(9) (14 points) Given the sequence $\left\{a_{n}\right\}$ satisfies $\frac{a_{n+1}+a_{n}-1}{a_{n+1}-a_{n}+1}=n\left(n \in \mathbf{N}^{*}\right)$, and $a_{2}=6$.
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Let $b_{n}=\frac{a_{n}}{n+c}\left(n \in \mathbf{N}^{*}\right)$, where $c$ is a non-zero constant. If the sequence $\left\{b_{n}\right\}$ is an arithmetic sequence, let $c_{n}=\frac{b_{n}}{2^{n}}$, and $S_{n}=c_{1}+c_{2}+\cdots+c_{n}$, find $S_{n}$.
9 (1) From the given,
$$
\begin{array}{l}
\frac{a_{n+1}+a_{n}-1}{a_{n+1}-a_{n}+1}=n\left(n \in \mathbf{N}^{*}\right) \\
\Rightarrow(n-1) a_{n+1}-(n+1) a_{n}=-(n+1) .
\end{array}
$$

When $n \geqslant 2, n \in \mathbf{N}^{*}$,
$$
\frac{a_{n+1}}{n+1}-\frac{a_{n}}{n-1}=-\frac{1}{n-1},
$$

Further, we get
$$
\frac{a_{n+1}}{(n+1) n}-\frac{a_{n}}{n(n-1)}=-\frac{1}{n(n-1)}=-\left(\frac{1}{n-1}-\frac{1}{n}\right) .
$$

Thus,
$$
\begin{aligned}
& \frac{a_{n+1}}{(n+1) n}-\frac{a_{2}}{2} \\
= & -\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{n-2}-\frac{1}{n-1}\right)+\left(\frac{1}{n-1}-\frac{1}{n}\right)\right] \\
= & \frac{1}{n}-1 .
\end{aligned}
$$

Given $a_{2}=6$, when $n \geqslant 2$, $a_{n+1}=(n+1)(2 n+1)$, i.e., when $n \geqslant 3$, $a_{n}=n(2 n-1) \cdot a_{2}=6$ fits the above formula.

Substituting $a_{2}=6$ and $(1-1) a_{2}-(1+1) a_{1}=-(1+1) \Rightarrow a_{1}=1$, which also fits the above formula.
Therefore, for any $n \in \mathbf{N}^{*}, a_{n}=n(2 n-1)$.
(2) From $b_{n}=\frac{a_{n}}{n+c}=\frac{n(2 n-1)}{n+c}$, we know $b_{1}=\frac{1}{1+c}, b_{2}=\frac{6}{2+c}, b_{3}=$ $\frac{15}{3+c}$. Since the sequence $\left\{b_{n}\right\}$ is an arithmetic sequence, we have $2 b_{2}=b_{1}+b_{3}, \frac{12}{2+c}=$ $\frac{1}{1+c}+\frac{15}{3+c}$, which gives $c=-\frac{1}{2}$. Thus, $b_{n}=2 n$, and further,
$$
\begin{array}{c}
c_{n}=\frac{b_{n}}{2^{n}}=\frac{2 n}{2^{n}}=\frac{n}{2^{n-1}}, \\
S_{n}=1+\frac{2}{2^{1}}+\frac{3}{2^{2}}+\cdots+\frac{n-1}{2^{n-2}}+\frac{n}{2^{n-1}}, \\
2 S_{n}=2+\frac{2}{2^{0}}+\frac{3}{2^{1}}+\frac{4}{2^{2}}+\cdots+\frac{n-1}{2^{n-3}}+\frac{n}{2^{n-2}} .
\end{array}
$$

Subtracting the two equations, we get
$$
S_{n}=3+\frac{1}{2}+\frac{1}{2^{2}}+\cdots+\frac{1}{2^{n-2}}-\frac{n}{2^{n-1}}=4-\frac{n+2}{2^{n-1}} .
$$