> [!NOTE] Theorem > Let $(y_{k})_{k=0}^{\infty}$ be a [[Sequences|sequence]] that satisfies the linear first order [[Recurrence Relation|recurrence relation]] $y_{k}=my_{k-1}+b, \quad \text{for all } k \geq 1$with $m,b\in \mathbb{R}$ and $y_{0}=\bar{y}\in \mathbb{R}.$ > > Then for all $k\geq 1,$ $y_{k} = \begin{cases} > m^{k} \left( \bar{y} - \frac{b}{1-m} \right) + \frac{b}{1-m} &\text{if } m \neq 1,\\ > y_{0} + kb & \text{otherwise.} > \end{cases}$ **Proof**: We can show inductively that for all $k\geq 1,$ $y_{k}=m^{k}\bar{y}+b\sum_{i=0}^{k-1}m^{i}.$Now if $m=1,$ then $\sum_{i=0}^{k-1}m^{i}=k.$ Thus $y_{k}=\bar{y}+kb.$ Otherwise by [[Geometric Series]], $\sum_{i=0}^{k-1} m^{i} = \frac{1-m^{k}}{1-m}$thus $y_{k}=m^{k}\bar{y}+ \frac{b(1-m^{k})}{1-m} = m^{k}\left( \bar{y} - \frac{b}{1-m} \right) + \frac{b}{1-m} .$