Tetrahedral Number as Sum of Squares

Theorem

 * $H_n = \displaystyle \sum_{k \mathop = 0}^{n / 2} \left({n - 2 k}\right)^2$

where $H_n$ denotes the $n$th tetrahedral number.

Proof
Let $n$ be even such that $n = 2 m$.

We have:

Let $n$ be odd such that $n = 2 m + 1$.

We have: