I am reading through Sierpinski's Elementary Theory of Numbers and i have come across the fact that all numbers of the form 21, 2211, 222111, ... are triangular. I recognize the pattern (6*7)/2, (66*67)/2, (666*667)/2, are all of the proper form, but Professor Sierpinski states:
"We leave the simple proof of this fact to the reader."
I'm not sure how to proceed with this simple proof. Any suggestions?
On
An integer $n$ is a triangular number if and only if $8n + 1$ is a square.
The numbers in question are of the form $n=2\cdot 10^k u +u$ where $u=\dfrac{10^k-1}{9}$.
This gives $n=18u^2+3u$ and so $8n+1=(12u+1)^2$ is a square.
Indeed, then $n$ is the $6u$-th triangular number, $\dfrac{6u(6u+1)}{2}$.
The numbers of the form $6666$ are $\frac 13(2\cdot 10^n-2)$ and the ones of the form $666667$ are $\frac 13(2\cdot 10^n+1)$ To form the triangular number from them you get $$\begin {align}\frac 12\cdot \frac 13(2\cdot 10^n-2)\cdot \frac 13(2\cdot 10^n+1)&=\frac 1{18}(4\cdot 10^{2n}-2\cdot 10^n-2)\\ &=\frac 19(2\cdot 10^{2n}-10^n-1)\\ &=2\frac{10^{2n}-1}9-\frac {10^n-1}9 \end {align}$$ The first gives you a run of $2n\ 2$s and the second subtracts $1$ from the last $n$ of them, giving a number with $n\ 2$s followed by $n\ 1$s
You should actually do the proof in the opposite order, but all the steps are reversible.