Example 15 Find the general term formula of the Fibonacci sequence $a_{0}=1, a_{1}=1, a_{n+2}=a_{n+1}+a_{n}(n \geqslant 1)$.
Let $f(x)=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{n+2} x^{n+2}+\cdots$
Then
$$
\begin{aligned}
-x f(x) & =-a_{0} x-a_{1} x^{2}-\cdots-a_{n+1} x^{n+2}-\cdots \\
-x^{2} f(x) & =-a_{0} x^{2}-\cdots-a_{n} x^{n+2}-\cdots
\end{aligned}
$$

Adding the above three equations and using $a_{n+2}-a_{n+1}-a_{n}=0, a_{0}=a_{1}=1$ we get
$$
\left(1-x-x^{2}\right) f(x)=1 \text {, }
$$

which means $f(x)=\frac{1}{1-x-x^{2}}=\frac{1}{(1-\alpha x)(1-\beta x)}$,
where $\alpha=\frac{1+\sqrt{5}}{2}, \beta=\frac{1-\sqrt{5}}{2}$. Thus,
$$
\begin{aligned}
f(x) & =\frac{1}{\alpha-\beta}\left(\frac{\alpha}{1-\alpha x}-\frac{\beta}{1-\beta x}\right)=\frac{1}{\alpha-\beta}\left(\alpha \sum_{n=0}^{\infty} \alpha^{n} x^{n}-\beta \sum_{n=0}^{\infty} \beta^{n} x^{n}\right) \\
& =\frac{1}{\alpha-\beta} \sum_{n=0}^{\infty}\left(\alpha^{n+1} \beta^{n+1}\right) x^{n},
\end{aligned}
$$

Therefore, $a_{n}=\frac{1}{\alpha-\beta}\left(\alpha^{n+1} \beta^{n+1}\right)=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right]$.