Hi funsters, We all know that 1^3 + 2^3 + ... + n^3 = (1+2+...+n)^2, and there's a pretty geometric dissection proof (eg on p.58 of the Book of Numbers here<http://books.google.com/books?id=0--3rcO7dMYC&lpg=PP1&ots=-aWHB2jOpz&dq=book%20of%20numbers%20conway%20guy&pg=PA58#v=onepage&q=&f=false>, though sadly the picture is redacted). Is there a geometric dissection for its generalizations? Let S(d) be the dth-degree polynomial that is sum( i^(d-1), i=1..n ) -- um, that must have a name, aside from my off-by-1 definition, but I can't think of what. Then we have: S(2d) = sum( (-1)^i binomial(2d-2, i-1) S(i) S(2n-i), i=2...d-1 ) + (-1)^d binomial(2d-3, d-1) S(d)^2 and there's another formulation where the S(d)^2 term can be folded into the sum(..., i=2..2d-2), but you have to divide by 2 in that case. Um, and d can't be 1; these start with d=2 where it gives the S(4) = S(2)^2 mentioned above, and then you get S(6) = 4 S(2) S(4) - 3 S(3)^2, S(8) = 6 S(2) S(6) - 15 S(3) S(5) + 10 S(4)^2, and so on. Do any of you know a pretty geometric dissection for this identity? Note that, aside from S(4) = S(2)^2, the RHSs above all have mixed sign, so presumably some shuffling around of terms would be required. --Michael -- Forewarned is worth an octopus in the bush.