Three, the sequence $\left\{a_{n}\right\}$ satisfies
$$
(n-1) a_{n+1}=(n+1) a_{n}-2(n-1) \text {, }
$$
$n=1,2, \cdots$, and $a_{100}=10098$. Find the general term formula of the sequence $\left\{a_{n}\right\}$.
Three, Solution 1: The given equation can be transformed into
$$
(n-1)\left(a_{n+1}-2 n\right)=(n+1)\left[a_{n}-2(n-1)\right] \text {. }
$$

Let $b_{n}=a_{n}-2(n-1)$, then we have
$(n-1) b_{n+1}=(n+1) b_{n}$.
When $n \geqslant 2$, $b_{n+1}=\frac{n+1}{n-1} \cdot b_{n}$;
When $n \geqslant 3$, $b_{n}=\frac{n}{n-2} \cdot b_{n-1}=\frac{n}{n-2} \cdot \frac{n-1}{n-3} \cdot b_{n-2}$
$$
=\cdots=\frac{n(n-1) \cdots 3}{(n-2)(n-3) \cdots 1} \cdot b_{2}=\frac{n(n-1)}{2} \cdot b_{2} \text {. }
$$

Since $b_{2}=a_{2}-2, b_{n}=a_{n}-2(n-1)$, then
$$
\begin{array}{l}
a_{n}=2(n-1)+\frac{n(n-1)}{2} \cdot\left(a_{2}-2\right) \\
=(n-1)\left(\frac{n}{2} \cdot a_{2}-n+2\right) .
\end{array}
$$

Given $a_{100}=10098$, then
$$
99\left(50 a_{2}-100+2\right)=10098 \text {, }
$$

Solving for $a_{2}=4$.
Thus, $a_{n}=(n-1)(n+2)(n \geqslant 3)$.
Substituting $n=1$ into the given equation, we find $a_{1}=0$.
Also, $a_{2}=4$, so, $a_{n}=(n-1)(n+2)$ holds for all positive integers $n$.
Therefore, the general term formula for $\left\{a_{n} \mid\right.$ is $a_{n}=(n-1)(n+2)$.
Solution 2: Let $c_{n}=\frac{a_{n}}{n-1}(n \geqslant 2)$, then
$$
(n-1) n c_{n+1}=(n+1)(n-1) c_{n}-2(n-1) \text {. }
$$

Thus, $n c_{n+1}=(n+1) c_{n}-2$,
$$
\begin{array}{l}
n\left(c_{n+1}-2\right)=(n+1)\left(c_{n}-2\right) . \\
\text { Hence } \frac{c_{n+1}-2}{n+1}=\frac{c_{n}-2}{n} .
\end{array}
$$

Therefore, when $n \geqslant 3$,
$$
\begin{array}{l}
\frac{c_{n}-2}{n}=\frac{c_{n-1}-2}{n-1}=\cdots=\frac{c_{2}-2}{2}, \\
c_{n}=\frac{n}{2} \cdot c_{2}-n+2, \\
a_{n}=(n-1) c_{n}=(n-1)\left(\frac{n}{2} \cdot c_{2}-n+2\right) .
\end{array}
$$

The rest follows the same as Solution 1.