Hexagonal Number as 4 times Triangular Number plus n
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Theorem
Let $H_n$ be the $n$th hexagonal number.
Then:
- $H_n = 4 T_{n - 1} + n$
where $T_{n - 1}$ is the $n - 1$th triangular number.
Proof
| \(\ds H_n\) | \(=\) | \(\ds n \paren {2 n - 1}\) | Closed Form for Hexagonal Numbers | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {2 n \paren {2 n - 1} } 2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {2 n \paren {2 n - 2 + 1} } 2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 4 \frac {n \paren {n - 1} } 2 + \frac {2 n} 2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 4 T_{n - 1} + n\) | Closed Form for Triangular Numbers |
Hence the result.
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $45$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $45$