Example 3.1 Let $f(n)=3^{n}$, find $\Delta^{k} f(n) \quad(k \geqslant 1)$.
Solution one:
$$
\begin{array}{l}
\Delta f(n)=f(n+1)-f(n) \\
=3^{n+1}-3^{n}=2 \cdot 3^{n} \text {. } \\
\Delta^{2} f(n)=\Delta f(n+1)-\Delta f(n) \\
=2 \cdot 3^{n+1}-2 \cdot 3^{n}=2^{2} \cdot 3^{n} \text {. } \\
\end{array}
$$

Let $\Delta^{s} f(n)=2^{s} 3^{n}$, then
$$
\begin{aligned}
\Delta^{s+1} f(n) & =\Delta\left(\Delta^{s} f(n)\right) \\
& =\Delta^{s} f(n+1)-\Delta^{s} f(n) \\
& =2^{s} \cdot 3^{n+1}-2^{s} \cdot 3^{n}=2^{s+1} \cdot 3^{n}
\end{aligned}
$$

By mathematical induction, we know:
$$
\begin{array}{l} 
\Delta^{k} f(n)=2^{k} 3^{n} \quad(k \geqslant 1) . \\
\Delta^{k} f(n)=\Delta^{k} 3^{n}=(E-I)^{k} 3^{n} \\
=\sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j} E^{j} 3^{n} \\
=\sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j} 3^{n+j} \\
= 3^{n} \sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j} 3^{j} \\
= 3^{n}(3-1)^{k}=2^{k} \cdot 3^{n} .
\end{array}
$$