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A*triangular number* is a number of the form
\[
1+2+\cdots+n=\frac{n(n+1)}{2}.
\]
A triangular number can be pictured by a triangle of dots, with $1$ dot in the first row, $2$ dots in the second row, ..., and
$n$ dots in the $n$th row.
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Triangular Numbers in Base 9

A

Prove that $1$, $11$, $111$, $1111$, ... are all triangular numbers in base $9$.

Solution
The proof is by mathematical induction. The first number, $1$,
is triangular: $1=\frac{1(1+1)}{2}$. Each successive number is $9$ times the previous number (which, since we are working
base $9$, puts a $0$ at the right of the number) plus $1$. We claim that the operation of multiplying a number by $9$ and
adding $1$ transforms *any* triangular number into another triangular number. Let's see:
\[
9 \frac{n(n+1)}{2}+1=\frac{9n^2+9n+2}{2}=\frac{(3n+1)(3n+2)}{2}.
\]
This is a triangular number. And that completes the inductive step and the proof.

The operation ``multiply by $9$ and add $1$,'' used to produce new triangular numbers, can be generalized. If $k$ is a positive integer, then multiplying any triangular number by $(2k+1)^2$ and adding $k(k+1)/2$ gives another triangular number. This fact is apparent from the identity \[ (2k+1)^2\frac{n(n+1)}{2}+\frac{k(k+1)}{2}=\frac{[(2k+1)n+k][(2k+1)n+k+1]}{2}. \]

Our solution corresponds to the identity when $k=1$. For $k=2$, the identity shows that multiplying any triangular number by $25$ and adding $3$ gives another triangular number. It follows that $3$, $33$, $333$, $3333$, ... are all triangular numbers in base $25$.