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
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $15$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $15$