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

[Math100]

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.

 

[fresh_42]

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.

There is a typo on the numerator of the first quotient, a missing +.

And why do $t_n$ and $t_{n+20}$ have the same last digit?

 

[Math100]

There is a typo on the numerator of the first quotient, a missing +.

And why do $t_n$ and $t_{n+20}$ have the same last digit?

I got the typo part. For the second part, is it because $t_{n+20}\equiv t_{n}\pmod {10}$?

 

[fresh_42]

I got the typo part. For the second part, is it because $t_{n+20}\equiv t_{n}\pmod {10}$?

Essentially, yes. I would approach the question otherwise, step by step. You can always take the definitions! We just have shown that
$$
t_n \equiv t_{n+20} \pmod {20}
$$
This means that $t_n -t_{n+20} \equiv 0 \pmod {20},$ i.e. $20\,|\,(t_n -t_{n+20}).$ Now if $20$ divides that number, then $10$ divides it for sure. So $t_n -t_{n+20}= 10\cdot q$ for some integer $q$. Now you can consider the last digit. E.g. write $t_n=a_m\cdot 10^m+a_{m-1}\cdot 10^{m-1}+\ldots+a_1\cdot 10+a_0$ and $t_{n+20}=b_k\cdot 10^k+b_{k-1}\cdot 10^{k-1}+\ldots+b_1\cdot 10+b_0$ and show us that $a_0=b_0.$

It might not be the shortest way to go down to the definitions but you can practice proof writing.

 

[Delta2]

Essentially, yes. I would approach the question otherwise, step by step. You can always take the definitions! We just have shown that
$$
t_n \equiv t_{n+20} \pmod {20}
$$
This means that $t_n -t_{n+20} \equiv 0 \pmod {20},$ i.e. $20\,|\,(t_n -t_{n+20}).$ Now if $20$ divides that number, then $10$ divides it for sure. So $t_n -t_{n+20}= 10\cdot q$ for some integer $q$. Now you can consider the last digit. E.g. write $t_n=a_m\cdot 10^m+a_{m-1}\cdot 10^{m-1}+\ldots+a_1\cdot 10+a_0$ and $t_{n+20}=b_k\cdot 10^k+b_{k-1}\cdot 10^{k-1}+\ldots+b_1\cdot 10+b_0$ and show us that $a_0=b_0.$

It might not be the shortest way to go down to the definitions but you can practice proof writing.

Er sorry it might be just because I 've just woke up but we have proved $t_{n+2k}=t_n \pmod k$ if we apply this for k=10 we get $t_{n+20}=t_n \pmod {10}$ and NOT $t_{n+20}=t_n \pmod {20}$...

 

[WWGD]

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.

It seems you did not expand the product $(n^2+4nk+4k^2)(n+2k)/2$, from the top line to the 2nd one.. Or did you simplify in some way?

 

[Delta2]

It seems you did not expand the product $(n^2+4nk+4k^2)(n+2k)/2$, from the top line to the 2nd one.. Or did you simplify in some way?

There is a typo there, it is a sum not a product of those two terms.

 

[fresh_42]

Er sorry it might be just because I 've just woke up but we have proved $t_{n+2k}=t_n \pmod k$ if we apply this for k=10 we get $t_{n+20}=t_n \pmod {10}$ and NOT $t_{n+20}=t_n \pmod {20}$...

Yes, you are right. For some reason I thought we had proven $t_{n}\equiv t_{n+2k}\pmod{2k}$ although I carefully read the proof. We have a proverb for this: I didn't see the forest due to all the trees.

Hence, sorry @Math100, your answer in post #3 was just perfect. I made a mistake. Forget post #4.