Repunit in Base 9 is Triangular

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $m$ be a repunit base $9$.

Then $m$ is a triangular number.


Proof

Let $m$ be a repunit base $9$ with $n$ digits.

We have:

\(\ds m\) \(=\) \(\ds \sum_{k \mathop = 0}^{n - 1} 9^k\) Basis Representation Theorem
\(\ds \) \(=\) \(\ds \dfrac {9^n - 1} {9 - 1}\) Sum of Geometric Sequence
\(\ds \) \(=\) \(\ds \dfrac {3^{2 n} - 1} 8\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {3^n - 1} \paren {3^n + 1} } 8\) Difference of Two Squares
\(\ds \) \(=\) \(\ds \dfrac {\paren {\paren {2 s + 1} - 1} \paren {\paren {2 s + 1} + 1} } 8\) $3^n$ is odd, and so $3^n = 2 s + 1$ for some integer $s$
\(\ds \) \(=\) \(\ds \dfrac {2 s \paren {2 s + 2} } 8\) simplifying
\(\ds \) \(=\) \(\ds \dfrac 1 2 \dfrac {2 s} 2 \dfrac {2 s + 2} 2\)
\(\ds \) \(=\) \(\ds \dfrac {s \paren {s + 1} } 2\)

The result follows from Closed Form for Triangular Numbers.

$\blacksquare$


Sources