Summation Formula for Polygonal Numbers

Theorem
Let $\map P {k, n}$ be the $n$th $k$-gonal number.

Then:
 * $\ds \map P {k, n} = \sum_{j \mathop = 1}^n \paren {\paren {k - 2} \paren {j - 1} + 1}$

Proof
We have that:

$\map P {k, n} = \begin{cases} 0 & : n = 0 \\ \map P {k, n - 1} + \paren {k - 2} \paren {n - 1} + 1 & : n > 0 \end{cases}$

Proof by induction:

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

Basis for the Induction
$\map \Pi 1$ is the statement that $\map P {k, 1} = 1$.

This follows directly from:

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map \Pi r$ is true, where $r \ge 1$, then it logically follows that $\map \Pi {r + 1}$ is true.

So this is our induction hypothesis:


 * $\ds \map P {k, r} = \sum_{j \mathop = 1}^r \paren {\paren {k - 2} \paren {j - 1} + 1}$

Then we need to show:


 * $\ds \map P {k, r + 1} = \sum_{j \mathop = 1}^{r + 1} \paren {\paren {k - 2} \paren {j - 1} + 1}$

Induction Step
This is our induction step:

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

Therefore:
 * $\ds \forall n \in \N: \map P {k, n} = \sum_{j \mathop = 1}^n \paren {\paren {k - 2} \paren {j - 1} + 1}$