Tetrahedral Number as Sum of Squares

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Theorem

$H_n = \ds \sum_{k \mathop = 0}^{n / 2} \paren {n - 2 k}^2$

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


Examples

\(\ds H_1\) \(=\) \(\, \ds 1 \, \) \(\, \ds = \, \) \(\ds 1^2\)
\(\ds H_2\) \(=\) \(\, \ds 4 \, \) \(\, \ds = \, \) \(\ds 2^2\)
\(\ds H_3\) \(=\) \(\, \ds 10 \, \) \(\, \ds = \, \) \(\ds 1^2 + 3^2\)
\(\ds H_4\) \(=\) \(\, \ds 20 \, \) \(\, \ds = \, \) \(\ds 2^2 + 4^2\)
\(\ds H_5\) \(=\) \(\, \ds 35 \, \) \(\, \ds = \, \) \(\ds 1^2 + 3^2 + 5^2\)
\(\ds H_6\) \(=\) \(\, \ds 56 \, \) \(\, \ds = \, \) \(\ds 2^2 + 4^2 + 6^2\)
\(\ds H_7\) \(=\) \(\, \ds 84 \, \) \(\, \ds = \, \) \(\ds 1^2 + 3^2 + 5^2 + 7^2\)

... and so on.


Proof

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

We have:

\(\ds \sum_{k \mathop = 0}^{n / 2} \paren {n - 2 k}^2\) \(=\) \(\ds \sum_{k \mathop = 0}^m \paren {2 m - 2 k}^2\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^m \paren {2 k}^2\) Permutation of Indices of Summation
\(\ds \) \(=\) \(\ds \frac {2 m \paren {m + 1} \paren {2 m + 1} } 3\) Sum of Sequence of Even Squares
\(\ds \) \(=\) \(\ds \frac {2 m \paren {2 m + 1} \paren {2 m + 2} } 6\)
\(\ds \) \(=\) \(\ds H_{2 m}\) Closed Form for Tetrahedral Numbers


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

We have:

\(\ds \sum_{k \mathop = 0}^{n / 2} \paren {n - 2 k}^2\) \(=\) \(\ds \sum_{k \mathop = 0}^m \paren {2 m + 1 - 2 k}^2\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^m \paren {2 k + 1}^2\) Permutation of Indices of Summation
\(\ds \) \(=\) \(\ds \frac {\paren {m + 1} \paren {2 m + 1} \paren {2 m + 3} } 3\) Sum of Sequence of Odd Squares: Formulation 1
\(\ds \) \(=\) \(\ds \frac {\paren {2 m + 1} \paren {2 m + 2} \paren {2 m + 3} } 6\)
\(\ds \) \(=\) \(\ds H_{2 m + 1}\) Closed Form for Tetrahedral Numbers

$\blacksquare$

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