I am trying to prove by induction that $F_{n-1}$ = $\frac{-1}{n}\sum_{a=0}^{n-2}{n \choose a} F_a$. The sequence $F_n$ is the Bernoulli numbers, with $F_0=1, F_2=-1/2$ etc. so the formula in question is a recurrence formula for the Bernoulli numbers. I understand how to take the base case, but am struggling with making the inductive step. Should it be something like this: Base Case: $F_{n-1}$ = $\frac{-1}{n}\sum_{a=0}^{n-3}{n \choose a} F_a$ Inductive Step: $F_{n} = \frac{-1}{n+1}(\sum_{a=0}^{n-1}{n\choose a}F_a + {n\choose n-1}F_{n-1})$. I'm then confused about where to go because of the $\frac{-1}{n+1}$ out in front. Do I try to modify it by algebra to look like $F_{n} = \frac{-1}{n+1}(\sum_{a=0}^{n-1}{{n+1}\choose a}F_a$ ? Please help!


closed as off-topic by Shailesh, steven gregory, hardmath, user91500, zhoraster Apr 12 '17 at 8:05

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  • $\begingroup$ Surely the index on $F$ under the summation is $a$ and not $n$. $\endgroup$ – Mark Viola Apr 12 '17 at 2:59
  • $\begingroup$ You are correct, sorry! $\endgroup$ – R. Collins Apr 12 '17 at 3:13
  • $\begingroup$ And what is $F_n$? $\endgroup$ – Mark Viola Apr 12 '17 at 3:19
  • $\begingroup$ $F_n$ is an element of a sequence going to infinity. $\endgroup$ – R. Collins Apr 12 '17 at 3:41
  • $\begingroup$ Is that all that you know of $F_n$? $\endgroup$ – Mark Viola Apr 12 '17 at 3:42