Editing Tetrahedral number (section)

== Formula ==
{{Pascal_triangle_simplex_numbers.svg}}

The formula for the {{mvar|n}}th tetrahedral number is represented by the 3rd [[rising factorial]] of {{mvar|n}} divided by the [[factorial]] of 3:
:<math>Te_n= \sum_{k=1}^n T_k = \sum_{k=1}^n \frac{k(k+1)}{2} = \sum_{k=1}^n \left(\sum_{i=1}^k i\right)=\frac{n(n+1)(n+2)}{6} = \frac{n^{\overline 3}}{3!}</math>

The tetrahedral numbers can also be represented as [[binomial coefficient]]s:
:<math>Te_n=\binom{n+2}{3}.</math>
Tetrahedral numbers can therefore be found in the fourth position either from left or right in [[Pascal's triangle]].

===Proofs of formula===
This proof uses the fact that the {{mvar|n}}th triangular number is given by
:<math>T_n=\frac{n(n+1)}{2}.</math>
It proceeds by [[Mathematical induction|induction]].

;Base case
:<math>Te_1 = 1 = \frac{1\cdot 2\cdot 3}{6}.</math>

;Inductive step
:<math>\begin{align} 
Te_{n+1} \quad 
&= Te_n + T_{n+1} \\
&= \frac{n(n+1)(n+2)}{6} + \frac{(n+1)(n+2)}{2}  \\
&= (n+1)(n+2)\left(\frac{n}{6}+\frac{1}{2}\right) \\
&= \frac{(n+1)(n+2)(n+3)}{6}.
\end{align}</math>

The formula can also be proved by [[Gosper's algorithm]].

===Recursive relation===
Tetrahedral and triangular numbers are related through the recursive formulas

:<math>\begin{align}
& Te_n = Te_{n-1} + T_n &(1)\\
& T_n = T_{n-1} + n &(2)
\end{align}</math>

The equation <math>(1)</math> becomes

:<math>\begin{align}
& Te_n = Te_{n-1} + T_{n-1} + n
\end{align}</math>

Substituting <math>n-1</math> for <math>n</math> in equation <math>(1)</math>

:<math>\begin{align}
& Te_{n-1} = Te_{n-2} + T_{n-1}
\end{align}</math>

Thus, the <math>n</math>th tetrahedral number satisfies the following recursive equation

:<math>\begin{align}
& Te_{n} = 2Te_{n-1} - Te_{n-2} + n
\end{align}</math>