I am trying to calculate how many triangles that can be found in an equilateral triangle with $2n$ lines starting at the bottom angles and ending at the opposite side, such that equally many lines start/end of either side.
This is rather hard to explain, so I drew that first 5 terms:
The 1st term has, of course, 1 triangle.
The 2nd term has 8 triangles.
The 3rd term has 27 triangles.
This problem is really killing me, so any help would be greatly appreciated.
Well, the way I see it; break down the lines into the two groups of $n$ lines on the left and right, and add the "left" and "right" side of the original triangle to make two groups of $n+1$ lines. There are then two cases I see:
One side of the triangle is from a line on the "left", one from the "right", and the bottom side of the original triangle. There are $(n+1)^2$ such triangles.
Two of the sides are from lines in one group, while the third side is from a line in the other. This gets us $2(n+1)\binom{n+1}{2}$ more.
That gets me $(n+1)^2 + 2(n+1)\binom{n+1}{2} = (n+1)^3$.