Closed Form for Pentatope Numbers

From ProofWiki
Jump to navigation Jump to search

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

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Closed_Form_for_Pentatope_Numbers&oldid=383030"