(1)\(q^{2}-4pr\ne0\land p+q+r\ne0\)のとき

\(a_{n}=a\)とおくと、
\[ pa+qa+ra=s \] となるので、
\[ a\left(p+q+r\right)=s \] となり、
\[ a=\frac{s}{p+q+r} \] となる。
これより、\(a_{n}=\frac{s}{p+q+r}\)は\(pa_{n+2}+qa_{n+1}+ra_{n}=s\)を満たす。
漸化式の一般解は、\(\alpha_{\pm}=\frac{-q\pm\sqrt{q^{2}-4pr}}{2p}\)とおくと、
\[ a_{n}=c_{1}\alpha_{+}^{n-1}+c_{2}\alpha_{-}^{n-1}+\frac{s}{p+q+r} \] となる。
\(c_{1},c_{2}\)は\(a_{n_{1}},a_{n_{2}}\)が既知なので、
\[ \left(\begin{array}{c} a_{n_{1}}\\ a_{n_{2}} \end{array}\right)=\left(\begin{array}{cc} \alpha_{+}^{n_{1}-1} & \alpha_{-}^{n_{1}-1}\\ \alpha_{+}^{n_{2}-1} & \alpha_{-}^{n_{2}-1} \end{array}\right)\left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right)+\left(\begin{array}{c} \frac{s}{p+q+r}\\ \frac{s}{p+q+r} \end{array}\right) \] となり
\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}}-\frac{s}{p+q+r}\\ a_{n_{2}}-\frac{s}{p+q+r} \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{s}{p+q+r}\\ a_{n_{2}}-\frac{s}{p+q+r} \end{array}\right) \end{align*} となる。

(2)\(q^{2}-4pr\ne0\land p+q+r=0\)のとき

\(a_{n}=an+b\)とおくと、
\begin{align*} s & =p\left\{ a\left(n+2\right)+b\right\} +q\left\{ a\left(n+1\right)+b\right\} +r\left(an+b\right)\\ & =a\left(n\left(p+q+r\right)+2p+q\right)+b\left(p+q+r\right)\\ & =a\left(2p+q\right)\\ & =a\left(p+p+q\right)\\ & =a\left(p-r\right) \end{align*} となるので
\[ a=\frac{s}{p-r} \] となり、\(a_{n}=\frac{s}{p-r}n\)は\(pa_{n+2}+qa_{n+1}+ra_{n}=s\)を満たす。
これより一般解は、
\begin{align*} a_{n} & =c_{1}\left(\frac{-q+\sqrt{q^{2}-4pr}}{2p}\right)^{n-1}+c_{2}\left(\frac{-q-\sqrt{q^{2}-4pr}}{2p}\right)^{n-1}+\frac{s}{p-r}n\\ & =c_{1}\left(\frac{-q+\sqrt{q^{2}+4\left(q+r\right)r}}{2p}\right)^{n-1}+c_{2}\left(\frac{-q-\sqrt{q^{2}+4\left(q+r\right)r}}{2p}\right)^{n-1}+\frac{s}{p-r}n\\ & =c_{1}\left(\frac{-q+\sqrt{\left(q+2r\right)^{2}}}{2p}\right)^{n-1}+c_{2}\left(\frac{-q-\sqrt{\left(q+2r\right)^{2}}}{2p}\right)^{n-1}+\frac{s}{p-r}n\\ & =c_{1}\left(\frac{-q+\left|q+2r\right|}{2p}\right)^{n-1}+c_{2}\left(\frac{-q-\left|q+2r\right|}{2p}\right)^{n-1}+\frac{s}{p-r}n\\ & =c_{1}\left(\frac{r}{p}\right)^{n-1}+c_{2}\left(\frac{-2q-2r}{2p}\right)^{n-1}+\frac{s}{p-r}n\cmt{p+2<0\rightarrow c_{1}\rightarrow c_{2}\land c_{2}\rightarrow c_{1}}\\ & =c_{1}\left(-\frac{p+q}{p}\right)^{n-1}+c_{2}\left(\frac{-\left(q+r\right)}{p}\right)^{n-1}+\frac{s}{p-r}n\\ & =c_{1}\left(\frac{r}{p}\right)^{n-1}+c_{2}+\frac{s}{p-r}n \end{align*} となる。
\(c_{1},c_{2}\)は\(a_{n_{1}},a_{n_{2}}\)が既知なので、
\[ \left(\begin{array}{c} a_{n_{1}}\\ a_{n_{2}} \end{array}\right)=\left(\begin{array}{cc} \left(\frac{r}{p}\right)^{n_{1}-1} & 1\\ \left(\frac{r}{p}\right)^{n_{2}-1} & 1 \end{array}\right)\left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right)+\left(\begin{array}{c} \frac{s}{p-r}n_{1}\\ \frac{s}{p-r}n_{2} \end{array}\right) \] となり
\begin{align*} \left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right) & =\left(\begin{array}{cc} \left(\frac{r}{p}\right)^{n_{1}-1} & 1\\ \left(\frac{r}{p}\right)^{n_{2}-1} & 1 \end{array}\right)^{-1}\left(\begin{array}{c} a_{n_{1}}-\frac{s}{p-r}n_{1}\\ a_{n_{2}}-\frac{s}{p-r}n_{2} \end{array}\right)\\ & =\frac{1}{\left(\frac{r}{p}\right)^{n_{1}-1}-\left(\frac{r}{p}\right)^{n_{2}-1}}\left(\begin{array}{cc} 1 & -1\\ -\left(\frac{r}{p}\right)^{n_{2}-1} & \left(\frac{r}{p}\right)^{n_{1}-1} \end{array}\right)\left(\begin{array}{c} a_{n_{1}}-\frac{s}{p-r}n_{1}\\ a_{n_{2}}-\frac{s}{p-r}n_{2} \end{array}\right) \end{align*} となる。

(3)\(q^{2}-4pr=0\land p+q+r\ne0\)のとき

\(a_{n}=\frac{s}{p+q+r}\)とおくと、\(pa_{n+2}+qa_{n+1}+ra_{n}=s\)を満たす。
漸化式の一般解は、
\[ a_{n}=\left(\frac{-q}{2p}\right)^{n-2}\left(c_{1}+c_{2}n\right)+\frac{s}{p+q+r} \] となる。
\(c_{1},c_{2}\)は\(a_{n_{1}},a_{n_{2}}\)が既知なので、
\[ \left(\begin{array}{c} a_{n_{1}}\\ a_{n_{2}} \end{array}\right)=\left(\begin{array}{cc} \left(\frac{-q}{2p}\right)^{n_{1}-2} & \left(\frac{-q}{2p}\right)^{n_{1}-2}n_{1}\\ \left(\frac{-q}{2p}\right)^{n_{2}-2} & \left(\frac{-q}{2p}\right)^{n_{2}-2}n_{2} \end{array}\right)\left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right)+\left(\begin{array}{c} \frac{s}{p+q+r}\\ \frac{s}{p+q+r} \end{array}\right) \] となり
\begin{align*} \left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right) & =\left(\begin{array}{cc} \left(\frac{-q}{2p}\right)^{n_{1}-2} & \left(\frac{-q}{2p}\right)^{n_{1}-2}n_{1}\\ \left(\frac{-q}{2p}\right)^{n_{2}-2} & \left(\frac{-q}{2p}\right)^{n_{2}-2}n_{2} \end{array}\right)^{-1}\left(\begin{array}{c} a_{n_{1}}-\frac{s}{p+q+r}\\ a_{n_{2}}-\frac{s}{p+q+r} \end{array}\right)\\ & =\frac{1}{\left(\frac{-q}{2p}\right)^{n_{1}-2}\left(\frac{-q}{2p}\right)^{n_{2}-2}n_{2}-\left(\frac{-q}{2p}\right)^{n_{1}-2}\left(\frac{-q}{2p}\right)^{n_{2}-2}n_{1}}\left(\begin{array}{cc} \left(\frac{-q}{2p}\right)^{n_{2}-2}n_{2} & -\left(\frac{-q}{2p}\right)^{n_{1}-2}n_{1}\\ -\left(\frac{-q}{2p}\right)^{n_{2}-2} & \left(\frac{-q}{2p}\right)^{n_{1}-2} \end{array}\right)\left(\begin{array}{c} a_{n_{1}}-\frac{s}{p+q+r}\\ a_{n_{2}}-\frac{s}{p+q+r} \end{array}\right) \end{align*} となる。

(4)\(q^{2}-4pr=0\land p+q+r=0\)のとき

\begin{align*} 0 & =q^{2}-4pr\\ & =q^{2}+4p\left(p+q\right)\\ & =q^{2}+4p^{2}+4pq\\ & =\left(2p+q\right)^{2} \end{align*} となるので\(2p+q=0\)となる。
\(a_{n}=an^{2}\)とおくと、
\[ pa_{n+2}+qa_{n+1}+ra_{n}=s \] \begin{align*} s & =pa_{n+2}+qa_{n+1}+ra_{n}\\ & =pa\left(n+2\right)^{2}+qa\left(n+1\right)^{2}+ran^{2}\\ & =\left(p+q+r\right)an^{2}+2\left(2p+q\right)an+\left(4p+q\right)a\\ & =\left(2p+2p+q\right)a\\ & =2pa \end{align*} となるので
\[ a=\frac{s}{2p} \] となり\(a_{n}=\frac{s}{2p}n^{2}\)は\(pa_{n+2}+qa_{n+1}+ra_{n}=s\)を満たす。
漸化式の一般解は、
\begin{align*} a_{n} & =\left(\frac{-q}{2p}\right)^{n-2}\left(c_{1}+c_{2}n\right)+\frac{s}{2p}n^{2}\\ & =\left(c_{1}+c_{2}n\right)+\frac{s}{2p}n^{2} \end{align*} となる。
\(c_{1},c_{2}\)は\(a_{n_{1}},a_{n_{2}}\)が既知なので、
\[ \left(\begin{array}{c} a_{n_{1}}\\ a_{n_{2}} \end{array}\right)=\left(\begin{array}{cc} 1 & n_{1}\\ 1 & n_{2} \end{array}\right)\left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right)+\left(\begin{array}{c} \frac{s}{2p}n_{1}^{2}\\ \frac{s}{2p}n_{2}^{2} \end{array}\right) \] となり
\begin{align*} \left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right) & =\left(\begin{array}{cc} 1 & n_{1}\\ 1 & n_{2} \end{array}\right)^{-1}\left(\begin{array}{c} a_{n_{1}}-\frac{s}{2p}n_{1}^{2}\\ a_{n_{2}}-\frac{s}{2p}n_{2}^{2} \end{array}\right)\\ & =\frac{1}{n_{2}-n_{1}}\left(\begin{array}{cc} n_{2} & -n_{1}\\ -1 & 1 \end{array}\right)\left(\begin{array}{c} a_{n_{1}}-\frac{s}{2p}n_{1}^{2}\\ a_{n_{2}}-\frac{s}{2p}n_{2}^{2} \end{array}\right) \end{align*} となる。