Example 3.4 Try to find a sequence $\{f(n)\}_{n \geqslant 0}$, whose first 5 terms are $1,3,7$, 13,21, and whose general term $f(n)$ is a polynomial in $n$ of the lowest degree.
Solution: Let the sequence $\{f(n)\}_{n \geqslant 0}$ be the one we are looking for, then its difference table is

By Theorem 3.6, $\Delta^{k} f(n) \equiv 0(k=3,4,5, \cdots)$. By Newton's formula
$$
\begin{array}{l}
\quad f(n)=E^{n} f(0)=\sum_{i=0}^{n}\binom{n}{i} \Delta^{i} f(0) \\
=\sum_{i=0}^{2}\binom{n}{i} \Delta^{i} f(0) \\
=f(0)+n \cdot \Delta f(0)+\binom{n}{2} \Delta^{2} f(0) \\
=1+2 n+n(n-1)=n^{2}+n+1 .
\end{array}
$$