Sum of Arithmetic-Geometric Sequence/Proof 1

Proof
Proof by induction:

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

Basis for the Induction
$P \left({1}\right)$ is the case:

demonstrating that $P(1)$ holds.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({m}\right)$ is true, where $m \ge 1$, then it logically follows that $P \left({m+1}\right)$ is true.

So this is our induction hypothesis:
 * $\displaystyle \sum_{k \mathop = 0}^{m - 1} \left({a + k d}\right) r^k = \frac {a \left({1 - r^m}\right)} {1 - r} + \frac {r d \left({1 - m r^{m - 1} + \left({m - 1}\right) r^m}\right)} {\left({1 - r}\right)^2}$

Then we need to show:
 * $\displaystyle \sum_{k \mathop = 0}^m \left({a + k d}\right) r^k = \frac {a \left({1 - r^{m+1} }\right)} {1 - r} + \frac {r d \left({1 - \left({m + 1}\right) r^m + m r^{m + 1}}\right)} {\left({1 - r}\right)^2}$

Induction Step
This is our induction step:

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

Therefore:
 * $\displaystyle \forall n \in \N_{> 0}: \sum_{k \mathop = 0}^{n - 1} \left({a + k d}\right) r^k = \frac {a \left({1 - r^n}\right)} {1 - r} + \frac {r d \left({1 - n r^{n - 1} + \left({n - 1}\right) r^n}\right)} {\left({1 - r}\right)^2}$