Sum of Consecutive Triangular Numbers

Theorem

The sum of two consecutive triangular numbers is a square number.

Proof

Let $T_{n-1}$ and $T_n$ be two consecutive triangular numbers.

From Closed Form for Triangular Numbers‎, we have:

• $\displaystyle T_{n-1} = \frac {\left({n-1}\right) n} 2$
• $\displaystyle T_n = \frac {n \left({n+1}\right)} 2$

So:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle T_{n-1} + T_n\)	\(=\)	\(\displaystyle \frac {\left({n-1}\right) n} 2 + \frac {n \left({n+1}\right)} 2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {\left({n-1 + n + 1}\right) n} 2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {2n^2} 2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle n^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$