9. (14 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\frac{a_{n+1}+a_{n}-1}{a_{n+1}-a_{n}+1}=n\left(n \in \mathbf{N}_{+}\right) \text {, and } a_{2}=6 \text {. }
$$
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$.
(2) Let $b_{n}=\frac{a_{n}}{n+c}\left(n \in \mathbf{N}_{+}, c\right.$ be a non-zero constant). If the sequence $\left\{b_{n}\right\}$ is an arithmetic sequence, denote
$$
c_{n}=\frac{b_{n}}{2^{n}}, S_{n}=c_{1}+c_{2}+\cdots+c_{n} \text {. }
$$

Find $S_{n}$.
(1) From the problem, we have
$$
\begin{array}{l}
\frac{a_{n+1}+a_{n}-1}{a_{n+1}-a_{n}+1}=n\left(n \in \mathbf{N}_{+}\right) \\
\Rightarrow(n-1) a_{n+1}-(n+1) a_{n}=-(n+1) .
\end{array}
$$

When $n \geqslant 2\left(n \in \mathbf{N}_{+}\right)$,
$$
\begin{array}{l}
\frac{a_{n+1}}{n+1}-\frac{a_{n}}{n-1}=-\frac{1}{n-1} \\
\Rightarrow \frac{a_{n+1}}{(n+1) n}-\frac{a_{n}}{n(n-1)}=-\left(\frac{1}{n-1}-\frac{1}{n}\right) \\
\Rightarrow \frac{a_{n+1}}{(n+1) n}-\frac{a_{2}}{2} \\
\quad=-\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{n-1}-\frac{1}{n}\right)\right] \\
\quad=\frac{1}{n}-1 .
\end{array}
$$

Given $a_{2}=6$, when $n \geqslant 2$,
$$
a_{n+1}=(n+1)(2 n+1) \text {, }
$$

which means when $n \geqslant 3$, $a_{n}=n(2 n-1), a_{2}=6$ fits the above formula.
Substituting $a_{2}=6$ into
$$
(1-1) a_{2}-(1+1) a_{1}=-(1+1)
$$
$\Rightarrow a_{1}=1$, which also fits the above formula.
Therefore, for any $n \in \mathbf{N}_{+}, a_{n}=n(2 n-1)$.
(2) Given $b_{n}=\frac{a_{n}}{n+c}=\frac{n(2 n-1)}{n+c}$, we know
$$
b_{1}=\frac{1}{1+c}, b_{2}=\frac{6}{2+c}, b_{3}=\frac{15}{3+c} .
$$

Since $\left\{b_{n}\right\}$ is an arithmetic sequence, we have
$$
\begin{array}{l}
2 b_{2}=b_{1}+b_{3} \Rightarrow \frac{12}{2+c}=\frac{1}{1+c}+\frac{15}{3+c} \\
\Rightarrow c=-\frac{1}{2} \Rightarrow b_{n}=2 n \\
\Rightarrow c_{n}=\frac{b_{n}}{2^{n}}=\frac{2 n}{2^{n}}=\frac{n}{2^{n-1}} .
\end{array}
$$

Thus, $S_{n}=1+\frac{2}{2^{1}}+\frac{3}{2^{2}}+\cdots+\frac{n}{2^{n-1}}$,
$$
2 S_{n}=2+\frac{2}{2^{0}}+\frac{3}{2^{1}}+\cdots+\frac{n}{2^{n-2}} \text {. }
$$

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