TITLE: Proving that 2 out of every 3 triangular numbers are divisible by 3
QUESTION [6 upvotes]: I am trying to prove the observation that the sequence of triangular numbers are divisible in the repeating pattern of not-divisible, divisible and divisible. I've never done proofs before and I'm also a long-time away from doing any maths. High-school dropout level of maths kind of thing. So I'm not confident about my thinking processes and I would like some help and feedback.

Observed pattern: not divisible, divisible, divisible, not divisible, divisible, divisible, etc
1           = 1   no
1+2         = 3   yes
1+2+3       = 6   yes
1+2+3+4     = 10  no
1+2+3+4+5   = 15  yes
1+2+3+4+5+6 = 21  yes

I reason that a number is divisible by three if it can be written as $3n$, where $n$ is an integer.
Since the triangular numbers, $T_1$, $T_2$ and $T_3$, or
1        = 1
1 2      = 3
1 2 3    = 6 
respectively,
can be represented as:

$T_1 = 3i + 1$ 

$T_2 = T_1 + 3i + 2$ 
$T_2 = 3i + 1 + 3i + 2 = 6i + 3 = 3(2i+1)$


$T_3 = T_1 + T_2 + 3i + 3$ 
$T_3 = 3i + 1 + 3i + 2 + 3i + 3 = 9i + 6 = 3(3i+2)$
where $i=0$,

I conclude that $T_1$ is not divisible by 3 but that $T_2$ and $T_3$ are because they can be expressed in the form $3n$.
If it is correct to conclude this for $T_1$, $T_2$ and $T_3$, where $i=0$, it would be correct to conclude the same for $T_4$, $T_5$ and $T_6$, where $i=1$, and for $T_7$, $T_8$ and $T_9$, where $i=2$ and so on and so on for all integer values of $i$.
Thus, triangular numbers do repeat the 
not divisible  divisible divisiblepattern forever because they repeat the

$3i + 1$ 
$3(2i + 1)$ 
$3(3i + 3)$ 

pattern forever for all integers $i$.

Is my reasoning valid? Thanks for your time.

REPLY [3 votes]: You have a really good insight there. If you learn about residue classes modulo $3$ some day, you might recognize your approach as a result of repeatedly adding the residue classes $1,$ $2,$ and $3$ (aka the residue class of $0$) to the sum.
The only thing you are missing in the sums is a suitable representation for the sum of all the terms before $3i+1.$. When $i=0$ this is not a problem since there are no previous terms and the sum of no terms is $0.$
But for any larger $i$ you should notice that after you add $3i+1$ you have a sum which is larger than just $3i+1.$
But you have shown that the sum of all the previous terms ($1$ to $3i$) is divisible by $3$, so it is $3k$ for some integer $k$.
If you just account for that, and make the induction part of your argument a little more explicit, I think you could have a very nice proof.
The key point is you recognized a useful pattern.