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watchmath
 4 years ago
Let \(\{F_n\}\) be the Fibonacci sequence: \(F_0=1, F_1=1\) and \(F_n=F_n+F_{n1}\) for \(n\geq 2\). Show that
\[\sum_{i=0}^n (ni)F_i=F_{n+3}n3\]
watchmath
 4 years ago
Let \(\{F_n\}\) be the Fibonacci sequence: \(F_0=1, F_1=1\) and \(F_n=F_n+F_{n1}\) for \(n\geq 2\). Show that \[\sum_{i=0}^n (ni)F_i=F_{n+3}n3\]

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Fn = Fn + Fn1 doesn't make any sense to me, first because that isn't a valid statement (it's like saying x = x + 1 where x = 2, thus 2 = 3 or 1 = 0) and also because the real Fibonacci sequence is: Fn = Fn1 + Fn2 unless i'm missing something and am just too tired to thoroughly think it out

watchmath
 4 years ago
Best ResponseYou've already chosen the best response.0yes, sorry it's a typo :\(F_n=F_{n1}+F_{n2}\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Induction should be sufficient to prove it. For n=0, you have that \[\sum_{i=0}^niF_i=0=303=F_3n3\] Now assume the claim for n=k. Then for n=k+1, you have that \[\sum_{i=0}^{k+1}(k+1i)F_i=\sum_{i=0}^{k}(k+1i)F_i\] \[=\sum_{i=0}^k\left((ki)F_i+F_i\right)=\sum_{i=0}^k(ki)F_i+\sum_{i=0}^k F_i\] \[=F_{k+3}k3+\sum_{i=0}^k F_i\] Using a common identity, which itself is a simple proof by induction, you have that \[\sum_{i=0}^k F_i=F_{k+2}1\] Thus, inserting that into the above equation gives \[=F_{k+3}k3+F_{k+2}1 = F_{k+4}k13\] \[=F_{(k+1)+3}(k+1)3\] and the claim is proven.

watchmath
 4 years ago
Best ResponseYou've already chosen the best response.0Just wondering. Is it possible to have a bijective proof for this?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What do you mean? Do you mean that if a_n is a sequence such that \[\sum_{i=0}^n(ni)a_i=a_{n+3}n3\] then it is the Fibonacci sequence? That seems to be a much harder question and I'm not even confident that it is true. There are a lot of sequences after all! I'd guess that (if it is true), using induction would also allow you to go in reverse, though the steps might be a bit harder. You also may need to put some constraints on the sequence to make this a true statement.

watchmath
 4 years ago
Best ResponseYou've already chosen the best response.0No, I mean there is a concrete object that we can count that gives the LHS and the RHS. For example the identity \[2^n=\sum_{k=0}^n {n\choose k}\] counts the number of subsets of \(\{1,2,\ldots,n\}\) in two ways.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ah, I'm not sure. Since the Fibonacci numbers often correspond to seashell type patterns, maybe it is possible that the sum over (ni)F_i has some sort of geometric interpretation that can be altered in some clever way to make it look like the RHS. If there is a counting argument that can show this, something along those lines would be my best guess.
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