Example 4.6 Let $a_{0}=3, a_{1}=9, a_{n}=4 a_{n-1}-3 a_{n-2}-4 n+2 \quad(n \geqslant$ 2 ), find the general term formula of the sequence $\left\{a_{n}\right\}_{n \geqslant 0}$.
Let $A(t)=\sum_{n=0}^{\infty} a_{n} t^{n}$, then
$$
\begin{aligned}
A(t)= & \sum_{n=2}^{\infty} a_{n} t^{n}+a_{1} t+a_{0} \\
= & \sum_{n=2}^{\infty}\left(4 a_{n-1}-3 a_{n-2}-4 n+2\right) t^{n}+9 t+3 \\
= & 4 t \sum_{n=2}^{\infty} a_{n-1} t^{n-1}-3 t^{2} \sum_{n=2}^{\infty} a_{n-2} t^{n-2}- \\
& 4 \sum_{n=2}^{\infty}(n+1) t^{n}+6 \sum_{n=2}^{\infty} t^{n}+9 t+3 \\
= & 4 t \sum_{n=1}^{\infty} a_{n} t^{n}-3 t^{2} \sum_{n=0}^{\infty} a_{n} t^{n}- \\
& 4 \sum_{n=0}^{\infty}(n+1) t^{n}+6 \sum_{n=0}^{\infty} t^{n}+11 t+1 \\
= & 4 t[A(t)-3]-3 t^{2} \cdot A(t)- \\
& \frac{4}{(1-t)^{2}}+\frac{6}{1-t}+11 t+1 \\
= & \left(4 t-3 t^{2}\right) A(t)-\frac{4}{(1-t)^{2}}+\frac{6}{1-t}+1-t,
\end{aligned}
$$

So
$$
\begin{array}{l}
\left(1-4 t+3 t^{2}\right) \cdot A(t)=\frac{2-6 t+(1-t)^{3}}{(1-t)^{2}}, \\
A(t)=\frac{2-6 t+(1-t)^{3}}{(1-t)^{3}(1-3 t)} \\
=\frac{2}{(1-t)^{3}}+\frac{1}{1-3 t} \\
=2 \sum_{n=0}^{\infty}\binom{n+2}{2} t^{n}+\sum_{n=0}^{\infty} 3^{n} t^{n} \\
=\sum_{n=0}^{\infty}\left(n^{2}+3 n+2+3^{n}\right) t^{n}, \\
a_{n}=3^{n}+n^{2}+3 n+2. \\
\end{array}
$$