Approximately 2 Giant Knowledge $a_{n+1}=\frac{q-a_{n} a_{n-1}}{2 p-a_{n}-a_{n-1}}(n=$ $1,2, \cdots)$, where $p, q \in \mathbf{R}, a_{0}=a_{1}=0$ and $p^{2} \neq$ $q$. Find $a_{n}$.
Given $\alpha=\frac{q-\alpha^{2}}{2 p-2 \alpha}$, we have
$$
\alpha^{2}-2 p \alpha+q=0 \text {. }
$$

When $p^{2} \neq q$, the roots are
$$
\begin{array}{l} 
\alpha_{1}=p+\sqrt{p^{2}-q}, \alpha_{2}=p-\sqrt{p^{2}-q} . \\
\text { Also, } \quad a_{n+1}-\alpha_{1} \\
= \frac{q-a_{n} a_{n-1}-2 p a_{1}+\alpha_{1} a_{n}+\alpha_{1} a_{n-1}}{2 p-a_{n}-a_{n-1}} .
\end{array}
$$

Using $q-2 p \alpha_{1}=-\alpha_{1}^{2}$, the above expression can be simplified to
$$
a_{n+1}-\alpha_{1}=\frac{\left(a_{n}-\alpha_{1}\right)\left(a_{n-1}-\alpha_{1}\right)}{2 p-a_{n}-a_{n-1}} \text {. }
$$

Similarly,
$$
\begin{aligned}
a_{n+1}-\alpha_{2}= & \frac{\left(a_{n}-\alpha_{2}\right)\left(a_{n-1}-\alpha_{2}\right)}{2 p-a_{n}-a_{n-1}} . \\
\therefore & \frac{a_{n+1}-\alpha_{1}}{a_{n+1}-\alpha_{2}}=\left(\frac{a_{n}-\alpha_{1}}{a_{n}-\alpha_{2}}\right)\left(\frac{a_{n-1}-\alpha_{1}}{a_{n-1}-\alpha_{2}}\right) . \\
\because & \frac{a_{0}-\alpha_{1}}{a_{0}-\alpha_{2}}=\frac{a_{1}-\alpha_{1}}{a_{1}-\alpha_{2}}=\frac{\alpha_{1}}{\alpha_{2}}, \\
\therefore & \frac{a_{2}-\alpha_{1}}{a_{2}-\alpha_{2}}=\left(\frac{\alpha_{1}}{\alpha_{2}}\right)^{2}, \\
& \frac{a_{3}-\alpha_{1}}{a_{3}-\alpha_{2}}=\left(\frac{\alpha_{1}}{\alpha_{2}}\right)^{3}, \\
& \cdots \cdots, \\
& \frac{a_{n}-\alpha_{1}}{a_{n}-\alpha_{2}}=\left(\frac{\alpha_{1}}{\alpha_{2}}\right)^{f_{n}} .
\end{aligned}
$$

Solving this, we get
$$
a_{n}=\frac{\alpha_{1} \cdot \alpha_{z^{n}}^{f_{n}}-\alpha_{2} \cdot \alpha_{1_{n}}^{f_{n}}}{\alpha_{z^{n}}^{f}-\alpha_{1_{n}^{n}}^{f_{n}}} .
$$

where $\left\{f_{n}\right\}$ is the Fibonacci sequence: $f_{k+1}=f_{k}+f_{k-1}$, $k \in \mathbf{N}$.