Product of Consecutive Triangular Numbers
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Theorem
- $2 T_n T_{n - 1} = T_{n^2 - 1}$
where $T_n$ denotes the $n$th triangular number.
Proof
| \(\ds 2 T_n T_{n - 1}\) | \(=\) | \(\ds 2 \paren {\frac {n \paren {n + 1} } 2} \paren {\frac {\paren {n - 1} n} 2}\) | Closed Form for Triangular Numbers | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\paren {n^2 + n} \paren {n^2 - n} } 2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {n^4 - n^2} 2\) | Difference of Two Squares | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {n^2 \paren {n^2 - 1} } 2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds T_{n^2 - 1}\) | Closed Form for Triangular Numbers |
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $15$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $15$