Example 2. In the set of non-negative integers $\overline{\mathbf{Z}}-1$: Define the function $\mathrm{f}(\mathrm{n})$:
$$
\left\{\begin{array}{l}
f(0)=2, f(1)=3, \\
f(k+1)=3 f(k)-2 f(k-1) . \text { where, } k \geqslant 1 .
\end{array}\right.
$$

Try to find the formula for $f(n)$ expressed in terms of $n$.
Given $f(0)=2, f(1)=3$, by the recursive formula: $f(k+1)=3 f(k)-2 f(k-1)$, we get
$$
\begin{array}{l}
f(2)=3 f(1)-2 f(0)=5, \\
f(3)=3 f(2)-2 f(1)=9, \\
f(4)=3 f(3)-2 f(2)=17,
\end{array}
$$

Since $2=2^{4}+1,3=2^{1}+1,5=2^{2}+1,9=2^{3}+1,17=2^{4}+1$, by incomplete induction, we have
$$
f(n)=2^{n}+1.
$$

Next, we prove (1) using mathematical induction. When $n=0, n=1$, it has already been verified.

Assume that for $n \leqslant k$, (1) is correct, i.e., $f(k-1)=2^{k-1}+1, f(k)=2^{k}+1$, then when $n=k+1$, we have
$$
\begin{aligned}
f(k+1) & =3 f(k)-2 f(k-1) \\
& =3\left(2^{k}+1\right)-2\left(2^{k-1}+1\right) \\
& =3 \cdot 2^{k}+3-2 \cdot 2^{k-1}-2 \\
& =3 \cdot 2^{k}-2^{k}+1 \\
& =(2+1) 2^{k}-2^{k}+1 \\
& =2^{k+1}+1.
\end{aligned}
$$

Thus, (1) is also correct when $n=k+1$. Therefore, (1) is correct for any non-negative integer $n$.