Five. (20 points) Given $a_{1}=1, a_{n+1}=2 a_{n}+(-1)^{n} n+1(n=1,2,3, \cdots)$. Find the general term formula $a_{n}$ (require it to be written as a finite sum, and the number of terms should be independent of $n$).


Let $a_{n}=2^{n}-1+b_{n}$, then $b_{1}=0$, and
$$
2^{n+1}-1+b_{n+1}=2^{n+1}-2+2 b_{n}+(-1)^{n} n+1 \text {, }
$$

i.e., $b_{1}=0, b_{n+1}=2 b_{n}+(-1)^{n} n$.
Let $b_{n}=(-1)^{n} c_{n}$, then
$$
c_{1}=0, c_{n+1}=-2 c_{n}-n \text {. }
$$

Adding $s(n+1)+t$ to both sides of the above equation, we get
$$
\begin{array}{l}
c_{n+1}+s(n+1)+t \\
=(-2)\left(c_{n}+\frac{1-s}{2} \cdot n-\frac{s+t}{2}\right) . \\
\text { Let } s=\frac{1-s}{2} \text { and } t=-\frac{s+t}{2},
\end{array}
$$

Solving, we get $s=\frac{1}{3}, t=-\frac{1}{9}$. Therefore,
$$
\begin{array}{l}
c_{n+1}+\frac{1}{3}(n+1)-\frac{1}{9} \\
=(-2)\left(c_{n}+\frac{1}{3} \times n-\frac{1}{9}\right)=\cdots \\
=(-2)^{n}\left(c_{1}+\frac{1}{3} \times 1-\frac{1}{9}\right)=(-2)^{n} \frac{2}{9},
\end{array}
$$

i.e., $c_{n}=-\frac{n}{3}+\frac{1}{9}+(-1)^{n+1} \frac{2^{n}}{9}$. 
Thus, $b_{n}=-\frac{n}{3}(-1)^{n}+\frac{1}{9}(-1)^{n}-\frac{2^{n}}{9}$. 
Hence, $a_{n}=2^{n}-1+b_{n}$
$$
=\frac{8}{9} \times 2^{n}+\left(\frac{1}{9}-\frac{n}{3}\right)(-1)^{n}-1 \text {. }
$$