a) Subtract (n-1) x n x (n+1) = n x (n+1) x (n-1) cuboid from one end of n x (n+1) x (n+2) cuboid, leaving n x (n+1) rectangle x 3 = n-triangle x 6 --- a visually appealing diagram would require some draughtsmanship I imagine, certainly more than is readily available in this text-file! c) More of a challenge --- looks as if 4 space dimensions would be required ... WFL On 12/31/07, Dan Asimov <dasimov@earthlink.net> wrote:
Can someone please point me to proofs without words for
a) nth tetrahedral number = n(n+1)(n+2)/6
b) sum of the first n squares = n(n+1)(2n+1)/6
c) sum of the first n cubes = (sum of first n numbers)^2 = (T_n)^2 (T_n = nth triangular #) ?
Thanks,
Dan
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun