Closed Form for Tetrahedral Numbers

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Theorem

The closed-form expression for the $n$th tetrahedral number is:

$H_n = \dfrac {n \paren {n + 1} \paren {n + 2} } 6$


Proof

\(\ds H_n\) \(=\) \(\ds \sum_{k \mathop = 1}^n T_k\) Definition of Tetrahedral Number
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n \frac {n \paren {n + 1} } 2\) Closed Form for Triangular Number‎s
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n \paren {\frac 1 2 n^2 + \frac 1 2 n}\)
\(\ds \) \(=\) \(\ds \frac 1 2 \sum_{k \mathop = 1}^n n^2 + \frac 1 2 \sum_{k \mathop = 1}^n n\) Summation is Linear
\(\ds \) \(=\) \(\ds \frac 1 2 \sum_{k \mathop = 1}^n n^2 + \frac 1 2 T_n\) Definition 2 of Triangular Number
\(\ds \) \(=\) \(\ds \frac 1 2 \frac {n \paren {n + 1} \paren {2 n + 1} } 6 + \frac 1 2 \frac {n \paren {n + 1} } 2\) Sum of Sequence of Squares, Closed Form for Triangular Numbers
\(\ds \) \(=\) \(\ds \frac {n \paren {n + 1} } {12} \paren {\paren {2 n + 1} + 3}\)
\(\ds \) \(=\) \(\ds \frac {n \paren {n + 1} \paren {n + 2} } 6\)

$\blacksquare$


Sources

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