If you start with $u_1 = 1$ and $u_2 = 1$, then the sequence can be generated using the formula $$u_{n+1} = u_n + u_{n-1}\ .$$ If $u_n = r^n$, what is r?
Can anyone figure this out? I am so stuck please help!
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See http://en.wikipedia.org/wiki/Fibonacci_number. – Dietrich Burde Jun 17 '14 at 18:36
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If $u_n = r^n$, then can you compute $u_{n + 1}$ as a power of $r$? – Jun 17 '14 at 18:44
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Well, you have an equation there. If $u_{n} = r^{n}$, then the relation $u_{n+1} = u_{n} + u_{n-1}$ implies $r^{n+1} = r^{n} + r^{n-1}$. Now can you solve for $r$? – Alex Wertheim Jun 17 '14 at 18:45
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Guys I think the question is "which numbers of the Fibonacci sequence have an integer n-th root, where n is their position in the sequence?" – AnalysisStudent0414 Jun 17 '14 at 18:48
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@user130520, Make question clear. – L.K. Jun 17 '14 at 19:05
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I'm still not really getting it. I tried solving for r, and it makes sense mathematically but really only if I use u_n= u_n-1 + u_n-2 but I need help with this proof. – user130520 Jun 19 '14 at 18:13
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If r is integer, then only solution is r=1. – Tahir Imanov Jan 23 '15 at 11:30
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$F_n$ is approximately equal to $\dfrac{\Phi^n}{\sqrt{5}}$, where $\Phi$ is the golden ratio, $\Phi = \dfrac{\sqrt{5}+1}{2}$.
If $F_{n+2} = F_{n+1} + F_n$ and $F_{n+1} = rF_n$, then $r^2F_n = rF_n + F_n$ and $r^2 = r + 1$ or $r^2 - r - 1 = 0$. This quadratic has roots $\Phi$ and $1 - \Phi = \dfrac{1 - \sqrt{5}}{2}$ which is negative.
NovaDenizen
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my problem is where does n+2 come in? and where did we get phi from? – user130520 Jun 19 '14 at 17:22
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It's better style to deal with $F_{n+2}, F_{n+1}, $ and $F_n$ than $F_{n+1}, F_n, $ and $ F_{n-1}$. And $\Phi$ is a root of the quadratic $r^2 - r - 1$, long associated with the Fibonacci sequence. – NovaDenizen Jun 24 '14 at 23:13