China Mathematical Competition

(1) By Vieta's theorem, we have $\alpha \times \beta = q \neq 0$, $\alpha + \beta = p$. Then
$$\begin{aligned} a_n &= p a_{n-1} - q a_{n-2} \\ &= (\alpha + \beta)a_{n-1} - \alpha\beta a_{n-2} \quad (n = 3, 4, \dots). \end{aligned}$$
This can be rewritten as
$$a_n - \beta a_{n-1} = \alpha (a_{n-1} - \beta a_{n-2}).$$
Let $b_n = a_{n+1} - \beta a_n$. Then $b_{n+1} = \alpha b_n$ ($n = 1, 2, \dots$). This means that $\{b_n\}$ is a geometric sequence with the common ratio $\alpha$. The first term of $\{b_n\}$ is
$$b_1 = a_2 - \beta a_1 = p^2 - q - \beta p = (\alpha + \beta)^2 - \alpha\beta - \beta(\alpha + \beta) = \alpha^2.$$
Therefore, $b_n = \alpha^2 \times \alpha^{n-1} = \alpha^{n+1}$. Then $a_{n+1} - \beta a_n = \alpha^{n+1}$. By rewriting
$$a_{n+1} = \alpha^{n+1} + \beta a_n \quad (n = 1, 2, \dots). \quad \textcircled{1}$$
When $\Delta = p^2 - 4q = 0$, we have $\alpha = \beta \neq 0$, $a_1 = p - 2\alpha$. The expression $\textcircled{1}$ becomes $a_{n+1} = \alpha^{n+1} + \alpha a_n$, i.e. $\frac{a_{n+1}}{\alpha^{n+1}} - \frac{a_n}{\alpha^n} = 1$.
Then $\left\{\frac{a_n}{\alpha^n}\right\}$ is an arithmetic sequence with the common difference $1$, whose first term is $\frac{a_1}{\alpha} = \frac{2\alpha}{\alpha} = 2$. Therefore,
$$\frac{a_n}{\alpha^n} = 2 + 1 \times (n - 1) = n + 1.$$
As a result, the general expression of $\{a_n\}$ is
$$a_n = (n + 1)\alpha^n. \quad \textcircled{2}$$
When $\Delta > 0$, $\alpha \neq \beta$, we have
$$\begin{aligned} a_{n+1} &= \alpha^{n+1} + \beta a_n \\ &= \beta a_n + \frac{\beta}{\beta - \alpha} \alpha^{n+1} - \frac{\alpha}{\beta - \alpha} \alpha^{n+1} \quad (n = 1, 2, \dots). \end{aligned}$$
By rewriting,
$$a_{n+1} + \frac{\alpha^{n+2}}{\beta - \alpha} = \beta \left(a_n + \frac{\alpha^{n+1}}{\beta - \alpha}\right) \quad (n = 1, 2, \dots).$$
Then $\left\{a_n + \frac{\alpha^{n+1}}{\beta - \alpha}\right\}$ becomes a geometric sequence with the common ratio $\beta$, whose first term is
$$a_1 + \frac{\alpha^2}{\beta - \alpha} = \alpha + \beta + \frac{\alpha^2}{\beta - \alpha} = \frac{\beta^2}{\beta - \alpha}.$$
Therefore,
$$a_n + \frac{\alpha^{n+1}}{\beta - \alpha} = \frac{\beta^2}{\beta - \alpha} \beta^{n-1}.$$
Then the general expression of $\{a_n\}$ is
$$a_n = \frac{\beta^{n+1} - \alpha^{n+1}}{\beta - \alpha} \quad (n = 1, 2, \dots). \qquad \textcircled{3}$$

(2) Given $p = 1$, $q = \frac{1}{4}$, we have $\Delta = p^2 - 4q = 0$. Then $\alpha = \beta = \frac{1}{2}$. By the expression (2), the general expression of $\{a_n\}$ is
$$a_n = (n+1)\left(\frac{1}{2}\right)^n = \frac{n+1}{2^n} \quad (n=1, 2, \dots).$$
Therefore, the sum of the first $n$ terms of $\{a_n\}$ is
$$S_n = \frac{2}{2} + \frac{3}{2^2} + \dots + \frac{n+1}{2^n}. \qquad \textcircled{4}$$
Then
$$\frac{1}{2} S_n = \frac{2}{2^2} + \frac{3}{2^3} + \dots + \frac{n+1}{2^{n+1}}. \qquad \textcircled{5}$$
$\textcircled{4} - \textcircled{5}$,
$$\frac{1}{2} S_n = \frac{3}{2} - \frac{n+3}{2^{n+1}}.$$
We finally get
$$S_n = 3 - \frac{n+3}{2^n}.$$

(1) By Vieta's theorem, we have $\alpha \times \beta = q \neq 0$, $\alpha + \beta = p$. Then
$$a_1 = \alpha + \beta, \quad a_2 = \alpha^2 + \alpha\beta + \beta^2. \qquad \textcircled{6}$$
The characteristic equation of $\{a_n\}$ is $\lambda^2 - p\lambda + q = 0$, which has roots $\alpha, \beta$. Then we can write down the general expression of $\{a_n\}$ according to the following different situations:

When $\alpha = \beta \neq 0$, $a_n = (A_1 + A_2 n) \alpha^n$. From (6), we have
$$(A_1 + A_2)\alpha = 2\alpha, \\ (A_1 + 2A_2)\alpha^2 = 3\alpha^2.$$
Then we get $A_1 = A_2 = 1$. Therefore,
$$a_n = (n + 1)\alpha^n.$$
When $\alpha \neq \beta$, $a_n = \Lambda_1 \alpha^n + \Lambda_2 \beta^n$. From (6), we have
$$A_1 \alpha + A_2 \beta = \alpha + \beta, \\ A_1 \alpha^2 + A_2 \beta^2 = \alpha^2 + \alpha\beta + \beta^2.$$
Then we get $A_1 = \frac{-\alpha}{\beta - \alpha}$, $A_2 = \frac{\beta}{\beta - \alpha}$. Therefore,
$$a_n = \frac{-\alpha^{n+1}}{\beta - \alpha} + \frac{\beta^{n+1}}{\beta - \alpha} = \frac{\beta^{n+1} - \alpha^{n+1}}{\beta - \alpha}.$$

(2) The solution is the same as Solution I.