15. Let the sequence $\left\{a_{n}\right\}$ have the first term $a_{1}$ (a constant), and $a_{n+1}=a a_{n}+\beta_{2} n^{2}+\beta_{1} n+\beta_{0}\left(n \in \mathbf{N}^{*}\right) \cdots(1)$, where $\alpha \neq 1, \alpha \beta_{2} \neq 0$, find $a_{n}$ and $S_{n}$.
15. Solution: Suppose the recursive formula of the sequence can be written as
$$
a_{n+1}-\left[x_{1}(n+1)^{2}+x_{2}(n+1)+x_{3}\right]=\alpha\left[a_{n}-\left(x_{1} n^{2}+x_{2} n+x_{3}\right)\right] \text {. }
$$

To find $x_{1}, x_{2}, x_{3}$, rearrange it into the same form as the original recursive formula:
$$
a_{n+1}=\alpha a_{n}+\left(x_{1}-\alpha x_{1}\right) n^{2}+\left(2 x_{1}+x_{2}-\alpha x_{2}\right) n+\left(x_{1}+x_{2}+x_{3}-\alpha x_{3}\right) \cdots \text { (2) }
$$

Comparing it with the original recursive formula, we get $\left\{\begin{array}{l}x_{1}-\alpha x_{1}=\beta_{2} \\ 2 x_{1}+(1-\alpha) x_{2}=\beta_{1} \\ x_{1}+x_{2}+(1-\alpha) x_{3}=\beta_{0}\end{array}\right.$.
Since $\alpha \neq 1$, this system of equations has a unique solution.
Therefore, when $x_{1}, x_{2}, x_{3}$ are the solutions to the above system of equations, the recursive formula (1) can be written as (2).
Let $b_{n}=a_{n}-\left(x_{1} n^{2}+x_{2} n+x_{3}\right)$, then $b_{1}=a_{1}-\left(x_{1}+x_{2}+x_{3}\right)$, and $b_{n+1}=\alpha b_{n}\left(n \in \mathbf{N}^{*}\right)$.
Thus, the sequence $\left\{b_{n}\right\}$ is a geometric sequence with common ratio $\alpha$ and first term $b_{1}$.
Hence, $b_{n}=b_{1} \alpha^{n-1}=\left[a_{1}-\left(x_{1}+x_{2}+x_{3}\right)\right] \alpha^{n-1}$.
So, $a_{n}-\left(x_{1} n^{2}+x_{2} n+x_{3}\right)=\left[a_{1}-\left(x_{1}+x_{2}+x_{3}\right)\right] \cdot \alpha^{n-1}$.
Therefore, the general term formula of the sequence $\left\{a_{n}\right\}$ is $a_{n}=\left[a_{1}-\left(x_{1}+x_{2}+x_{3}\right)\right] \cdot \alpha^{n-1}+x_{1} n^{2}+x_{2} n+x_{3}\left(n \in \mathbf{N}^{*}\right)$.
Thus, the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$ is
$$
\begin{array}{l}
S_{n}=\sum_{k=1}^{n} a_{k}=\left[a_{1}-\left(x_{1}+x_{2}+x_{3}\right)\right] \sum_{k=1}^{n} \alpha^{k-1}+x_{1} \sum_{k=1}^{n} k^{2}+x_{2} \sum_{k=1}^{n} k+x_{3} n \\
=\left[a_{1}-\left(x_{1}+x_{2}+x_{3}\right)\right] \frac{1-\alpha^{n}}{1-\alpha}+\frac{x_{1} n(n+1)(2 n+1)}{6}+\frac{x_{2} n(n+1)}{2}+n x_{3}\left(n \in \mathbf{N}^{*}\right) .
\end{array}
$$