On Sun, Apr 17, 2011 at 2:12 PM, Bill Gosper <billgosper@gmail.com> wrote:
Can anybody prove: Sum[k/n*Binomial[-n, n - k]*Fibonacci[k + 1], {k, 1, n}] == 1 ? I was almost surprised when it defeated Julian last nite. --rwg
He wasn't really defeated--just lazy. He typed Sum[k/n*Binomial[-n, n - k]*Fibonacci[k + 1], {k, 1, n}] and got 1 ----------------- Binomial[-n, -1 + n] (5 + 3 Sqrt[5]) n 1 ((7 + 3 Sqrt[5]) Hypergeometric2F1[2, 1 - n, 2 - 2 n, - (-1 - Sqrt[5])] - 2 2 2 Hypergeometric2F1[2, 1 - n, 2 - 2 n, -----------]) 1 + Sqrt[5] and said "Use A=B". I did, and after a major struggle, found this expression to satisfy In[561]:= Equal[x[n + 2], (((17*n^2 + 37*n + 18)*x[n + 1] - 4*(2*n + 1)*(2*n + 3)*x[n])/((n + 2)*(n + 3)))] Out[561]= 2 -4 (1 + 2 n) (3 + 2 n) x[n] + (18 + 37 n + 17 n ) x[1 + n] x[2 + n] == ---------------------------------------------------------- (2 + n) (3 + n) which is solved by x[n]->1. As marvelously inelegant as JPGrossman's solution. --rwg