Proving that 2 out of every 3 triangular numbers are divisible by 3 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.
 A: Your proof is specific to $1,2,3$.  What you should do is use the fact that $T_k=\frac 12k(k+1)$.  Now you can work $\bmod 3$ and just point out that if $k \equiv 1 \pmod 3, T_k \equiv 1 \pmod 3$ as well because $k+1 \equiv 2 \pmod 3$.  If $k \equiv 2$ or $3 \pmod 3$, one of the factors is a multiple of $3$ so $T_k$ is.
A: 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.
A: You are quite correct. You can simply summarize it as:
The $n_{th}$ triangular number is defined as $T_n=\dfrac{n(n+1)}{2}$
Now for any number $n$, there can be only three cases, that is, $n\equiv 0\space\text {(mod 3)},\space n\equiv 1\space\text {(mod 3) and } n\equiv 2\space\text{(mod 3)}$
Now in only one of these cases, that is, $n\equiv 1\space\text {(mod 3)}$, we have $T_n\not\equiv 0\space\text{(mod 3)}$
$\therefore$ Every $2$ of $3$ consecutive triangular numbers will be divisible by $3$.
A: You could try and pin down your observation by observing that a formula for triangular numbers is,
$$T_n=\frac{1}{2}(n)(n+1)$$
such that $T_1=1, T_2=3, T_3=6, T_4=10, ...$
Proving the formula for triangular numbers is not difficult. Ask if you want it.
All numbers give a remainder of either 0, 1 or 2 when divided by 3.
That is, they are of one of the forms $3m$ or $3m+1$ or $3m+2$ for integer $m$.
$ T_{3m}=\frac{1}{2}(3m)(3m+1)$ which is divisible by 3
$ T_{3m+1}=\frac{1}{2}(3m+1)(3m+2)$ which is NOT divisible by 3
$ T_{3m+2}=\frac{1}{2}(3m+2)(3m+3)$ which is divisible by 3
Comments or requests for clarification welcome...
A: Your proof is not quite correct, or perhaps not complete. Particularly, you need a lot more justification in this step:

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 $i$.

For example, in the case $i = 2$, $3 \cdot 2 + 1$ is not a triangular number. This is perhaps in part due to some confusion with how you've phrased things. I suspect you meant to use a mixture of "for all" and "there exists".
Here's how you could give a more complete proof from the inductive definition of $T_i$, not using the formula.
Claim: for all $i \ge 0$, $T_{3i + 1} = 3k + 1$ for some $k$, $T_{3i + 2} = 3n$ for some $n$, and $T_{3i + 3} = 3m$ for some $m$.
Proof: By induction on $i$. In the case $i = 0$, this is true, by taking $k = 0$, $n = 1$, $m = 2$.
Now suppose that the claim holds for $i$, consider the claim for $i + 1$.
We have that
$T_{3(i + 1) + 1} = T_{3i + 3} + 3i + 4 = 3m' + 3i + 4 = 3(m' + i + 1) + 1$ for some $m'$. So take $k = m' + i + 1$ to fulfil the first part of the claim.
Continuing on,
$T_{3(i + 1) + 2} = T_{3i + 4} + 3i + 5 = 3k + 1 + 3i + 5 = 3(k + i + 2)$. So take $n = k + i + 2$ to fulfil the second part of the claim.
Lastly,
$T_{3(i + 1) + 3} = T_{3i + 5} + 3i + 6 = 3n + 3i + 6 = 3(n + i + 2)$. So take $m = n + i + 2$ to fulfil the last part of the claim.
This proves exactly what you wanted to show.
