With tetrahedra, at first I was expecting the n dimensional analog of taking a triangle, area n^2/2, plus the half-squares, area n (1/2), for the sum 1+…+n to e

With tetrahedra, at first I was expecting the n dimensional analog of taking a triangle, area n^2/2, plus the half-squares, area n (1/2), for the sum 1+…+n to equal n^2/2+n/2. This would require 4 dimensions for sum of cubes, though.