(1)

\(q^{2}-4r\ne0\)のとき

特性方程式
\[ pa^{2}+qa+r=0 \] の解は
\[ a=\frac{-q\pm\sqrt{q^{2}-4r}}{2p} \] となるので、\(\alpha_{\pm}=\frac{-q\pm\sqrt{q^{2}-4r}}{2p}\)とおく。
これより、元の漸化式は、
\[ \begin{cases} a_{n+2}-\alpha_{-}a_{n+1}=\alpha_{+}\left(a_{n+1}-\alpha_{-}a_{n}\right)\\ a_{n+2}-\alpha_{+}a_{n+1}=\alpha_{-}\left(a_{n+1}-\alpha_{+}a_{n}\right) \end{cases} \] となり、各漸化式を解くと、
\[ \begin{cases} a_{n+1}-\alpha_{-}a_{n}=\alpha_{+}^{n-1}\left(a_{2}-\alpha_{-}a_{1}\right)\\ a_{n+1}-\alpha_{+}a_{n}=\alpha_{-}^{n-1}\left(a_{2}-\alpha_{+}a_{1}\right) \end{cases} \] となる。
辺々引くと、
\begin{align*} \left(\alpha_{+}-\alpha_{-}\right)a_{n} & =\alpha_{+}^{n-1}\left(a_{2}-\alpha_{-}a_{1}\right)-\alpha_{-}^{n-1}\left(a_{2}-\alpha_{+}a_{1}\right) \end{align*} となるので、
\begin{align*} a_{n} & =\frac{1}{\alpha_{+}-\alpha_{-}}\left\{ \alpha_{+}^{n-1}\left(a_{2}-\alpha_{-}a_{1}\right)-\alpha_{-}^{n-1}\left(a_{2}-\alpha_{+}a_{1}\right)\right\} \end{align*} となる。

\(a_{n_{1}},a_{n_{2}}\)が既知のとき

一般項は
\[ a_{n}=c_{1}\alpha_{+}^{n-1}+c_{2}\alpha_{-}^{n-1} \] と表され、\(n=n_{1}\)のとき\(a_{n}=a_{n_{1}}\)で\(n=n_{2}\)のとき\(a_{n}=a_{n_{2}}\)なので、
\[ \left(\begin{array}{c} a_{n_{1}}\\ a_{n_{2}} \end{array}\right)=\left(\begin{array}{cc} \alpha_{+}^{n_{1}-1} & \alpha_{-}^{n_{1}-1}\\ \alpha_{+}^{n_{2}-1} & \alpha_{-}^{n_{2}-1} \end{array}\right)\left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right) \] となり、これを\(c_{1},c_{2}\)について解くと、
\begin{align*} \left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right) & =\left(\begin{array}{cc} \alpha_{+}^{n_{1}-1} & \alpha_{-}^{n_{1}-1}\\ \alpha_{+}^{n_{2}-1} & \alpha_{-}^{n_{2}-1} \end{array}\right)^{-1}\left(\begin{array}{c} a_{n_{1}}\\ a_{n_{2}} \end{array}\right)\\ & =\frac{1}{\alpha_{+}^{n_{1}-1}\alpha_{-}^{n_{2}-1}-\alpha_{-}^{n_{1}-1}\alpha_{+}^{n_{2}-1}}\left(\begin{array}{cc} \alpha_{-}^{n_{2}-1} & -\alpha_{-}^{n_{1}-1}\\ -\alpha_{+}^{n_{2}-1} & \alpha_{+}^{n_{1}-1} \end{array}\right)\left(\begin{array}{c} a_{n_{1}}\\ a_{n_{2}} \end{array}\right) \end{align*} となる。
従って、
\[ a_{n}=c_{1}\alpha_{+}^{n-1}+c_{2}\alpha_{-}^{n-1} \] \[ \left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right)=\frac{1}{\alpha_{+}^{n_{1}-1}\alpha_{-}^{n_{2}-1}-\alpha_{-}^{n_{1}-1}\alpha_{+}^{n_{2}-1}}\left(\begin{array}{cc} \alpha_{-}^{n_{2}-1} & -\alpha_{-}^{n_{1}-1}\\ -\alpha_{+}^{n_{2}-1} & \alpha_{+}^{n_{1}-1} \end{array}\right)\left(\begin{array}{c} a_{n_{1}}\\ a_{n_{2}} \end{array}\right) \] となる。

\(q^{2}-4r=0\)のとき

特性方程式
\[ pa^{2}+qa+r=0 \] の解は
\[ a=-\frac{q}{2p} \] となるので\(\alpha=-\frac{q}{2p}\)とおく。
これより、元の漸化式は、
\[ a_{n+2}-\alpha a_{n+1}=\alpha\left(a_{n+1}-\alpha a_{n}\right) \] この漸化式を\(a_{n+1}-\alpha a_{n}\)について解くと、
\[ a_{n+1}-\alpha a_{n}=\left(a_{2}-\alpha a_{1}\right)a^{n-1} \] となる。
この漸化式は
\[ \frac{a_{n+1}}{a^{n+1}}-\frac{a_{n}}{a^{n}}=\frac{a_{2}-\alpha a_{1}}{a^{2}} \] となるので一般解は
\begin{align*} a_{n} & =\alpha^{n}\left\{ \frac{a_{1}}{\alpha}+\sum_{k=1}^{n-1}\left(\frac{a_{k+1}}{\alpha^{k+1}}-\frac{a_{k}}{\alpha^{k}}\right)\right\} \\ & =\alpha^{n}\left\{ \frac{a_{1}}{\alpha}+\sum_{k=1}^{n-1}\frac{a_{2}-\alpha a_{1}}{\alpha^{2}}\right\} \\ & =\alpha^{n}\left\{ \frac{a_{1}}{\alpha}+\left(n-1\right)\cdot\frac{a_{2}-\alpha a_{1}}{\alpha^{2}}\right\} \\ & =\alpha^{n-2}\left\{ a_{1}\alpha+\left(n-1\right)\left(a_{2}-\alpha a_{1}\right)\right\} \end{align*}

\(a_{n_{1}},a_{n_{2}}\)が既知のとき

一般項は
\[ a_{n}=\alpha^{n-2}\left(c_{1}+c_{2}n\right) \] と表され、\(n=n_{1}\)のとき\(a_{n}=a_{n_{1}}\)で\(n=n_{2}\)のとき\(a_{n}=a_{n_{2}}\)なので、
\[ \left(\begin{array}{c} a_{n_{1}}\\ a_{n_{2}} \end{array}\right)=\left(\begin{array}{cc} \alpha^{n_{1}-2} & n_{1}\alpha^{n_{1}-2}\\ \alpha^{n_{2}-2} & n_{2}\alpha^{n_{2}-2} \end{array}\right)\left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right) \] となり、これを\(c_{1},c_{2}\)について解くと、
\begin{align*} \left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right) & =\left(\begin{array}{cc} \alpha^{n_{1}-2} & n_{1}\alpha^{n_{1}-2}\\ \alpha^{n_{2}-2} & n_{2}\alpha^{n_{2}-2} \end{array}\right)^{-1}\left(\begin{array}{c} a_{n_{1}}\\ a_{n_{2}} \end{array}\right)\\ & =\frac{1}{n_{2}\alpha^{n_{1}-2}\alpha^{n_{2}-2}-n_{1}\alpha^{n_{1}-2}\alpha^{n_{2}-2}}\left(\begin{array}{cc} n_{2}\alpha^{n_{2}-2} & -n_{1}\alpha^{n_{1}-2}\\ -\alpha^{n_{2}-2} & \alpha^{n_{1}-2} \end{array}\right)\left(\begin{array}{c} a_{n_{1}}\\ a_{n_{2}} \end{array}\right) \end{align*} となる。
従って、
\[ a_{n}=\alpha^{n-2}\left(c_{1}+c_{2}n\right) \] \begin{align*} \left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right) & =\frac{1}{n_{2}\alpha^{n_{1}-2}\alpha^{n_{2}-2}-n_{1}\alpha^{n_{1}-2}\alpha^{n_{2}-2}}\left(\begin{array}{cc} n_{2}\alpha^{n_{2}-2} & -n_{1}\alpha^{n_{1}-2}\\ -\alpha^{n_{2}-2} & \alpha^{n_{1}-2} \end{array}\right)\left(\begin{array}{c} a_{n_{1}}\\ a_{n_{2}} \end{array}\right) \end{align*} となる。

(2)

\(q^{2}-4r\ne0\)のとき

特性方程式
\[ pa^{2}+qa+r=0 \] の解は
\[ a=\frac{-q\pm\sqrt{q^{2}-4r}}{2p} \] となるので、\(\alpha_{\pm}=\frac{-q\pm\sqrt{q^{2}-4r}}{2p}\)とおく。
これより、元の漸化式は、
\[ a_{n+2}-\alpha_{\mp}a_{n+1}=\alpha_{\pm}\left(a_{n+1}-\alpha_{\mp}a_{n}\right)+f\left(n\right) \] となり、両辺を\(\alpha_{\mp}^{n+1}\)で割ると、
\[ \frac{a_{n+2}-\alpha_{\mp}a_{n+1}}{\alpha_{\pm}^{n+1}}=\frac{a_{n+1}-\alpha_{\mp}a_{n}}{\alpha_{\pm}^{n}}+\frac{f\left(n\right)}{\alpha_{\pm}^{n+1}} \] となり、この漸化式を解くと、
\begin{align*} \frac{a_{n+1}-\alpha_{\mp}a_{n}}{\alpha_{\pm}^{n}} & =\frac{a_{2}-\alpha_{\mp}a_{1}}{\alpha_{\pm}}+\sum_{k=1}^{n-1}\left(\frac{a_{k+2}-\alpha_{\mp}a_{k+1}}{\alpha_{\pm}^{k+1}}-\frac{a_{k+1}-\alpha_{\mp}a_{k}}{\alpha_{\pm}^{k}}\right)\\ & =\frac{a_{2}-\alpha_{\mp}a_{1}}{\alpha_{\pm}}+\sum_{k=1}^{n-1}\left(\frac{f\left(k\right)}{\alpha_{\pm}^{k+1}}\right) \end{align*} となる。
両辺を\(a_{\mp}^{n}\)で割ると、
\begin{align*} \frac{a_{n+1}-\alpha_{\mp}a_{n}}{\alpha_{\pm}^{n}\alpha_{\mp}^{n}} & =\frac{a_{2}-\alpha_{\mp}a_{1}}{\alpha_{\pm}\alpha_{\mp}^{n}}+\frac{1}{\alpha_{\mp}^{n}}\sum_{k=1}^{n-1}\left(\frac{f\left(k\right)}{\alpha_{\pm}^{k+1}}\right) \end{align*} となる。
辺々引くと、
\begin{align*} \frac{\alpha_{+}-\alpha_{-}}{\alpha_{+}^{n}\alpha_{-}^{n}}a_{n} & =\frac{1}{\alpha_{+}\alpha_{-}^{n}}\left(a_{2}-\alpha_{-}a_{1}\right)-\frac{1}{\alpha_{-}\alpha_{+}^{n}}\left(a_{2}-\alpha_{+}a_{1}\right)+\frac{1}{\alpha_{-}^{n}}\sum_{k=1}^{n-1}\left(\frac{f\left(k\right)}{\alpha_{+}^{k+1}}\right)-\frac{1}{\alpha_{+}^{n}}\sum_{k=1}^{n-1}\left(\frac{f\left(k\right)}{\alpha_{-}^{k+1}}\right) \end{align*} となるので、\(a_{n}\)について解くと、
\begin{align*} a_{n} & =\frac{\alpha_{+}^{n}\alpha_{-}^{n}}{\alpha_{+}-\alpha_{-}}\left\{ \frac{1}{\alpha_{+}\alpha_{-}^{n}}\left(a_{2}-\alpha_{-}a_{1}\right)-\frac{1}{\alpha_{-}\alpha_{+}^{n}}\left(a_{2}-\alpha_{+}a_{1}\right)+\frac{1}{\alpha_{-}^{n}}\sum_{k=1}^{n-1}\left(\frac{f\left(k\right)}{\alpha_{+}^{k+1}}\right)-\frac{1}{\alpha_{+}^{n}}\sum_{k=1}^{n-1}\left(\frac{f\left(k\right)}{\alpha_{-}^{k+1}}\right)\right\} \\ & =\frac{1}{\alpha_{+}-\alpha_{-}}\left\{ \alpha_{+}^{n-1}\left(a_{2}-\alpha_{-}a_{1}\right)-\alpha_{-}^{n-1}\left(a_{2}-\alpha_{+}a_{1}\right)+\alpha_{+}^{n}\sum_{k=1}^{n-1}\left(\frac{f\left(k\right)}{\alpha_{+}^{k+1}}\right)-\alpha_{-}^{n}\sum_{k=1}^{n-1}\left(\frac{f\left(k\right)}{\alpha_{-}^{k+1}}\right)\right\} \end{align*} となる。

\(a_{n_{1}},a_{n_{2}}\)が既知のとき

一般項は
\[ a_{n}=c_{1}\alpha_{+}^{n-1}+c_{2}\alpha_{-}^{n-1}+\frac{\alpha_{+}^{n}}{\alpha_{+}-\alpha_{-}}\sum_{k=1}^{n-1}\left(\frac{f\left(k\right)}{\alpha_{+}^{k+1}}\right)-\frac{a_{-}^{n}}{\alpha_{+}-\alpha_{-}}\sum_{k=1}^{n-1}\left(\frac{f\left(k\right)}{\alpha_{-}^{k+1}}\right) \] と表されるので
\[ a_{n}'=a_{n}-\frac{\alpha_{+}^{n}}{\alpha_{+}-\alpha_{-}}\sum_{k=1}^{n-1}\left(\frac{f\left(k\right)}{\alpha_{+}^{k+1}}\right)+\frac{\alpha_{-}^{n}}{\alpha_{+}-\alpha_{-}}\sum_{k=1}^{n-1}\left(\frac{f\left(k\right)}{\alpha_{-}^{k+1}}\right) \] とおくと、
\[ a_{n}'=c_{1}\alpha_{+}^{n-1}+c_{2}\alpha_{-}^{n-1} \] となる。
\(n=n_{1}\)のとき\(a_{n}'=a_{n_{1}}'\)で\(n=n_{2}\)のとき\(a_{n}'=a_{n_{2}}'\)なので、
\[ \left(\begin{array}{c} a_{n_{1}}'\\ a_{n_{2}}' \end{array}\right)=\left(\begin{array}{cc} \alpha_{+}^{n_{1}-1} & \alpha_{-}^{n_{1}-1}\\ \alpha_{+}^{n_{2}-1} & \alpha_{-}^{n_{2}-1} \end{array}\right)\left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right) \] となり、これを\(c_{1},c_{2}\)について解くと、
\begin{align*} \left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right) & =\left(\begin{array}{cc} \alpha_{+}^{n_{1}-1} & \alpha_{-}^{n_{1}-1}\\ \alpha_{+}^{n_{2}-1} & \alpha_{-}^{n_{2}-1} \end{array}\right)^{-1}\left(\begin{array}{c} a_{n_{1}}'\\ a_{n_{2}}' \end{array}\right)\\ & =\frac{1}{\alpha_{+}^{n_{1}-1}\alpha_{-}^{n_{2}-1}-\alpha_{-}^{n_{1}-1}\alpha_{+}^{n_{2}-1}}\left(\begin{array}{cc} \alpha_{-}^{n_{2}-1} & -\alpha_{-}^{n_{1}-1}\\ -\alpha_{+}^{n_{2}-1} & \alpha_{+}^{n_{1}-1} \end{array}\right)\left(\begin{array}{c} a_{n_{1}}'\\ a_{n_{2}}' \end{array}\right)\\ & =\frac{1}{\alpha_{+}^{n_{1}-1}\alpha_{-}^{n_{2}-1}-\alpha_{-}^{n_{1}-1}\alpha_{+}^{n_{2}-1}}\left(\begin{array}{cc} \alpha_{-}^{n_{2}-1} & -\alpha_{-}^{n_{1}-1}\\ -\alpha_{+}^{n_{2}-1} & \alpha_{+}^{n_{1}-1} \end{array}\right)\left(\begin{array}{c} a_{n_{1}}-\frac{\alpha_{+}^{n_{1}}}{\alpha_{+}-\alpha_{-}}\sum_{k=1}^{n_{1}-1}\left(\frac{f\left(k\right)}{\alpha_{+}^{k+1}}\right)+\frac{\alpha_{-}^{n_{1}}}{\alpha_{+}-\alpha_{-}}\sum_{k=1}^{n_{1}-1}\left(\frac{f\left(k\right)}{\alpha_{-}^{k+1}}\right)\\ a_{n_{2}}-\frac{\alpha_{+}^{n_{2}}}{\alpha_{+}-\alpha_{-}}\sum_{k=1}^{n_{2}-1}\left(\frac{f\left(k\right)}{\alpha_{+}^{k+1}}\right)+\frac{\alpha_{-}^{n_{2}}}{\alpha_{+}-\alpha_{-}}\sum_{k=1}^{n_{2}-1}\left(\frac{f\left(k\right)}{\alpha_{-}^{k+1}}\right) \end{array}\right) \end{align*} となる。
従って、
\[ a_{n}=c_{1}\alpha_{+}^{n-1}+c_{2}\alpha_{-}^{n-1}+\frac{\alpha_{+}^{n}}{\alpha_{+}-\alpha_{-}}\sum_{k=1}^{n-1}\left(\frac{f\left(k\right)}{\alpha_{+}^{k+1}}\right)-\frac{\alpha_{-}^{n}}{\alpha_{+}-\alpha_{-}}\sum_{k=1}^{n-1}\left(\frac{f\left(k\right)}{\alpha_{-}^{k+1}}\right) \] \[ \left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right)=\frac{1}{\alpha_{+}^{n_{1}-1}\alpha_{-}^{n_{2}-1}-\alpha_{-}^{n_{1}-1}\alpha_{+}^{n_{2}-1}}\left(\begin{array}{cc} \alpha_{-}^{n_{2}-1} & -\alpha_{-}^{n_{1}-1}\\ -\alpha_{+}^{n_{2}-1} & \alpha_{+}^{n_{1}-1} \end{array}\right)\left(\begin{array}{c} a_{n_{1}}-\frac{\alpha_{+}^{n_{1}}}{\alpha_{+}-\alpha_{-}}\sum_{k=1}^{n_{1}-1}\left(\frac{f\left(k\right)}{\alpha_{+}^{k+1}}\right)+\frac{\alpha_{-}^{n_{1}}}{\alpha_{+}-\alpha_{-}}\sum_{k=1}^{n_{1}-1}\left(\frac{f\left(k\right)}{\alpha_{-}^{k+1}}\right)\\ a_{n_{2}}-\frac{\alpha_{+}^{n_{2}}}{\alpha_{+}-\alpha_{-}}\sum_{k=1}^{n_{2}-1}\left(\frac{f\left(k\right)}{\alpha_{+}^{k+1}}\right)+\frac{\alpha_{-}^{n_{2}}}{\alpha_{+}-\alpha_{-}}\sum_{k=1}^{n_{2}-1}\left(\frac{f\left(k\right)}{\alpha_{-}^{k+1}}\right) \end{array}\right) \] となる。

\(q^{2}-4r=0\)のとき

特性方程式
\[ pa^{2}+qa+r=0 \] の解は
\[ a=-\frac{q}{2p} \] となるので\(\alpha=-\frac{q}{2p}\)とおく。
これより、元の漸化式は、
\[ a_{n+2}-\alpha a_{n+1}=\alpha\left(a_{n+1}-\alpha a_{n}\right)+f\left(n\right) \] この漸化式を\(a_{n+1}-\alpha a_{n}\)について解くと、
\begin{align*} a_{n+1}-\alpha a_{n} & =\alpha^{n}\left\{ \frac{a_{2}-\alpha a_{1}}{\alpha}+\sum_{k=1}^{n-1}\left(\frac{a_{k+2}-\alpha a_{k+1}}{\alpha^{k+1}}-\frac{a_{k+1}-\alpha a_{k}}{\alpha^{k}}\right)\right\} \\ & =\alpha^{n}\left\{ \frac{a_{2}-\alpha a_{1}}{\alpha}+\sum_{k=1}^{n-1}\frac{f\left(k\right)}{\alpha^{k+1}}\right\} \\ & =\alpha^{n-1}\left\{ a_{2}-\alpha a_{1}+\sum_{k=1}^{n-1}\frac{f\left(k\right)}{\alpha^{k}}\right\} \end{align*} となる。
これを\(a_{n}\)について解くと、
\begin{align*} a_{n} & =\alpha^{n}\left\{ \frac{a_{1}}{\alpha}+\sum_{j=1}^{n-1}\left(\frac{a_{j+1}}{\alpha^{j+1}}-\frac{a_{j}}{\alpha^{j}}\right)\right\} \\ & =\alpha^{n}\left\{ \frac{a_{1}}{\alpha}+\sum_{j=1}^{n-1}\left(\frac{1}{\alpha^{2}}\left(a_{2}-\alpha a_{1}+\sum_{k=1}^{j-1}\frac{f\left(k\right)}{\alpha^{k}}\right)\right)\right\} \\ & =\alpha^{n}\left\{ \frac{a_{1}}{\alpha}+\frac{1}{\alpha^{2}}\left(\left(a_{2}-\alpha a_{1}\right)\left(n-1\right)+\sum_{j=1}^{n-1}\sum_{k=1}^{j-1}\frac{f\left(k\right)}{\alpha^{k}}\right)\right\} \\ & =a_{1}\alpha^{n-1}+\alpha^{n-2}\left(\left(a_{2}-\alpha a_{1}\right)\left(n-1\right)+\sum_{j=1}^{n-1}\sum_{k=1}^{j-1}\frac{f\left(k\right)}{\alpha^{k}}\right) \end{align*} となる。

\(a_{n_{1}},a_{n_{2}}\)が既知のとき

一般項は
\[ a_{n}=\alpha^{n-2}\left(c_{1}+c_{2}n\right)+\alpha^{n-2}\sum_{j=1}^{n-1}\sum_{k=1}^{j-1}\frac{f\left(k\right)}{\alpha^{k}} \] と表され、\(n=n_{1}\)のとき\(a_{n}=a_{n_{1}}\)で\(n=n_{2}\)のとき\(a_{n}=a_{n_{2}}\)なので、
\[ \left(\begin{array}{c} a_{n_{1}}\\ a_{n_{2}} \end{array}\right)=\left(\begin{array}{cc} \alpha^{n_{1}-2} & n_{1}\alpha^{n_{1}-2}\\ \alpha^{n_{2}-2} & n_{2}\alpha^{n_{2}-2} \end{array}\right)\left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right)+\left(\begin{array}{c} \alpha^{n_{1}-2}\sum_{j=1}^{n_{1}-1}\sum_{k=1}^{j-1}\frac{f\left(k\right)}{\alpha^{k}}\\ \alpha^{n_{2}-2}\sum_{j=1}^{n_{2}-1}\sum_{k=1}^{j-1}\frac{f\left(k\right)}{\alpha^{k}} \end{array}\right) \] となり、これを\(c_{1},c_{2}\)について解くと、
\begin{align*} \left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right) & =\left(\begin{array}{cc} \alpha^{n_{1}-2} & n_{1}\alpha^{n_{1}-2}\\ \alpha^{n_{2}-2} & n_{2}\alpha^{n_{2}-2} \end{array}\right)^{-1}\left(\begin{array}{c} a_{n_{1}}-\alpha^{n_{1}-2}\sum_{j=1}^{n_{1}-1}\sum_{k=1}^{j-1}\frac{f\left(k\right)}{\alpha^{k}}\\ a_{n_{2}}-\alpha^{n_{2}-2}\sum_{j=1}^{n_{2}-1}\sum_{k=1}^{j-1}\frac{f\left(k\right)}{\alpha^{k}} \end{array}\right)\\ & =\frac{1}{n_{2}\alpha^{n_{1}-2}\alpha^{n_{2}-2}-n_{1}\alpha^{n_{1}-2}\alpha^{n_{2}-2}}\left(\begin{array}{cc} n_{2}\alpha^{n_{2}-2} & -n_{1}\alpha^{n_{1}-2}\\ -\alpha^{n_{2}-2} & \alpha^{n_{1}-2} \end{array}\right)\left(\begin{array}{c} a_{n_{1}}-\alpha^{n_{1}-2}\sum_{j=1}^{n_{1}-1}\sum_{k=1}^{j-1}\frac{f\left(k\right)}{\alpha^{k}}\\ a_{n_{2}}-\alpha^{n_{2}-2}\sum_{j=1}^{n_{2}-1}\sum_{k=1}^{j-1}\frac{f\left(k\right)}{\alpha^{k}} \end{array}\right) \end{align*} となる。
従って、
\[ a_{n}=\alpha^{n-2}\left(c_{1}+c_{2}n\right)+\alpha^{n-2}\sum_{j=1}^{n-1}\sum_{k=1}^{j-1}\frac{f\left(k\right)}{\alpha^{k}} \] \begin{align*} \left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right) & =\frac{1}{n_{2}\alpha^{n_{1}-2}\alpha^{n_{2}-2}-n_{1}\alpha^{n_{1}-2}\alpha^{n_{2}-2}}\left(\begin{array}{cc} n_{2}\alpha^{n_{2}-2} & -n_{1}\alpha^{n_{1}-2}\\ -\alpha^{n_{2}-2} & \alpha^{n_{1}-2} \end{array}\right)\left(\begin{array}{c} a_{n_{1}}-\alpha^{n_{1}-2}\sum_{j=1}^{n_{1}-1}\sum_{k=1}^{j-1}\frac{f\left(k\right)}{\alpha^{k}}\\ a_{n_{2}}-\alpha^{n_{2}-2}\sum_{j=1}^{n_{2}-1}\sum_{k=1}^{j-1}\frac{f\left(k\right)}{\alpha^{k}} \end{array}\right) \end{align*} となる。