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Thread: Recurrence and Fibonacci sequence

  1. #1
    lpd
    lpd is offline
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    Recurrence and Fibonacci sequence

    Hi. I have this problem that im stuck on.

    Define a sequence $\displaystyle (G_n)_{n \ge0} $ by the recurrence $\displaystyle G_n = G_{n-1}+G_{n-2}$ for $\displaystyle n \ge2$. Subject to the initial values $\displaystyle G_0=2$, $\displaystyle G_1=1$. Let $\displaystyle (F_n)_{n \ge0} $ denote the Fibonacci sequence.

    a) Write out explicitly $\displaystyle (G_n)_{n=0,...,10} $
    I think this is easy.
    {2,1,3,4,7,11,18,29,47,76}

    b) Prove that $\displaystyle G_n=F_{n-1}+F_{n+1}$, $\displaystyle n\ge1$
    How do I go about it?
    I know $\displaystyle F_0=0$ and $\displaystyle F_1=1$ and $\displaystyle F_2=1$ and $\displaystyle F_3=2$
    So if $\displaystyle n=1$, $\displaystyle G_1=F_0+F_3=1+2=3$, so it looks like its true.But how do I prove it explicitly?\

    c) Let $\displaystyle \tau= \frac{1+ \sqrt{5}}{2}$ denote the golden ratio. Show $\displaystyle G_n= \tau^n+(-\tau)^{-n}$
    I'm just a bit lost in this one...

    d) The Fibonacci sequence counts pavings by monomers and dimers of an n-board. Conjecture what sort of pavings the sequence $\displaystyle (G_n)_{n \ge1} $ counts?Draw the objects corresponding to $\displaystyle G_3$.
    $\displaystyle G_3=G_2+G_1$
    $\displaystyle G_3=3+1=4$
    But I'm not quite sure what it counts...

    Any help would be great. Thank-you so much!!
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  2. #2
    MHF Contributor chisigma's Avatar
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    In...

    http://www.mathhelpforum.com/math-he...ce-154056.html

    ... it has been demonstrated that the general solution of the difference equation...

    $\displaystyle x_{n} = x_{n-1} + x_{n-2} $ (1)

    ... is...

    $\displaystyle \displaystyle x_{n}= c_{1} (\frac{1-\sqrt{5}}{2})^{n} + c_{2} (\frac{1+\sqrt{5}}{2})^{n} $ (2)

    If You start setting $\displaystyle x_{0}= 0$ and $\displaystyle x_{1}=1$ You obtain the 'Fibonacci sequence'. If You set $\displaystyle x_{0}=2$ and $\displaystyle x_{1}=1$ and You obtain the 'G sequence'...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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