If n is Triangular then so is 9n + 1

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Theorem

Let $n$ be a triangular number.


Then $9 n + 1$ is also triangular.


Proof

Let $n$ be triangular.

Then:

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

So:

\(\ds 9 n + 1\) \(=\) \(\ds 9 \frac {k \paren {k + 1} } 2 + 1\)
\(\ds \) \(=\) \(\ds \frac {9 k^2 + 9 k + 2} 2\)
\(\ds \) \(=\) \(\ds \frac {\paren {3 k + 1} \paren {3 k + 2} } 2\)

which is triangular.

$\blacksquare$


Historical Note

David M. Burton, in his Elementary Number Theory, revised ed. of $1980$, reports that the result If $n$ is Triangular then so is $9 n + 1$ was published by Leonhard Paul Euler in $1775$.

He published this along with If $n$ is Triangular then so is $25 n + 3$ and If $n$ is Triangular then so is $49 n + 6$.

There is an obvious pattern.


Sources