Pentagonal Number as Sum of Triangular Numbers

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Theorem

Let $P_n$ be the $n$th pentagonal number.

Then:

$P_n = T_n + 2 T_{n - 1}$

where $T_n$ is the $n$th triangular number.


Proof

\(\ds T_n + 2 T_{n - 1}\) \(=\) \(\ds \frac {n \paren {n + 1} } 2 + 2 \frac {\paren {n - 1} n} 2\) Closed Form for Triangular Numbers
\(\ds \) \(=\) \(\ds \frac {n^2 + n + 2 \paren {n^2 - n} } 2\)
\(\ds \) \(=\) \(\ds \frac {3 n^2 - n} 2\)
\(\ds \) \(=\) \(\ds \frac {3 n \paren {n - 1} } 2\)
\(\ds \) \(=\) \(\ds P_n\) Closed Form for Pentagonal Numbers

$\blacksquare$


Illustration

PentagonNumberAsSumOfTriangles.png


Sources