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
can be represented as:
$T_1 = 3i + 1$
$T_2 = T_1 + 3i + 2$
$T_2 = 3i + 1 + 3i + 2 = 6i + 3 = 3(2i+1)$
$T_3 = T_1 + T_2 + 3i + 3$
$T_3 = 3i + 1 + 3i + 2 + 3i + 3 = 9i + 6 = 3(3i+2)$
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
pattern 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.