How many triangles will there be after n number of sequences?

I drew two more iterations of the pattern and added 9, then 12 more triangles. Your first two drawings showed 3 then 6. So the first 4 iterations go 3, 6, 9, 12… if that continues, it seems the pattern is adding 3n each time. But you’re asking what the total number of triangles is after n iterations? So then the first four iterations are 4, 10, 19, 31… and you’ll have to give me a moment to come up with the formula for that sequence.

Okay, I’m pretty sure the formula is (3/2)(n^2 + n) + 1. Let me know if I’m wrong and I’ll try again!

@Mariah
I kind of just want the explanation behind it but I’ll give it a look over in a bit to see if it helps me.

Okay, you see what I did though? I determined that I was adding 3n each time, so the series I was trying to find a formula for was 4, 10, 19, 31, 46, 63…etc. It’s difficult to explain how I came up with the formula for it, it’s mostly trial and error, but I’ll try to tell you my thought process. I subtracted 1 from each number and noticed that the resulting series (3, 9, 18, 30, 45, 63) takes the form 3•1, 3•3, 3•6, 3•10, 3•15, 3•21. I decided to focus on finding the pattern in those factors. So the equation will take the form 3( ) + 1, and once I figure out the pattern in 1, 3, 6, 10, 15, 21, that formula will go in the blank. Notice that in that sequence of numbers, for each subsequent number you’re adding 2, then 3, then 4, etc. To the previous number. It took me a while to notice that (n^2 + n)/2 yields the correct results. Figuring that out was mosttly trial and error. So I stuck that into the parenthesis above and simplfied. Hope that helps at all.

And this had better not be homework or the Fluther gods will strike me down.

@Mariah 1, 3, 6, 10… are so-called triangular numbers, usually representing the sum of successive integers:
1 = 1
3 = 1 + 2
6 = 1 + 2 + 3
10 = 1 + 2 + 3 + 4
etc.
The sum of integers from 1 to N is N(N+1)/2 – as you already figured.
For N = 100 the sum is 5,050.

@gasman Ha, would’ve helped to have known that going into this problem! Thanks.

@sunssi Could you please clarify: Is the original triangle isosceles right angle, with a horizontal hypotenuse? (This creates bilateral symmetry about a vertical axis.) Are all triangles congruent?

I’m not sure what you mean by, ”...triangles will start taking place of other triangles…” It looks like the next iteration (as with preceding) adds another 6 triangles; while the one after that adds 8. Do I have the correct method?

I am finding this problem a bit tricky, but here is a way of looking at it. The number of triangles added each time depends on the number of sides that are exposed. The first time there are 3 sides exposed, so 3 new triangles are added. These 3 triangles each have 2 of their sides exposed so the second time 6 triangles are added. At this point, it is clear how things get messy. The 6 triangles have 12 exposed sides, but 3 pairs of them will be used for the same triangle, so that the third time we get 6 + 3 = 9 new triangles. Unless there is a pattern to the number of shared sides, I do not see an immediate solution.