Think about Question 2 Given $x_{1}=1, x_{2}=6$, and $x_{n+1}=6 x_{n}-9 x_{n-1}+3^{n}(n=2,3, \cdots)$, find the general term formula of the sequence $\left\{x_{n}\right\}$.
Given the recurrence relation, we have $x_{n+1}-3 x_{n}=3\left(x_{n}-3 x_{n-1}\right)+3^{n}$. Dividing both sides by $3^{n}$, we get $\frac{x_{n+1}-3 x_{n}}{3^{n}}=\frac{x_{n}-3 x_{n-1}}{3^{n-1}}+1$. Therefore, the sequence $\left\{\frac{x_{n+1}-3 x_{n}}{3^{n}}\right\}$ is an arithmetic sequence with the first term 1 and common difference 1. So, $\frac{x_{n+1}-3 x_{n}}{3^{n}}=1+(n-1)=n$, which means
$$
\frac{x_{n}}{3^{n-1}}=\frac{x_{1}}{3^{0}}+\sum_{k=1}^{n-1}\left(\frac{x_{k+1}}{3^{k}}-\frac{x_{k}}{3^{k-1}}\right)=1+\sum_{k=1}^{n-1} k=1+\frac{n(n-1)}{2}, x_{n}=\frac{3^{n-1}}{2}\left(n^{2}-n+2\right) .
$$