}^{0}+C_{n}^{1} q^{2}+C_{n}^{2} q+C_{n}^{3}+C_{n}^{1} q^{2}+\cdots
\end{array}
$$

Multiplying (2) by $q^{2}$, (3) by $q$, and adding to (1), we get
$$
\begin{aligned}
& 3 C_{n}^{1}+3 C_{n}^{4}+3 C_{n}^{7}+\cdots+3 C_{n}^{3 m+1} \\
= & 2^{n}+q^{2}(1+q)^{n}+q\left(1+q^{2}\right)^{n} \\
= & 2^{n}+q^{2}\left(-q^{2}\right)^{n}+q(-q)^{n} .
\end{aligned}
$$

It is easy to see that
$$
q^{2}\left(-q^{2}\right)^{n}+q(-q)^{n}=2 \cos \frac{n-2}{3} \pi .
$$

Thus,
$$
C_{n}^{1}+C_{n}^{4}+C_{n}^{n}+\cdots+C_{n}^{3 m+1}=\frac{1}{3}\left(2^{n}+2 \cos \frac{n-2}{3} \pi\right) .
$$