Example 4 Let the terms of the sequence $\left\{a_{n}\right\}$ be
$$
3,3,7,7, \cdots,(4 n-1),(4 n-1), \cdots \text {. }
$$

Find $a_{n}$ and $S_{n}$.
Solution: This problem involves an equal-segment increment sequence, where the number of items in each segment (repeated items) is equal. Using the formula for the sum of an arithmetic sequence, the sum of the first $(m-1)$ segments of the sequence $\left\{a_{n}\right\}$ is
$$
\underbrace{2+2+\cdots+2}_{(m-1) \uparrow}=2(m-1) \text {. }
$$

It is clear that when and only when
$$
2(m-1)+1 \leqslant n \leqslant 2 m
$$

then, $a_{n}=4 m-1$. Solving the inequality (4) for $m$, we get
$\frac{n}{2} \leqslant m \leqslant \frac{n+1}{2}$, and the same item appears 2 times.
$$
\because 0<\frac{n+1}{2}-\frac{n}{2}<1 \text {, }
$$
$\therefore m$ is the integer part of $\frac{n+1}{2}$, denoted as
$$
\begin{array}{l}
m=\left[\frac{n+1}{2}\right] . \\
\therefore a_{n}=4\left[\frac{n+1}{2}\right]-1 .
\end{array}
$$

Thus, the sum of the first $n$ segments of the sequence $\left\{a_{n}\right\}$ is
$$
\begin{array}{l}
S_{n \in}=2[3+7+11+\cdots+(4 n-1)] \\
=2 \cdot \frac{n[3+(4 n-1)]}{2}=2 n(2 n+1) .
\end{array}
$$

Substituting $n$ with $m-1$, the sum of the first $(m-1)$ segments is
$$
\begin{array}{l}
S_{2(m-1)}=2(m-1)(2 m-1) . \\
\because a_{n}=4 m-1,
\end{array}
$$
$\therefore$ the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$ is
$$
\begin{aligned}
S_{n}= & S_{2(m-1)}+[n-2(m-1)](4 m-1) \\
= & 2(m-1)(2 m-1)+4 m n-n \\
& -(4 m-1) \cdot 2(m-1) \\
= & 4 m n-n-4 m(m-1),
\end{aligned}
$$

where $m=\left[\frac{n+1}{2}\right]$.