(1)

両辺を\(p^{n+1}\)で割ると、
\[ \frac{a_{n+1}}{p^{n+1}}=\frac{a_{n}}{p^{n}}+\frac{q}{p^{n+1}} \] これを解くと、
\begin{align*} a_{n} & =p^{n}\left(\frac{a_{1}}{p}+\sum_{k=1}^{n-1}\left(\frac{a_{k+1}}{p^{k+1}}-\frac{a_{k}}{p^{k}}\right)\right)\\ & =p^{n-1}a_{1}+p^{n-2}q\sum_{k=1}^{n-1}\left(\frac{1}{p^{k-1}}\right)\\ & =p^{n-1}a_{1}+q\frac{p^{n-1}-1}{p-1} \end{align*} \(p=1\)のときは等比数列になるので、
\begin{align*} a_{n} & =a_{1}+\sum_{j=1}^{n-1}\left(a_{j+1}-a_{j}\right)\\ & =a_{1}+\left(n-1\right)q \end{align*}

\(a_{c}\)が既知のとき

\begin{align*} a_{n} & =p^{n}\left(\frac{a_{c}}{p^{c}}+\sum_{k=c}^{n-1}\left(\frac{a_{k+1}}{p^{k+1}}-\frac{a_{k}}{p^{k}}\right)\right)\\ & =p^{n}\left(\frac{a_{c}}{p^{c}}+\sum_{k=c}^{n-1}\left(\frac{q}{p^{k+1}}\right)\right)\\ & =p^{n-c}a_{c}+p^{n-1}q\sum_{k=c}^{n-1}\frac{1}{p^{k}}\\ & =p^{n-c}a_{c}+p^{n-1}\frac{q}{p^{c}}\frac{1-\left(\frac{1}{p}\right)^{n-c}}{1-\frac{1}{p}}\\ & =p^{n-c}a_{c}+q\frac{p^{n-c}-1}{p-1} \end{align*} \(p=1\)のときは等比数列になるので、
\begin{align*} a_{n} & =a_{c}+\sum_{j=c}^{n-1}\left(a_{j+1}-a_{j}\right)\\ & =a_{c}+\left(n-c\right)q \end{align*}

(1)別解

特性方程式の解は\(\frac{q}{1-p}\)なので\(p\ne1\)のとき、
\[ a_{n+1}-\frac{q}{1-p}=p\left(a_{n}-\frac{q}{1-p}\right) \] これを解くと、
\begin{align*} a_{n} & =\left(a_{1}-\frac{q}{1-p}\right)p^{n-1}+\frac{q}{1-p}\\ & =p^{n-1}a_{1}+q\frac{p^{n-1}-1}{p-1} \end{align*} \(p=1\)のときも同様にする。

(2)

両辺を\(p^{n+1}\)で割ると、
\[ \frac{a_{n+1}}{p^{n+1}}=\frac{a_{n}}{p^{n}}+\frac{q\left(n\right)}{p^{n+1}} \] これを解くと、
\begin{align*} a_{n} & =p^{n}\left(\frac{a_{1}}{p}+\sum_{k=1}^{n-1}\left(\frac{a_{k+1}}{p^{k+1}}-\frac{a_{k}}{p^{k}}\right)\right)\\ & =p^{n-1}a_{1}+\sum_{k=1}^{n-1}p^{n-1-k}q\left(k\right) \end{align*}

\(a_{c}\)が既知のとき

\begin{align*} a_{n} & =p^{n}\frac{a_{c}}{p^{c}}+p^{n}\sum_{k=c}^{n-1}\left(\frac{a_{k+1}}{p^{k+1}}-\frac{a_{k}}{p^{k}}\right)\\ & =p^{n-c}a_{c}+p^{n}\sum_{k=c}^{n-1}\frac{q\left(k\right)}{p^{k+1}}\\ & =p^{n-c}a_{c}+\sum_{k=c}^{n-1}p^{n-k-1}q\left(k\right) \end{align*}

(3)

両辺に\(\prod_{i=1}^{n}p^{-1}(i)\)をかけると、
\[ a_{n+1}\prod_{i=1}^{n}p^{-1}\left(i\right)=a_{n}\prod_{i=1}^{n-1}p^{-1}\left(i\right)+q\prod_{i=1}^{n}p^{-1}\left(i\right) \] これを解くと、
\begin{align*} a_{n} & =\left(\prod_{i=1}^{n-1}p\left(i\right)\right)\left(a_{1}+\sum_{j=1}^{n-1}\left(a_{j+1}\prod_{i=1}^{j}p^{-1}\left(i\right)-a_{j}\prod_{i=1}^{j-1}p^{-1}\left(i\right)\right)\right)\\ & =\left(\prod_{i=1}^{n-1}p\left(i\right)\right)\left(a_{1}+q\sum_{j=1}^{n-1}\left(\prod_{i=1}^{j}p^{-1}\left(i\right)\right)\right)\\ & =a_{1}\prod_{i=1}^{n-1}p\left(i\right)+q\sum_{j=1}^{n-1}\left(\prod_{i=j+1}^{n-1}p\left(i\right)\right) \end{align*}

\(a_{c}\)が既知のとき

\begin{align*} a_{n} & =\left(\prod_{i=1}^{n-1}p\left(i\right)\right)\left(a_{c}\prod_{i=1}^{c-1}p^{-1}\left(i\right)+\sum_{j=c}^{n-1}\left(a_{j+1}\prod_{i=1}^{j}p^{-1}\left(i\right)-a_{j}\prod_{i=1}^{j-1}p^{-1}\left(i\right)\right)\right)\\ & =\left(\prod_{i=1}^{n-1}p\left(i\right)\right)\left(a_{c}\prod_{i=1}^{c-1}p^{-1}\left(i\right)+q\sum_{j=c}^{n-1}\left(\prod_{i=1}^{j}p^{-1}\left(i\right)\right)\right)\\ & =a_{c}\prod_{i=c}^{n-1}p\left(i\right)+q\sum_{j=c}^{n-1}\left(\prod_{i=j+1}^{n-1}p\left(i\right)\right) \end{align*}

(4)

両辺に\(\prod_{i=1}^{n}p^{-1}\left(i\right)\)をかけると、
\[ a_{n+1}\prod_{i=1}^{n}p^{-1}\left(i\right)=a_{n}\prod_{i=1}^{n-1}p^{-1}\left(i\right)+q\left(n\right)\prod_{i=1}^{n}p^{-1}\left(i\right) \] これを解くと、
\begin{align*} a_{n} & =\left(\prod_{i=1}^{n-1}p\left(i\right)\right)\left(a_{1}+\sum_{j=1}^{n-1}\left(a_{j+1}\prod_{i=1}^{j}p^{-1}\left(i\right)-a_{j}\prod_{i=1}^{j-1}p^{-1}\left(i\right)\right)\right)\\ & =\left(\prod_{i=1}^{n-1}p\left(i\right)\right)\left(a_{1}+\sum_{j=1}^{n-1}q\left(j\right)\prod_{i=1}^{j}p^{-1}\left(i\right)\right)\\ & =a_{1}\prod_{i=1}^{n-1}p\left(i\right)+\sum_{j=1}^{n-1}q\left(j\right)\prod_{i=j+1}^{n-1}p\left(i\right) \end{align*}

\(a_{c}\)が既知のとき

\begin{align*} a_{n} & =\left(\prod_{i=1}^{n-1}p\left(i\right)\right)\left(a_{c}\prod_{i=1}^{c-1}p^{-1}\left(i\right)+\sum_{j=c}^{n-1}\left(a_{j+1}\prod_{i=1}^{j}p^{-1}\left(i\right)-a_{j}\prod_{i=1}^{j-1}p^{-1}\left(i\right)\right)\right)\\ & =\left(\prod_{i=1}^{n-1}p\left(i\right)\right)\left(a_{c}\prod_{i=1}^{c-1}p^{-1}\left(i\right)+\sum_{j=c}^{n-1}q\left(j\right)\prod_{i=1}^{j}p^{-1}\left(i\right)\right)\\ & =\left(a_{c}\prod_{i=c}^{n-1}p\left(i\right)+\sum_{j=c}^{n-1}q\left(j\right)\prod_{i=j+1}^{j}p\left(i\right)\right) \end{align*}