I asked: << 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 for the nice answers. (Alas, Richard, my beloved copy of TBoN is still locked up in storage along with most of my math books.) * Fred's instantly clear induction step for a) immediately leads to a proof that generalizes the formulae for triangular and tetrahedral numbers to all "K-simplicial numbers" S_K(n): S_K(n) = (n+K-1)!/(K! (n-1)!). Which suggests the possibility of a proof directly based on this being the binomial coefficient n+K-1 choose K. (Or n-K+1 choose n-1: take your choice.) * Both a) and b), at least, look like discrete versions of the geometrical facts that a pyramid with base an equilateral right triangle (resp. unit square) and height 1 has volume = 1/6 of that of a unit cube (resp. 2x1x1 rectangular solid). (Perhaps this can be made to fit into an instant continuous-to-discrete theorem?) * I just read somewhere that for odd p, the (formula for the) sum of the first n pth powers is always a polynomial in the nth triangular number. QUESTION: For which (p,q) is the (formula for the) sum of the first n pth powers a polynomial in (the formula for) the sum of the first n qth powers? --Dan P.S. A formula I once found for the sum of the first n reciprocal sth powers (at least for real s >= 1) is: Sum_{k = 1 to n} 1/k^s = (1/Gamma(s))*Integral_{0 to oo} ((1-u^n)/(1-u))) (ln(1/u))^(s-1) du. Letting n -> oo gives an integral formula for the product of the Gamma and Zeta functions. I was crestfallen when in 1964 my freshman advisor (Henry McKean) showed me this limiting formula in Whittaker & Watson (1902).