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I would appreciate if somebody could help me with the following problem

Q: $f_1=f_2=1, f_{n+2}=f_{n+1}+f_n(n\geq 2)$

Show that combinatoric identity (using by combinatorial proof)

$$1 f_1+2 f_2+\cdots+n f_n=n f_{n + 2} - f_{n + 3} + 2$$

Young
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  • A "combinatorial proof" means a proof that works by counting something (as opposed to a proof by just algebraic fiddling). So, what do the Fibonacci numbers count? – Gerry Myerson Jul 05 '13 at 12:48
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    I don't know about combinatorial proof, but it follows from $f_1+f_2+\cdots+f_n=f_{n + 2} - 1$. – lhf Jul 05 '13 at 13:01
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    @GerryMyerson From the historical orgin, the Fibonacci numbers count rabbits ;) – Hagen von Eitzen Jul 05 '13 at 13:31
  • Follow lhf's comment, I think that's what you are looking for. – newbie Jul 05 '13 at 20:19
  • @Hagen, yes, I forgot about that. Now let's see whether we can apply that to this question.... – Gerry Myerson Jul 05 '13 at 23:28
  • This problem is mentioned in chapter 1 of the well-known "Proofs that really count" as an "uncounted identity", and other problems in that section suggest to me this section is a list of open problems. (Fibonacci numbers are interpreted as tilings that use tiles of size 1 and 2.) lhf`s identity does have a combinatorial interpretation. – sdcvvc Jul 09 '13 at 23:17
  • See also http://math.stackexchange.com/questions/15469/combinatorial-proof-of-a-fibonacci-identity-n-f-1-n-1f-2-cdots-f-n?rq=1 – lhf Jul 10 '13 at 14:13

1 Answers1

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Here is a proof, though not a combinatorial one.

Let $s_n = f_1+f_2+\cdots+f_n$. Then $s_n=f_{n + 2} - 1$, which is easily proved by induction.*

The sum in question is $S=1 f_1+2 f_2+\cdots+n f_n= \sum_{k=0}^{n-1} (s_n-s_k)$, where $s_0=0$. Hence $$S=n s_n - \sum_{k=0}^{n-1} s_k = n s_n - \sum_{k=0}^{n-1} (f_{k+2}-1) = n f_{n+2}-n - s_{n+1}+f_1+n = n f_{n+2} - f_{n+3} + 2$$

*You can also proceed as follows: sum all equations $f_k= f_{k-1}+f_{k-2}$ for $k=2,\dots n+2$ and get $s_{n+2}-f_1 = s_{n+1} + s_n$, from which you get $f_{n+2}-f_1 = s_n$, as required.

lhf
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