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
- 1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $1$: Some Preliminary Considerations: $1.3$ Early Number Theory: Problems $1.3$: $1 \ \text {(d)}$