Closed Form for Polygonal Numbers

Theorem

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

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

Hence the closed-form expression for $P \left({k, n}\right)$ is:

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

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:

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

Basis for the Induction

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

This follows directly from:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle P \left({k, 1}\right)\)	\(=\)	\(\displaystyle P \left({k, 0}\right) + \left({k-2}\right) \left({0}\right) + 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle P \left({k, r+1}\right)\)	\(=\)	\(\displaystyle P \left({k, r}\right) + \left({k-2}\right) r + 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{j=1}^r \left({\left({k-2}\right)\left({j-1}\right) + 1}\right) + \left({k-2}\right) r + 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from the induction hypothesis
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{j=1}^{r+1} \left({\left({k-2}\right)\left({j-1}\right) + 1}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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

Therefore:

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

$\blacksquare$

Proof of Closed Form

$\displaystyle \sum_{j=1}^n \left({\left({k-2}\right)\left({j-1}\right) + 1}\right)$ is the sum of the arithmetic progression where:

• The initial term $a$ is $1$
• the common difference $d$ is $k-2$

The result follows immediately.

$\blacksquare$