Proof by induction: For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$\displaystyle \sum_{k \mathop = 0}^{n - 1} \paren {a + k d} r^k = \frac {a \paren {1 - r^n} } {1 - r} + \frac {r d \paren {1 - n r^{n - 1} + \paren {n - 1} r^n} } {\paren {1 - r}^2}$ === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^{1 - 1} \paren {a + k d} r^k        | r = \frac {a \paren {1 - r^1} } {1 - r} + \frac {r d \paren {1 - 1 r^{1 - 1} + \paren {1 - 1} r^1} } {\paren {1 - r}^2}       | c =  }} {{eqn | r = a + \frac {r d \paren {1 - 1 + \paren {1 - 1} } } {\paren {1 - r}^2}       | c =  }} {{eqn | r = a       | c =  }} {{end-eqn}} demonstrating that $\map P 1$ holds. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P m$ is true, where $m \ge 1$, then it logically follows that $\map P {m + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \sum_{k \mathop = 0}^{m - 1} \paren {a + k d} r^k = \frac {a \paren {1 - r^m} } {1 - r} + \frac {r d \paren {1 - m r^{m - 1} + \paren {m - 1} r^m} } {\paren {1 - r}^2}$ Then we need to show: :$\displaystyle \sum_{k \mathop = 0}^m \paren {a + k d} r^k = \frac {a \paren {1 - r^{m + 1} } } {1 - r} + \frac {r d \paren {1 - \paren {m + 1} r^m + m r^{m + 1} } } {\paren {1 - r}^2}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^m \paren {a + k d} r^k       | r = \sum_{k \mathop = 0}^{m - 1} \paren {a + k d} r^k + \paren {a + m d} r^m       | c =  }} {{eqn | r = \frac {a \paren {1 - r^m} } {1 - r} + \frac {r d \paren {1 - m r^{m - 1} + \paren {m - 1} r^m} } {\paren {1 - r}^2} + \paren {a + m d} r^m       | c = Induction Hypothesis }} {{eqn | r = \frac {a \paren {1 - r^m} } {1 - r} + \frac {r d \paren {1 - m r^{m - 1} + \paren {m - 1} r^m} } {\paren {1 - r}^2}       | c =  }} {{eqn | o =        | ro= +       | r = \frac {a r^m \paren {1 - r} } {1 - r} + \frac {m d r^m \paren {1 - r}^2} {\paren {1 - r}^2}       | c = common denominator (2 instances) }} {{eqn | r = \frac {a \paren {1 - r^m + r^m \paren {1 - r} } } {1 - r}       | c =  }} {{eqn | o =        | ro= +       | r = \frac {r d \paren {1 - m r^{m - 1} + \paren {m - 1} r^m} + r d \paren {m r^{m - 1} \paren {1 - r}^2} } {\paren {1 - r}^2}       | c = simplifying }} {{eqn | r = \frac {a \paren {1 - r^m + r^m - r^{m + 1} } } {1 - r}       | c =  }} {{eqn | o =        | ro= +       | r = \frac {r d \paren {1 - m r^{m - 1} + m r^m - r^m + m r^{m - 1} - 2 m r^m + m r^{m + 1} } } {\paren {1 - r}^2}       | c = multiplying out }} {{eqn | r = \frac {a \paren {1 - r^{m + 1} } } {1 - r} + \frac {r d \paren {1 - m r^m - r^m + m r^{m + 1} } } {\paren {1 - r}^2}       | c = cancelling out terms }} {{eqn | r = \frac {a \paren {1 - r^{m + 1} } } {1 - r} + \frac {r d \paren {1 - \paren {m + 1} r^m + m r^{m + 1} } } {\paren {1 - r}^2}       | c = simplification }} {{end-eqn}} So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \N_{> 0}: \sum_{k \mathop = 0}^{n - 1} \paren {a + k d} r^k = \frac {a \paren {1 - r^n} } {1 - r} + \frac {r d \paren {1 - n r^{n - 1} + \paren {n - 1} r^n} } {\paren {1 - r}^2}$ {{qed}}