Example 17 (2004 National College Entrance Examination, Science Question 22) Given that the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$, $S_{n}$, satisfies:
$$
S_{n}=2 a_{n}+(-1)^{n}(n \geqslant 1) \text {. }
$$
(1) Write down the first three terms of the sequence $\left\{a_{n}\right\}$, $a_{1}, a_{2}, a_{3}$;
(2) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(3) Prove that for any integer $n>4$, $\frac{1}{a_{4}}+\frac{1}{a_{5}}+\cdots+\frac{1}{a_{m}}<\frac{7}{8}$.
The solutions for (1) and (3) are omitted. Here we only provide the solution for (2).

In the standard answer of the college entrance examination, the term $a_{n}=\frac{2}{3}\left[2^{n-2}+(-1)^{n-1}\right]\left(n \in \mathbf{N}_{+}\right)$ is obtained using the "recurrence" method. Below, we use the "superposition" method to find $a_{n}$.
Solution: First, from $a_{k}=\left\{\begin{array}{l}S_{1}, \text { when } k=1, \\ S_{k}-S_{k-1}, \text { when } k \geqslant 2,\end{array}\right.$
we can get $a_{k}=2 a_{k-1}+2(-1)^{k-1}$, which means $a_{k}-2 a_{k-1}=2(-1)^{k-1}(k \geqslant 2)$. (*)
In equation (*), let $k=2,3, \cdots, n$, then we have
$$
\begin{array}{l}
2^{n-2} a_{2}-2^{n-1} a_{1}=2^{n-1}(-1), \\
2^{n-3} a_{3}-2^{n-2} a_{2}=2^{n-2}(-1)^{2}, \\
\cdots \cdots \\
2 a_{n-1}-2^{2} a_{n-2}=2^{2}(-1)^{n-2}, \\
a_{n}-2 a_{n-1}=2(-1)^{n-1} .
\end{array}
$$
$(n-1)$
By adding both sides of the above $n-1$ inequalities, we get
$$
\begin{array}{l}
a_{n}-2^{n-1} a_{1}=2^{n-1}(-1)+2^{n-2}(-1)^{2}+\cdots+2(-1)^{n-1}, \\
\therefore a_{n}=2^{n-1}-(-1)^{n} \frac{2}{3}\left[1-(-2)^{n-1}\right]=\frac{2}{3}\left[2^{n-2}+(-1)^{n-1}\right] .
\end{array}
$$

Upon verification, when $n=1$, $a_{1}=1$ satisfies the above formula, hence
$$
a_{n}=\frac{2}{3}\left[2^{n-2}+(-1)^{n-1}\right], n \in \mathbf{N}^{*} \text {. }
$$