Product of Consecutive Triangular Numbers

Theorem

$2 T_n T_{n - 1} = T_{n^2 - 1}$

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

$\ds 2 T_n T_{n - 1}$

$=$

$\ds 2 \paren {\frac {n \paren {n + 1} } 2} \paren {\frac {\paren {n - 1} n} 2}$

$\ds $

$=$

$\ds \frac {\paren {n^2 + n} \paren {n^2 - n} } 2$

$\ds $

$=$

$\ds \frac {n^4 - n^2} 2$

$\ds $

$=$

$\ds \frac {n^2 \paren {n^2 - 1} } 2$

$\ds $

$=$

$\ds T_{n^2 - 1}$

$\blacksquare$

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