Se definirmos a sequência $x_n$ de forma que:
\[ \begin{cases} x_1 = k \\~\\ x_n = c\cdot x_{n-1} +b^{n-1}, n\geq2 \end{cases} \]Com $k,c,b\in\mathbb{R}$ e $c\neq 0$.
Temos que:
\[ \begin{align*} &x_n = c\cdot x_{n-1} +b^{n-1} \\~\\ \iff &\dfrac{x_n}{c^n} = \dfrac{x_{n-1}}{c^{n-1}} + \dfrac{1}{c}\cdot\left(\dfrac{b}{c}\right)^{n-1} \end{align*}\]Definindo $ a_n = \dfrac{x_n}{c^n}$, temos que:
\[ \begin{cases} a_1 = \dfrac{x_1}{c} = \dfrac{k}{c} \\~\\ a_n = a_{n-1} + \dfrac{1}{c}\cdot\left(\dfrac{b}{c}\right)^{n-1}, n\geq 2 \end{cases} \]Listando mais elementos dessa sequência:
\[ \begin{cases} a_1 = \dfrac{k}{c} \\~\\ a_2 = a_1 + \dfrac{1}{c}\cdot\left(\dfrac{b}{c}\right) \\~\\ \cdots \\~\\ a_{n-1} = a_{n-2} + \dfrac{1}{c}\cdot\left(\dfrac{b}{c}\right)^{n-2} \\~\\ a_n = a_{n-1} + \dfrac{1}{c}\cdot\left(\dfrac{b}{c}\right)^{n-1} \end{cases} \]Somando todas as linhas:
\[ \begin{cases} \cancel{a_1} = \dfrac{k}{c} \\~\\ \cancel{a_2} = \cancel{a_1} + \dfrac{1}{c}\cdot\left(\dfrac{b}{c}\right) \\~\\ \cancel{\cdots} \\~\\ \cancel{a_{n-1}} = \cancel{a_{n-2}} + \dfrac{1}{c}\cdot\left(\dfrac{b}{c}\right)^{n-2} \\~\\ a_n = \cancel{a_{n-1}} + \dfrac{1}{c}\cdot\left(\dfrac{b}{c}\right)^{n-1} \end{cases} \]Temos que :
\[ \begin{align*} a_n &= \dfrac{k}{c} +\dfrac{1}{c}\cdot\displaystyle\sum_{i=1}^{n-1} \left(\dfrac{b}{c}\right)^{i} \\~\\ &= \dfrac{k}{c} +\dfrac{1}{c}\cdot\left[\dfrac{c\cdot\left(\frac{b}{c}\right)^n -b}{b-c}\right] \end{align*} \]Por definição $x_n = a_n\cdot c^n$, portanto:
\[ \begin{align*} x_n &= \dfrac{k}{c} +\dfrac{1}{c}\cdot\left[\dfrac{c\cdot\left(\frac{b}{c}\right)^n -b}{b-c}\right] \end{align*} \]