Sum of Consecutive Triangular Numbers is Square

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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:

$T_{n - 1} = \dfrac {\paren {n - 1} n} 2$
$T_n = \dfrac {n \paren {n + 1} } 2$


So:

\(\ds T_{n - 1} + T_n\) \(=\) \(\ds \frac {\paren {n - 1} n} 2 + \frac {n \paren {n + 1} } 2\)
\(\ds \) \(=\) \(\ds \frac {\paren {n - 1 + n + 1} n} 2\)
\(\ds \) \(=\) \(\ds \frac {2 n^2} 2\)
\(\ds \) \(=\) \(\ds n^2\)

$\blacksquare$


Visual Demonstration

SumOfConsecutiveTriangleNumbers.png

$\blacksquare$


Historical Note

According to David M. Burton, in his Elementary Number Theory, revised ed. of $1980$, this result has been attributed to Plutarch, circa $100$ C.E.


Sources

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