Here's a brute-force solution that uses essentially the same induction mechanism twice. To save typing I'll use B = Binomial and F = Fibonacci. First, rewrite the expression as Sum[(-1)^(n-k) * (k/n) * B[2n-k-1, n-1] * F[k+1], {k, 0, n}] == 1 Proceed by induction. Verify that it holds for n = 1, then: Sum[(-1)^(n+1-k) * (k/(n+1)) * B[2n-k+1, n] * F[k+1], {k, 0, n+1}] = Sum[(-1)^(n+1-k) * (k/(n+1)) * B[2n-k+1, n] * (F[k] + F[k-1]), {k, 0, n+1}] = Sum[(-1)^(n-k) * ((k+1)/(n+1)) * B[2n-k, n] * F[k+1], {k, 0, n}] + Sum[(-1)^(n-k-1) * ((k+2)/(n+1)) * B[2n-k-1, n] * F[k+1], {k, 0, n}] = Sum[(-1)^(n-k) * F[k+1]/(n+1) * ((k+1) * B[2n-k, n] - (k+2) * B[2n-k-1, n]), {k, 0, n}] = Sum[(-1)^(n-k) * F[k+1]/(n+1) * ((k+1) * B[2n-k-1, n-1] - B[2n-k-1, n]), {k, 0, n}] = Sum[(-1)^(n-k) * F[k+1]/(n+1) * (k * B[2n-k-1, n-1] + B[2n-k-1, n-1]- B[2n-k-1, n]), {k, 0, n}] = n/(n+1) + Sum[(-1)^(n-k) * F[k+1]/(n+1) * (B[2n-k-1, n-1]- B[2n-k-1, n]), {k, 0, n}] where in the last step we apply the induction hypothesis, and throughout we're a bit sloppy with summation bounds because the "extra" terms are always zero. Now it suffices to prove that Sum[(-1)^(n-k) * F[k+1] * (B[2n-k-1, n-1] - B[2n-k-1, n]), {k, 0, n}] == 1 Again, we proceed by induction. Verify that it holds for n = 1, then: Sum[(-1)^(n+1-k) * F[k+1] * (B[2n-k+1, n] - B[2n-k+1, n+1]), {k, 0, n+1}] = Sum[(-1)^(n+1-k) * (F[k] + F[k-1]) * (B[2n-k+1, n] - B[2n-k+1, n+1]), {k, 0, n+1}] = Sum[(-1)^(n-k) * F[k+1] * (B[2n-k, n] - B[2n-k, n+1]), {k, 0, n}] + Sum[(-1)^(n-k-1) * F[k+1] * (B[2n-k-1, n] - B[2n-k-1, n+1]), {k, 0, n}] = Sum[(-1)^(n-k) * F[k+1] * (B[2n-k, n] - B[2n-k-1, n] - B[2n-k, n+1] + B[2n-k-1, n+1]), {k, 0, n}] = Sum[(-1)^(n-k) * F[k+1] * (B[2n-k-1, n-1] - B[2n-k-1, n]), {k, 0, n}] = 1 J.P. On Sun, Apr 17, 2011 at 7:54 PM, quad <quadricode@gmail.com> wrote:
Parens around k/n is correct.
On Apr 17, 2011, at 5:03 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Bill G. wrote:
<< Sum[k/n*Binomial[-n, n - k]*Fibonacci[k + 1], {k, 1, n}] == 1 ?
I'm not entirely familiar with Mma syntax. May we assume there are parentheses around the k/n ?
Thanks,
Dan
Sometimes the brain has a mind of its own.
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