China Mathematical Competition

(1) By Vieta's theorem, we have $\alpha \times \beta =q\ne 0$, $\alpha +\beta =p$. Then $$\begin{array}{rl}{a}_n& =p{a}_{n-1}-q{a}_{n-2}\\ & =(\alpha +\beta )a_{n-1}-\alpha \beta a_{n-2}(n=3,4,\dots ).\end{array}$$ 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 $$a_{n+1}={\alpha }^{n+1}+\beta a_n(n=1,2,\dots ).\stackrel{◯}{1}$$ When $\Delta =p^2-4q=0$, we have $\alpha =\beta \ne 0$, $a_1=p-2\alpha$. The expression $\stackrel{◯}{1}$ 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 $$a_n=(n+1){\alpha }^n.\stackrel{◯}{2}$$ When $\Delta >0$, $\alpha \ne \beta$, we have $$\begin{array}{rl}{a}_{n+1}& ={\alpha }^{n+1}+\beta a_n\\ & =\beta a_n+\frac{\beta }{\beta -\alpha }{\alpha }^{n+1}-\frac{\alpha }{\beta -\alpha }{\alpha }^{n+1}(n=1,2,\dots ).\end{array}$$ By rewriting, $$a_{n+1}+\frac{{\alpha }^{n+2}}{\beta -\alpha }=\beta \left(a_n+\frac{{\alpha }^{n+1}}{\beta -\alpha }\right)(n=1,2,\dots ).$$ 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 $$a_n=\frac{{\beta }^{n+1}-{\alpha }^{n+1}}{\beta -\alpha }(n=1,2,\dots ).\stackrel{◯}{3}$$

(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 $$a_n=(n+1){\left(\frac{1}{2}\right)}^n=\frac{n+1}{2^n}(n=1,2,\dots ).$$ Therefore, the sum of the first $n$ terms of $\{a_n\}$ is $$S_n=\frac{2}{2}+\frac{3}{2^2}+\cdots +\frac{n+1}{2^n}.\stackrel{◯}{4}$$ Then $$\frac{1}{2}{S}_n=\frac{2}{2^2}+\frac{3}{2^3}+\cdots +\frac{n+1}{2^{n+1}}.\stackrel{◯}{5}$$ $\stackrel{◯}{4}-\stackrel{◯}{5}$, $$\frac{1}{2}{S}_n=\frac{3}{2}-\frac{n+3}{2^{n+1}}.$$ We finally get $$S_n=3-\frac{n+3}{2^n}.$$

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Alternative solution.

(1) By Vieta's theorem, we have $\alpha \times \beta =q\ne 0$, $\alpha +\beta =p$. Then $$a_1=\alpha +\beta ,a_2={\alpha }^2+\alpha \beta +{\beta }^2.\stackrel{◯}{6}$$ 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 $\alpha =\beta \ne 0$, $a_n=(A_1+A_{2}n){\alpha }^n$. From (6), we have $$\begin{gathered} (A_1+A_2)\alpha =2\alpha , \\ (A_1+2A_2){\alpha }^2=3{\alpha }^2. \end{gathered}$$ Then we get $A_1=A_2=1$. Therefore, $$a_n=(n+1){\alpha }^n.$$ When $\alpha \ne \beta$, $a_n={\Lambda }_1{\alpha }^n+{\Lambda }_2{\beta }^n$. From (6), we have $$\begin{gathered} A_1\alpha +A_2\beta =\alpha +\beta , \\ {A}_1{\alpha }^2+A_2{\beta }^2={\alpha }^2+\alpha \beta +{\beta }^2. \end{gathered}$$ 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.