Square of Odd Multiple of 3 is Difference between Triangular Numbers
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Theorem
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $T_n$ denote the $n$th triangular number.
Let $m = 2 n + 1$ be an odd integer
Then:
- $\paren {3 m}^2 = T_{9 n + 4} - T_{3 n + 1}$
Proof
\(\ds T_{9 n + 4} - T_{3 n + 1}\) | \(=\) | \(\ds \dfrac {\paren {9 n + 4} \paren {9 n + 5} } 2 - \dfrac {\paren {3 n + 1} \paren {3 n + 2} } 2\) | Closed Form for Triangular Numbers | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {81 n^2 + 81 n + 20} - \paren {9 n^2 + 9 n + 2} } 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {72 n^2 + 72 n + 18} 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 36 n^2 + 36 n + 9\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 9 \paren {4 n^2 + 4 n + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 9 \paren {2 n + 1}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 9 m^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {3 m}^2\) |
$\blacksquare$
Sources
- 1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $1$: Some Preliminary Considerations: $1.3$ Early Number Theory: Problems $1.3$: $4$