Dan Asimov c) sum of the first n cubes = (sum of first n numbers)^2

Yow! This reminded me I'd forgotten to forward this note a former funster sent back on 6/14: ======== From: Shel Kaphan Subject: How to visualize that a squared triangular number is the sum of cubes Hi Marc, Perhaps you could forward this to math-fun for me, if it still exists. Maybe someone there will enjoy it. Hopefully the fixed width font I "drew" this in will survive email transport. If not let me know and I'll try something different. thanks, Shel Here is how to visualize that a squared triangular number is a sum of cubes: A squared triangular number is a triangle of triangles of the same size, e.g. for n=3: x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x Tri(3)^2 = 36 (where Tri(n) = n(n+1)/2) To go from Tri(n) to Tri(n+1), we add a new row of triangles and increase the size of each existing triangle. E.g., to go from n=3 to n=4: x x x x x x o o o o x x x x x x x x x x x x o o o o o o o o x x x x x x x x x x x x x x x x x x o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o Notice that the last row of triangles can be stacked up as a triangular prism: o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o (this is supposed to be a 3D drawing...) This prism contains Tri(n+1) x (n+1) blobs. To fill out the cube, we would need an additional prism of Tri(n) x (n+1) blobs. This is the number of blobs added to the bottoms of the triangles in the diagram for Tri(n)^2 when we increased it to Tri(n+1)^2 (i.e., the o's added to the bottoms of the triangles of x's in the diagram above). So, Tri(n+1)^2 = Tri(n)^2 + (n+1)^3