Example 5. Let $a_{n}=5 n^{4}+2 n^{3}+4 n^{2}+3 n-3$, find $S_{\mathrm{n}}$.
$$
\begin{array}{l}
\text { Let } S_{n}=c_{1} n^{5}+c_{2} n^{4}+c_{3} n^{3}+c_{4} n^{2} \\
\text { + } c_{5} n \text {. } \\
\end{array}
$$

Then
$$
\left\{\begin{array}{c}
C_{5}^{1} c_{1}=5, \\
-C_{5}^{2} c_{1}+C{ }_{4}^{1} c_{2}=2, \\
C_{6}^{3} c_{1}-C_{4}^{2} c_{2}+C_{8}^{1} c_{3}=4, \\
-C_{5}^{4} c_{1}+C_{4}^{3} c_{2}-C_{3}^{2} c_{3}+C \frac{1}{2} c_{4}=3, \\
c_{1}-c_{2}+c_{3}-c_{4}+c_{3}=-3 .
\end{array}\right.
$$

Solving it, $c_{1}=1, c_{2}=3, c_{3}=4, c_{4}=4$,
$$
\begin{array}{l}
c_{5}=-1 \\
\therefore \quad S_{n}=n^{5}+3 n^{4}+4 n^{3}+4 n^{2}-n .
\end{array}
$$