Repeating question - did not approve of previous answers?

Question

Imagine an equlateral triangle that is built out of smaller eqalateral triangles. Let's say that the length of 1 side of the smallest triangle is 1. If the length of one side of the larger triangle were 2, then there would be a total of 5 triangles (4 (1) + 1 (2)). If length of one side of large triangle were 3, then there would be 13 total triangles (9 (1) + 3 (2) +1 (3)). Can anyone find a common formula for when the length of one side of the largest triangle is n that determines total number of triangles?

Answer 1

I don't have time to explain how I got there, but:

Sigma (n(3n+2)-n(mod 2))/4

Answer 2

I think the best way to proceed is to separate right-side-up and upside-down triangles.

Right side up:
side 1 triangles: 1 + 2 + 3 + ... + n = n(n + 1)/2
side 2 triangles: every right-side-up triangle above the bottom row can be extended to this, so 1 + 2 + 3 + ... + (n - 1) = n(n - 1)/2
side 3 triangles: every right-side-up triangle above the bottom 2 rows can be extended to this, so 1 + 2 + 3 + ... + (n - 2) = (n - 2)(n - 1)/2
...
side n triangles: just the top 1, 1

So, for right-side-up, there are n(n + 1)(n + 2)/(1*2*3) triangles.

Now look at upside-down.
Side 1: there are n(n - 1)/2
Side 2: There are every upside-down triangle a distance 2 from the outside can be extended into an upside-down triangle of side 2. For n < 4 this is 0, otherwise it is (n - 3)(n - 2)/2
Side 3: There are every upside-down triangle a distance 3 from the outside can be extended into an upside-down triangle of side 3. For n < 6 this is 0, otherwise it is (n - 5)(n - 4)/2

This is all the time I can spend on this now. There is a closed formula for the right-side-up triangles and an easily computable series for the upside-down ones. If I have time, and no one else gets it, I will look later to see if there is a closed formula for the upside-down ones.