1. (16 points) Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{n+1}+(-1)^{n} a_{n}=2 n-1$.
Find the sum of the first $4 n$ terms of the sequence $\left\{a_{n}\right\}$, $S_{4 n}$.
Given the problem, we have:

When $n=2 k-1, 2 k, 2 k+1$, respectively,
$$
\begin{array}{l}
a_{2 k}-a_{2 k-1}=4 k-3, \\
a_{2 k+1}+a_{2 k}=4 k-1, \\
a_{2 k+2}-a_{2 k+1}=4 k+1 .
\end{array}
$$
Subtracting (1) from (2) and adding (2) and (3) respectively, we get
$$
a_{2 k-1}+a_{2 k+1}=2, \quad a_{2 k}+a_{2 k+2}=8 k \text {. }
$$

Adding the above two equations, we get
$$
a_{2 k-1}+a_{2 k}+a_{2 k+1}+a_{2 k+2}=8 k+2 \text {. }
$$

Therefore, $S_{4 n}=\sum_{i=1}^{4 n} a_{i}$
$$
=\sum_{k=1}^{2 n-1}\left(a_{2 k-1}+a_{2 k}+a_{2 k+1}+a_{2 k+2}\right) \text {. }
$$

Let $b_{k}=a_{2 k-1}+a_{2 k}+a_{2 k+1}+a_{2 k+2}$. Then
$$
S_{4 n}=b_{1}+b_{3}+\cdots+b_{2 n-1} \text {. }
$$

Since $\left\{b_{2 k-1}\right\}$ is an arithmetic sequence with the first term $b_{1}=10$ and a common difference of 16, we have
$$
S_{4 n}=82+2 n
$$