10 Given the sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=\frac{1}{4}, a_{2}=\frac{3}{4}, a_{n+1}=2 a_{n}-a_{n-1}(n \geqslant$
2), the sequence $\left\{b_{n}\right\}$ satisfies: $b_{1} \neq \frac{1}{4}, 3 b_{n}-b_{n-1}=n(n \geqslant 2)$, and the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$ is $S_{n}$.
(1) Prove: The sequence $\left\{b_{n}-a_{n}\right\}$ is a geometric sequence;
(2) If $b_{1}=\frac{11}{12}$, find the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$, $S_{n}$.
10 (1) Since $a_{n+1}=2 a_{n}-a_{n-1}(n \geqslant 2)$, we have
$$
a_{n+1}-a_{n}=a_{n}-a_{n-1}(n \geqslant 2).
$$

Therefore, $\left\{a_{n}\right\}$ is an arithmetic sequence.
Given $a_{1}=\frac{1}{4}, a_{2}=\frac{3}{4}$, we have $a_{n}=\frac{2 n-1}{4}$.
From $3 b_{n}-b_{n-1}=n$, we get $b_{n}=\frac{b_{n-1}+n}{3}$. Thus,
$$
\begin{aligned}
b_{n}-a_{n} & =\frac{1}{3} b_{n-1}+\frac{n}{3}-\left(a_{n-1}+\frac{1}{2}\right) \\
& =\frac{1}{3}\left(b_{n-1}-a_{n-1}\right)-\frac{2}{3} a_{n-1}+\frac{2 n-3}{6} \\
& =\frac{1}{3}\left(b_{n-1}-a_{n-1}\right)-\frac{2}{3} \cdot \frac{2(n-1)-1}{4}+\frac{2 n-3}{6} \\
& =\frac{1}{3}\left(b_{n-1}-a_{n-1}\right).
\end{aligned}
$$

Since $b_{1}-a_{1}=b_{1}-\frac{1}{4} \neq 0$, the sequence $\left\{b_{n}-a_{n}\right\}$ is a geometric sequence.
(2) From (1), we have $b_{n}=\frac{2 n-1}{4}+\left(b_{1}-\frac{1}{4}\right) \cdot\left(\frac{1}{3}\right)^{n-1}$
$$
\begin{array}{l}
=\frac{2 n-1}{4}+\left(\frac{11}{12}-\frac{1}{4}\right) \cdot\left(\frac{1}{3}\right)^{n-1} \\
=\frac{2 n-1}{4}+2 \cdot\left(\frac{1}{3}\right)^{n},
\end{array}
$$

Therefore,
$$
S_{n}=\frac{1}{4} n^{2}+2 \cdot \frac{\frac{1}{3}\left(1-\left(\frac{1}{3}\right)^{n}\right)}{1-\frac{1}{3}}=\frac{1}{4} n^{2}-\left(\frac{1}{3}\right)^{n}+1
$$