(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}+(n-1)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、
\[ \frac{a_{n+1}}{p^{n+1}}=\frac{a_{n}}{p^{n}}+\frac{q(n)}{p^{n+1}} \]

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(n) \end{align*}

(3)

両辺に\(\prod_{i=1}^{n}p^{-1}(i)\)をかけると、

\[ a_{n+1}\prod_{i=1}^{n}p^{-1}(i)=a_{n}\prod_{i=1}^{n-1}p^{-1}(i)+q\prod_{i=1}^{n}p^{-1}(i) \]

これを解くと、
\begin{align*} a_{n} & =\left(\prod_{i=1}^{n-1}p(i)\right)\left(a_{1}+\sum_{j=1}^{n-1}\left(a_{j+1}\prod_{i=1}^{j}p^{-1}(i)-a_{j}\prod_{i=1}^{j-1}p^{-1}(i)\right)\right)\\ & =\left(\prod_{i=1}^{n-1}p(i)\right)\left(a_{1}+q\sum_{j=1}^{n-1}\left(\prod_{i=1}^{j}p^{-1}(i)\right)\right)\\ & =a_{1}\prod_{i=1}^{n-1}p(i)+q\sum_{j=1}^{n-1}\left(\prod_{i=j+1}^{n-1}p(i)\right) \end{align*}

(4)

両辺に\(\prod_{i=1}^{n}p^{-1}(i)\)をかけると、
\[ a_{n+1}\prod_{i=1}^{n}p^{-1}(i)=a_{n}\prod_{i=1}^{n-1}p^{-1}(i)+q(n)\prod_{i=1}^{n}p^{-1}(i) \]

これを解くと、
\begin{align*} a_{n} & =\left(\prod_{i=1}^{n-1}p(i)\right)\left(a_{1}+\sum_{j=1}^{n-1}\left(a_{j+1}\prod_{i=1}^{j}p^{-1}(i)-a_{j}\prod_{i=1}^{j-1}p^{-1}(i)\right)\right)\\ & =\left(\prod_{i=1}^{n-1}p(i)\right)\left(a_{1}+\sum_{j=1}^{n-1}q(j)\prod_{i=1}^{j}p^{-1}(i)\right)\\ & =a_{1}\prod_{i=1}^{n-1}p(i)+\sum_{j=1}^{n-1}q(j)\prod_{i=j+1}^{n-1}p(i) \end{align*}