Example 2 Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=\frac{1}{2}, a_{n+1}=\left\{\begin{array}{ll}2 a_{n}+2 n-2, & n \text { is odd; } \\ -a_{n}-n, & n\end{array}\right.$ is even. Let the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$ be $S_{n}, b_{n}=a_{2 n}$, where, $n \in \mathbf{Z}_{+}$.
(1) Find the value of $a_{2}+a_{3}$;
(2) Prove: The sequence $\left\{b_{n}\right\}$ is a geometric sequence;
(3) For all $n \in \mathbf{Z}_{+}$, compare $S_{2 n+1}-\frac{41}{2}$ with $b_{2 n}$.
(1) Solving easily, we get $a_{2}=1, a_{3}=-3$.

Therefore, $a_{2}+a_{3}=-2$.
(2) Proof Note that,
$$
\begin{array}{l}
b_{n+1}=a_{2 n+2}=2 a_{2 n+1}+4 n \\
=2\left(-a_{2 n}-2 n\right)+4 n=-2 a_{2 n}=-2 b_{n}, \\
b_{1}=a_{2}=2 a_{1}=1 .
\end{array}
$$

Hence, the sequence $\left\{b_{n}\right\}$ is a geometric sequence with the first term 1 and common ratio -2.
(3) Solution From (2), we know
$$
b_{n}=(-2)^{n-1} \Rightarrow b_{2 n}=(-2)^{2 n-1}=-2^{2 n-1} \text {. }
$$

Let $c_{n}=a_{2 n}+a_{2 n+1}\left(n \in \mathbf{Z}_{+}\right)$. Then
$$
\begin{array}{l}
c_{n}=-2 n . \\
\text { Also, } S_{2 n+1}=a_{1}+\sum_{i=1}^{n}\left(a_{2 i}+a_{2 i+1}\right)=a_{1}+\sum_{i=1}^{n} c_{i} \\
=-n^{2}-n+\frac{1}{2},
\end{array}
$$

Then, from $S_{2 n+1}-\frac{41}{2}=b_{2 n}$, we get
$$
2 n^{2}+2 n+40=4^{n} \text {. }
$$

Construct the sequence $d_{n}=\frac{2 n^{2}+2 n+40}{4^{n}}$.
Thus, $d_{n+1}-d_{n}=\frac{2}{4^{n+1}}\left(-3 n^{2}-n-58\right)<0$
$$
\Rightarrow d_{n+1}<d_{n}<\cdots<d_{3}=1<d_{2}<d_{1} \text {. }
$$
$\Rightarrow$ There exists a unique $n=3$, such that $S_{2 n+1}-\frac{41}{2}=b_{2 n}$ holds.

[Note] In the summation process of $S_{2 n+1}$, combining terms in pairs, the handling is quite flexible; constructing a new sequence and studying the monotonicity of the sequence to demonstrate the uniqueness of the solution of the equation is truly ingenious.