Closed Form for Polygonal Numbers

Theorem
Let $$P \left({k, n}\right)$$ be the $$n$$th $k$-gonal number.

Then $$P \left({k, n}\right) = \sum_{j=1}^n \left({\left({k-2}\right)\left({j-1}\right) + 1}\right)$$.

Hence $$P \left({k, n}\right) = \frac {n \left({2 + \left({n-1}\right)\left({k-2}\right)}\right)} 2$$.

Proof
We have that:

$$P \left({k, n}\right) = \begin{cases} 0 & : n = 0 \\ P \left({k, n-1}\right) + \left({k-2}\right) \left({n-1}\right) + 1 & : n > 0 \end{cases}$$

Proof of Summation Formula
Proof by induction:

For all $$n \in \N^*$$, let $$P \left({n}\right)$$ be the proposition $$P \left({k, n}\right) = \sum_{j=1}^n \left({\left({k-2}\right)\left({j-1}\right) + 1}\right)$$.

Basis for the Induction

 * $$P(1)$$ is true, as this just says $$P \left({k, 1}\right) = 1$$.

This follows directly from:

$$ $$

This is our basis for the induction.

Induction Hypothesis

 * Now we need to show that, if $$P \left({r}\right)$$ is true, where $$r \ge 2$$, then it logically follows that $$P \left({r+1}\right)$$ is true.

So this is our induction hypothesis:

$$P \left({k, r}\right) = \sum_{j=1}^r \left({\left({k-2}\right)\left({j-1}\right) + 1}\right)$$.

Then we need to show:

$$P \left({k, r+1}\right) = \sum_{j=1}^{r+1} \left({\left({k-2}\right)\left({j-1}\right) + 1}\right)$$.

Induction Step
This is our induction step:

$$ $$ $$

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

Therefore $$\forall n \in \N: P \left({k, n}\right) = \sum_{j=1}^n \left({\left({k-2}\right)\left({j-1}\right) + 1}\right)$$.

Proof of Closed Form
$$\sum_{j=1}^n \left({\left({k-2}\right)\left({j-1}\right) + 1}\right)$$ is the sum of arithmetic progression, where:
 * The initial term $$a$$ is $$1$$;
 * the common difference $$d$$ is $$k-2$$.

The result follows immediately.