If n is Triangular then so is 9n + 1

Theorem

Let $n$ be a triangular number.

Then $9 n + 1$ is also triangular.

Proof

Let $n$ be triangular.

Then $\exists k \in \Z: n = \dfrac {k \left({k+1}\right)} 2$.

So:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 9 n + 1\)	\(=\)	\(\displaystyle 9 \frac {k \left({k+1}\right)} 2 + 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {9 k^2 + 9 k + 2} 2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {\left({3 k + 1}\right) \left({3 k + 2}\right)} 2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

which is triangular.

$\blacksquare$