Example 4. Let $a_{\mathrm{n}}=n^{2}+2 n-3$ : Find $S_{\mathrm{s}}$.

Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
Solve
$$
\left\{\begin{array}{l}
3 b=1 \\
-3 a+2 b=2 \\
a-b+c=-3
\end{array}\right.
$$

to get
$$
\left\{\begin{array}{l}
a=\frac{1}{3} \\
b=\frac{3}{2} \\
c=-\frac{11}{6}
\end{array}\right.
$$

Thus,
$$
S_{n}=\frac{1}{3} n^{3}+\frac{3}{2} n^{2}-\frac{11}{6} n .
$$

In general, if
$$
a_{\mathrm{n}}=b_{1} n^{\mathrm{p}}+b_{2} n^{\mathrm{p}-1}+\cdots+b_{\mathrm{p}}
$$

is a polynomial of degree $p$ in $n$, then $S_{n}$ is a polynomial of degree $p+1$ in $n$:
$$
S_{n}=c_{1} n^{\mathrm{P}+1}+c_{2} n^{\mathrm{D}}+\cdots+c_{\mathrm{p}} n .
$$
Hence, we have
$$
\left\{\begin{array}{c}
C_{\mathrm{p}+1}^{1} c_{1}=b_{1}, \\
-C_{\mathrm{p}+1}^{2} c_{1}+C_{\mathrm{p}}^{1} c_{2}=b_{2}, \\
C_{\mathrm{p}+1}^{\mathrm{s}} c_{1}-C_{\mathrm{p}}^{2} c_{2}+C_{\mathrm{p}-1}^{1} c_{3}=b_{3}, \\
\cdots \cdots \\
(-1)^{p} c_{1}+(-1)^{\mathrm{D}-1} c_{2}+\cdots+c_{\mathrm{D}-2} \\
-c_{p-1}+c_{p}=b_{p} .
\end{array}\right.
$$

Solving this system of equations yields $c_{1}, c_{2}, \cdots, c_{p}$, which allows us to find the expression for $S_{\mathbf{n}}$.