# Closed Form for Pentatope Numbers

## Theorem

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

$P_n = \dfrac {n \paren {n + 1} \paren {n + 2} \paren {n + 3} } {24}$

## Proof

 $\ds P_n$ $=$ $\ds \sum_{k \mathop = 1}^n T_k$ Definition of Pentatope Number $\ds$ $=$ $\ds \sum_{k \mathop = 1}^n \frac {n \paren {n + 1} \paren {n + 2} } 6$ Closed Form for Tetrahedral Number‎s $\ds$ $=$ $\ds \sum_{k \mathop = 1}^n \frac {\paren {n^3 + 3 n^2 + 2 n} } 6$ $\ds$ $=$ $\ds \frac 1 6 \sum_{k \mathop = 1}^n n^3 + \frac 1 2 \sum_{k \mathop = 1}^n n^2 + \frac 1 3 \sum_{k \mathop = 1}^n n$ Summation is Linear $\ds$ $=$ $\ds \frac 1 6 \sum_{k \mathop = 1}^n n^3 + \frac 1 2 \sum_{k \mathop = 1}^n n^2 + \frac 1 3 \frac {n \paren {n + 1} } 2$ Closed Form for Triangular Numbers $\ds$ $=$ $\ds \frac 1 6 \sum_{k \mathop = 1}^n n^3 + \frac 1 2 \frac {n \paren {n + 1} \paren {2 n + 1} } 6 + \frac 1 3 \frac {n \paren {n + 1} } 2$ Sum of Sequence of Squares $\ds$ $=$ $\ds \frac 1 6 \dfrac {n^2 \paren {n + 1}^2} 4 + \frac 1 2 \frac {n \paren {n + 1} \paren {2 n + 1} } 6 + \frac 1 3 \frac {n \paren {n + 1} } 2$ Sum of Sequence of Cubes $\ds$ $=$ $\ds \dfrac {n^2 \paren {n + 1}^2 + 2 n \paren {n + 1} \paren {2 n + 1} + 4 n \paren {n + 1} } {24}$ putting over a common denominator $\ds$ $=$ $\ds \dfrac {n \paren {n + 1} \paren {n \paren {n + 1} + 2 \paren {2 n + 1} + 4} } {24}$ extracting $n \paren {n + 1}$ as a factor $\ds$ $=$ $\ds \dfrac {n \paren {n + 1} \paren {n^2 + 5 n + 6} } {24}$ simplifying $\ds$ $=$ $\ds \dfrac {n \paren {n + 1} \paren {n + 2} \paren {n + 3} } {24}$ factorising

$\blacksquare$