- #1

Math100

- 783

- 220

- Homework Statement
- If ## t_{n} ## denotes the nth triangular number, show that ## t_{n+2k}\equiv t_{n}\pmod {k} ##; hence, ## t_{n} ## and ## t_{n+20} ## must have the same last digit.

- Relevant Equations
- None.

Proof:

Let ## t_{n} ## be the nth triangular number.

Then ## t_{n}=\frac{n^{2}+n}{2} ## for ## n\geq 1 ##.

This means

\begin{align*}

&t_{n+2k}=\frac{(n+2k)^{2}(n+2k)}{2}\\

&=\frac{n^{2}+4kn+4k^{2}+n+2k}{2}\\

&=\frac{n^{2}+n}{2}+(2kn+2k^{2}+k)\\

&\equiv \frac{n^{2}+n}{2}\pmod {k}.\\

\end{align*}

Thus ## t_{n+2k}\equiv t_{n}\pmod {k} ##.

Therefore, ## t_{n+2k}\equiv t_{n}\pmod {k} ## if ## t_{n} ## denotes the nth triangular number.

Let ## t_{n} ## be the nth triangular number.

Then ## t_{n}=\frac{n^{2}+n}{2} ## for ## n\geq 1 ##.

This means

\begin{align*}

&t_{n+2k}=\frac{(n+2k)^{2}(n+2k)}{2}\\

&=\frac{n^{2}+4kn+4k^{2}+n+2k}{2}\\

&=\frac{n^{2}+n}{2}+(2kn+2k^{2}+k)\\

&\equiv \frac{n^{2}+n}{2}\pmod {k}.\\

\end{align*}

Thus ## t_{n+2k}\equiv t_{n}\pmod {k} ##.

Therefore, ## t_{n+2k}\equiv t_{n}\pmod {k} ## if ## t_{n} ## denotes the nth triangular number.

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