10. (15 points) In the sequence $\left\{a_{n}\right\}$, $a_{n}=n^{3} \cdot \prod_{i=1}^{99}\left(n^{2} - i^{2}\right)^{2}$, find the sum of the first $n$ terms $S_{n}$.


10. Solution: Since
$$
\begin{array}{l}
a_{n}=n^{3} \cdot \prod_{i=1}^{99}\left(n^{2}-i^{2}\right)^{2}=(n+99)^{2}(n+98)^{2} \cdots \\
(n+1)^{2} n^{2}(n-1)^{2} \cdots(n-98)^{2}(n-99)^{2}=\frac{1}{400}[(n+ \\
100)^{2}-(n-100)^{2}] \cdot(n+99)^{2} \cdot(n+98)^{2} \cdots(n- \\
99)^{2}, \\
\quad \text { so } S_{n}-S_{n-1}=\frac{1}{400}(n+100)^{2} \cdot(n+99)^{2} \cdot \\
(n+98)^{2} \cdots(n-99)^{2}, \\
\frac{1}{400}(n+99)^{2} \cdot(n+98)^{2} \cdots(n-99)^{2}(n- \\
100)^{2} .
\end{array}
$$

So $S_{n}-S_{n-1}=\frac{1}{400}(n+100)^{2} \cdot(n+99)^{2}$.
$$
\begin{array}{l}
(n+98)^{2} \cdots(n-99)^{2}, \\
\quad \frac{1}{400}(n+99)^{2} \cdot(n+98)^{2} \cdots(n-99)^{2}(n- \\
100)^{2} \text {. }
\end{array}
$$

Thus, we have $S_{n}-\frac{1}{400}(n+100)^{2} \cdot(n+99)^{2} \cdot(n+$
$$
\begin{array}{l}
98)^{2} \cdots(n-99)^{2}=S_{n-1}-\frac{1}{400}(n+99)^{2} \cdot(n+98)^{2} \\
\cdots(n-99)^{2}(n-100)^{2},
\end{array}
$$

Therefore, the new sequence
$$
\left\{S_{n}-\frac{1}{400}(n+100)^{2} \cdot(n+99)^{2} \cdot(n+98)^{2} \cdots(n-99)^{2}\right\}
$$

is a constant sequence, with the first term being $S_{1}-\frac{1}{400} \times 0=0$. So $S_{n}=$
$$
\frac{1}{400}(n+100)^{2} \cdot(n+99)^{2} \cdot(n+98)^{2} \cdots(n-99)^{2} \text {, }
$$

which means $S_{n}=\frac{n^{2}}{400}(n+100)^{2} \cdot \prod_{i=1}^{99}\left(n^{2}-i^{2}\right)^{2}$.