Question 21. Let the sequence $\left\{a_{n}\right\}$ be defined as
$$
a_{n+1}=2 a_{n}+2 a_{n-1} \quad(n=2,3,4, \cdots)
$$
(1) If the sequence $\left\{a_{n+1}-\alpha a_{n}\right\}$ is a geometric sequence with common ratio $\beta$ $(\alpha, \beta$ are real numbers), then $\alpha$, $\beta$ are the roots of the equation $x^{2}-2 x-2=0$.
(2) If $a_{1}=1, a_{2}=2$, try to find the general term formula $a_{n}$.
(1) From $\left\{a_{n}+1-\alpha a_{n}\right\}$ being a geometric sequence with common ratio $\beta$, we get
$$
a_{n+1}-\alpha a_{n}=\beta\left(a_{u}-\alpha: 1_{n-1}\right) .
$$

That is, $2 a_{n}+2 a_{n-1}-\alpha a_{n}=\beta a_{n}-\alpha \beta a_{n-1}$.
By comparing the coefficients on both sides of the equation, we have
$$
\alpha+\beta=2, \alpha \beta=-2 \text {. }
$$

Thus, $\alpha, \beta$ are the two roots of $x^{2}-2 x-2=0$.
(2) First, we prove the following proposition:
If $\left\{\mathrm{a}_{n}\right\}$ is a sequence, defined by
$$
a_{n+1}=2 a_{n}+2 a_{n-1}, \quad(n=2,3, \cdots) \text { and }
$$
$a, \beta$ are the two roots of the equation $x^{2}-2 x-2=0$, then the sequence $\left\{\mathrm{a}_{n+1}-\alpha \mathrm{a}_{n}\right\}$ is a geometric sequence with common ratio $\beta$.

This proposition is actually the inverse of (1). From the proof of (1), it can be seen that the inverse of (1) is also true.
Next, solving the equation $x^{2}-2 x-2=0$,
we get
$a=1-\sqrt{3}, \beta=1+\sqrt{3}$. Since $\left\{a_{n+1}-\alpha a_{n}\right\}$ is a geometric sequence with common ratio $\beta$, we have

From two
$$
\begin{array}{l}
a_{2}-\alpha a_{1}=2-\boldsymbol{\alpha}=\beta . \\
a_{n}+1-\alpha a_{n}=\boldsymbol{\beta}^{n} .
\end{array}
$$

Similarly, $\left\{\mathrm{a}_{\mathrm{n}+1}-\beta \mathrm{a}_{\mathrm{n}}\right\}$ is a geometric sequence with common ratio $\alpha$, so $\mathrm{a}_{2}-\beta \mathrm{a}_{1}=2-\beta=\alpha$,
$$
\mathrm{a}_{n}+1-\beta \mathrm{a}_{n}=\alpha^{\mathrm{n}}
$$
(2) - (1), we get $(\alpha-\beta) a_{n}=\alpha^{n}-\beta^{n}$. Therefore,
$$
a_{n}=-\frac{1}{2 \sqrt{3}}\left[(1-\sqrt{3})^{n}-(1+\sqrt{3})^{n}\right] .
$$