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 $a_{2}=6$.
When $n \geq 2, n \in \mathrm{N}^{+}$, $\frac{a_{n+1}}{n+1}-\frac{a_{n}}{n-1}=-\frac{1}{n-1}$.
Further, $\frac{a_{n+1}}{(n+1) n}-\frac{a_{n}}{n(n-1)}=-\frac{1}{n(n-1)}=-\left(\frac{1}{n-1}-\frac{1}{n}\right)$
From the above, $\frac{a_{\text {net }}}{(n+1) n}-\frac{a_{2}}{2}=-\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{n-2}-\frac{1}{n-1}\right)+\left(\frac{1}{n-1}-\frac{1}{n}\right)\right]$ $a_{2}=$ some value noted
Given $a_{2}=6$, substitute and get $(1-1) a_{2}-(1+1) a_{1}=-(1+1) \Rightarrow a_{1}=1$ satisfies the equation.
Thus, for any $n \in \mathrm{N}^{*}, a_{n}=n(2 n-1)$
(II) Given $b_{n}=\frac{a_{0}}{n+c}=\frac{n(2 n-1)}{n+c}$ or $b_{1}=\frac{1}{1+c}, b_{2}=\frac{6}{2+c}, b_{3}=\frac{15}{3+c}$, and since the sequence $\left\{b_{n}\right\}$ is an arithmetic sequence, we have $2 b_{2}=b_{1}+b_{1}, \frac{12}{2+c}=\frac{1}{1+c}+\frac{15}{3+c}$, solving gives $c=-\frac{1}{2}$, thus $b_{n}=2 n$,
Further, $c_{n}=\frac{b_{n}}{2^{n}}=\frac{2 n}{2^{n}}=\frac{n}{2^{n-1}}$.
$S_{n}=1+\frac{2}{2^{1}}+\frac{3}{1} \times+\frac{2}{2^{n-2}}+\frac{n}{2^{n-1}}$
(1) $\left.2 S_{n}=2+\frac{2}{2^{6}}+\frac{3}{2^{1}}+\frac{4}{2^{2}}+\frac{3}{7}\right) \frac{n=1}{2^{2}-3}+\frac{1 n}{2^{n-2}}$
From the above, $S_{n}=3+\frac{1}{2}+\frac{1}{2^{2}}+\cdots+\frac{1}{2^{n-2}}-\frac{n}{2^{n-1}}=4-\frac{n+?}{2^{n}}$ ? $\qquad$ 14 points