3.1.11 ** Let $\left\{u_{n}\right\}$ be a sequence with the sum of the first $n$ terms denoted as $S_{n}$. If $S_{1}=1$, and $S_{n+1}-S_{n}+(5 n+$ 1) $a^{n}, n=1,2,3, \cdots(|a| \neq 1)$. Find $S_{n}$.
From the conditions, we have $a_{n+1}=S_{n+1}-S_{n}=(5 n+1) a^{n}, n=1,2,3, \cdots$. Also, $a_{1}=$ $S_{1}=1$, so
$$
\begin{array}{l}
S_{n}=1+6 a+11 a^{2}+\cdots+(5 n-4) a^{n-1}, \\
a S_{n}=a+6 a^{2}+\cdots+(5 n-9) a^{n-1}+(5 n-4) a^{n} .
\end{array}
$$
(1) - (2) gives
$$
\begin{aligned}
(1-a) S_{n} & =1+5 a+5 a^{2}+\cdots+5 a^{n-1}-(5 n-4) a^{n} \\
& =5+5 a+5 a^{2}+\cdots+5 a^{n-1}-4-(5 n-4) a^{n} \\
& =\frac{5\left(1-a^{n}\right)}{1-a}-4-(5 n-4) a^{n}
\end{aligned}
$$

Since $|a| \neq 1$, we have $S_{n}=\frac{5\left(1-a^{n}\right)}{(1-a)^{2}}-\frac{4+(5 n-4) a^{n}}{1-a}$.