Sum of Arithmetic-Geometric Sequence/Proof 1

Proof
Proof by induction:

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
 * $\ds \sum_{k \mathop = 0}^{n - 1} \paren {a + k d} r^k = \frac {a \paren {1 - r^n} } {1 - r} + \frac {r d \paren {1 - n r^{n - 1} + \paren {n - 1} r^n} } {\paren {1 - r}^2}$

Basis for the Induction
$\map P 1$ is the case:

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:
 * $\ds \sum_{k \mathop = 0}^{m - 1} \paren {a + k d} r^k = \frac {a \paren {1 - r^m} } {1 - r} + \frac {r d \paren {1 - m r^{m - 1} + \paren {m - 1} r^m} } {\paren {1 - r}^2}$

Then we need to show:
 * $\ds \sum_{k \mathop = 0}^m \paren {a + k d} r^k = \frac {a \paren {1 - r^{m + 1} } } {1 - r} + \frac {r d \paren {1 - \paren {m + 1} r^m + m r^{m + 1} } } {\paren {1 - r}^2}$

Induction Step
This is our induction step:

So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\ds \forall n \in \N_{> 0}: \sum_{k \mathop = 0}^{n - 1} \paren {a + k d} r^k = \frac {a \paren {1 - r^n} } {1 - r} + \frac {r d \paren {1 - n r^{n - 1} + \paren {n - 1} r^n} } {\paren {1 - r}^2}$