# Closed Form for Tetrahedral Numbers

## 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 $\ds$ $=$ $\ds \frac {n \paren {n + 1} } {12} \paren {\paren {2 n + 1} + 3}$ $\ds$ $=$ $\ds \frac {n \paren {n + 1} \paren {n + 2} } 6$

$\blacksquare$