Triangular numbers divisible by $3$

I can't understand any of sentences from the images below. Since I don't understand almost every possible lines, I'm very troubled for what I should even ask. But I'll try to.

Firstly, how is it possible for the triangular number with 3k+1 as the last number to be followed by the one with 3k+3 as its last number? As you can see on the example above, 3k = 6, then 3k+1 = 7, which makes sense since the next triangular number for 21 is 21+7, 28. But how could 30, 21+9, where 9 come from 3k+3, be the second next triangular number? Is it perhaps that 3k in 3k+1 is a different number with 3k in 3k+3?

Secondly, how can they be sure they can show the pattern continues just by showing 3 cases?

And, in the last pictures, how does it make sense to say the first two figures are triangular numbers? They just aren't, judging from the look of them? It's not stair-like figure at all. Simply put, I can't believe the last figures are consecutive triangular numbers.  • Please state the source. It will help to understand better. Nov 19 '21 at 20:13

I agree: the sentence "second number after $$3k$$: $$3k + 1 + 2$$, or $$3k + 3$$" is extremely misleading. They change the meaning of $$k$$ mid-sentence.

I think to understand what is going on we should make as clear as possible the difference between the small numbers we add up in each step and the big triangular numbers.

The situation they start from is that for some small number (such as 6) that is itself divisible by 3 we find a big triangular number (in the example: 21) that is also divisible by 3. Now we want to introduce the number $$k$$ to write numbers that are divisible by 3 as 3k. The author of the picture does not specify which of the two they rewrite in this way and this remains unclear forever so let's do it better ourselves.

Let's say the small number (6 in our example) is denoted 3k (so k = 2 in our example) and the big number (21 in our example) I will write 3K (so the big number K is 7 in our example).

Now the next triangular number is formed by adding the next small number (3k + 1) to the big number 3K, so we get $$3K + 3k + 1$$. This can be rewritten as $$3(K + k) + 1$$ so this number is one more than a multiple of 3 and hence not a multiple of 3 itself. It is of the form $$3m + 1$$ (with m = K + k) but saying that it is 3k + 1 as the picture does is extremely misleading.

Now the next triangular number we get by adding the next small number: $$3k + 2$$ and so the next triangular number is $$3K + (3k + 1) + (3k + 2)$$. This can be expanded to $$3(K + k + k) + (1 + 2)$$, a number of the form $$3n + 3$$ (just take $$n = K + k + k$$) and hence divisible by $$3$$ as the picture argues. However calling this number $$3k + 3$$ is serious abuse of notation.

The next triangular number we get by adding the next small number $$3k + 3$$ to the total we already had and hence we end up with $$3(K + k + k) + 3 + (3k + 3) = 3(K + k + k + k) + 6 = 3(K + k + k + k + 2)$$ so again this number is divisible by 3.

I hope this helps!

The final thing you need to understand is that after three steps we are back in the same situation we started with:

The last small number we added (3k + 3) is divisible by 3 and the large triangular number we arrived at, 3(K + k + k + k + 2), also.

Since these two properties was what defined our starting position we can repeat the same reasoning over and over.

the triangular number $$T_n = \frac 12 n(n+1)$$

As n increases, $$T_n$$ will be divisible by 3 when n or (n+1) is a multiple of 3, and not when (n-1) is. Therefore the no-yes-yes pattern.

The formula $$\sum_{j=1}^n j=\frac{n(n+1)}{2}$$ shows that the sum is divisible by $$3$$ , if and only if either $$n$$ or $$n+1$$ is divisible by $$3$$. Only if $$n$$ is of the form $$3k+1$$ , this is not the case. This explains the pattern.