If n is Triangular then so is (2m+1)^2 n + m(m+1)/2

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Theorem

Let $n$ be a triangular number.

Let $m \in \Z_{\ge 0}$ be a positive integer.


Then $\paren {2 m + 1}^2 n + \dfrac {m \paren {m + 1} } 2$ is also a triangular number.


Proof

Let $n$ be a triangular number.

Then:

$\exists k \in \Z: n = \dfrac {k \paren {k + 1} } 2$

So:

\(\ds \paren {2 m + 1}^2 n + \dfrac {m \paren {m + 1} } 2\) \(=\) \(\ds \paren {2 m + 1}^2 \dfrac {k \paren {k + 1} } 2 + \dfrac {m \paren {m + 1} } 2\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {2 m + 1}^2 k^2 + \paren {2 m + 1}^2 k + m \paren {m + 1} } 2\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {\paren {2 m + 1} k + m} \paren {\paren {2 m + 1} k + m + 1} } 2\)

which is of the form:

$\dfrac {t \paren {t + 1} } 2$

where $t = \paren {2 m + 1} k + m$.

$\blacksquare$