Triangle base division math problem. I might go crazy.?

Yes, This could drive one crazy. My approach is to
count separately the triangle of height 1,2,3,...,n,
where n is the number of base divisions, which also
corresponds to the number of vertical 'layers' of triangles.

With triangles of height 1, each layer has 2 more height 1
triangles than the previous layer. So for the triangle with
3 base divisions there are 1 + 3 + 5 = 9 height 1 triangles.
For a triangle with n base divisions, there will be
1 + 3 + 5 + .... + (2n - 1) = n^2 height 1 triangles.

For triangles of height 2, we'll look at the 3 base division
triangle for some guidance. The bottom 2 layers contain
2 height 2 triangles, and the top 2 layers contains just 1.
For an n base division triangle, there will be
1 + 2 + .... + (n - 1) = n*(n-1)/2 height 2 triangles.

For triangles of height 3, the 4 base division triangle
will have 2 height 3 triangles in the bottom 2 layers
and 1 in the top 2 layers.

The 4 base division triangle contains just 1 height 4 triangle,
namely itself.

So the 4 base division triangle will have
4^2 + [4*(4-1)/2] + 3 + 1 = 16 + 6 + 3 + 1 = 26 triangles.

The 5 base division triangle will have
5^2 + [5*(5-1)/2] + 6 + 3 + 1 = 25 + 10 + 6 + 3 + 1 = 45 triangles.

Ughhhh.... I almost have a pattern. I'll be back shortly.....