A triangular number is a number of the form $n(n+1)/2$, where $n$ is a positive integer. Prove that $1$, $11$, $111$, $1111$, $\ldots$ 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 puts a $0$ at the right) 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 proof.

The operation ``multiply by $9$ and add $1$,'' used to produce new triangular numbers, can be generalized. If $k$ is a positive integer, multiplying any triangular number by $(2k+1)^2$ and adding $k(k+1)/2$ gives another triangular number. This is because of the identity
\[
(2k+1)^2\frac{n(n+1)}{2}+\frac{k(k+1)}{2}=\frac{[(2k+1)n+k][(2k+1)n+k+1]}{2}.
\]
For example (with $k=2$), multiplying any triangular number by $25$ and adding $3$ gives another triangular number. It follows that $3$, $33$, $333$, $3333$, $\ldots$ are all triangular numbers in base $25$.

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