Theorem: Sum of Arithmetic-Geometric Sequence

Let $\sequence {a_k}$ be an arithmetic-geometric sequence defined as:
:$a_k = \paren {a + k d} r^k$ for $k = 0, 1, 2, \ldots, n - 1$
Then its closed-form expression is:
:$\ds \sum_{k \mathop = 0}^{n - 1} \paren {a + k d} r^k = \frac {a \paren {1 - r^n} } {1 - r} + \frac {r d \paren {1 - n r^{n - 1} + \paren {n - 1} r^n} } {\paren {1 - r}^2}$

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