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Math Help - Lucas Number Proof By Induction

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    Lucas Number Proof By Induction

    I am having some trouble proving the relationship

    F_n L_n= F_{2n}

    where  F_n,  L_n are the Fibronachi and Lucas numbers respectively.

    Oh, and how should I have written this in Laytex?
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    Can anyone give this a go? I am expecting to solve it by induction. But I can see no way of making the leap from numbers like F_n all the way up to F_2n.

    The inductive hypothesis doesn't apply to F_(2n+1) so I can see no way to proceed.
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  3. #3
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    Hello,

    Try strong induction Mathematical induction - Wikipedia, the free encyclopedia

    And it's Fibonacci, not Fibronachi
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    Quote Originally Posted by Kiwi_Dave View Post
    I am having some trouble proving the relationship

    F_n L_n= F_{2n}

    where  F_n,  L_n are the Fibronachi and Lucas numbers respectively.
    well, it's not a very straightforward induction! it's clearly true for n = 1, 2, 3. so assuming

    that n > 3 and that the claim is true for all 1 \leq k \leq n, we'll have:

    F_{n+1}L_{n+1}=(F_n+F_{n-1})(L_n+L_{n-1})=F_nL_n+F_nL_{n-1}+F_{n-1}L_n+F_{n-1}L_{n-1}

    =F_{2n}+F_{2n-2}+F_nL_{n-1}+F_{n-1}L_n=F_{2n}+F_{2n-2}+ F_n(L_n - L_{n-2}) + F_{n-1}(L_{n-1}+L_{n-2})

    =2F_{2n}+2F_{2n-2}-L_{n-2}(F_n-F_{n-1})=2F_{2n}+2F_{2n-2}-L_{n-2}F_{n-2}=2F_{2n}+2F_{2n-2}-F_{2n-4}

    =2F_{2n}+2F_{2n-2}-(F_{2n-2}-F_{2n-3})=2F_{2n}+F_{2n-2}+F_{2n-3}=2F_{2n}+F_{2n-1}=F_{2n}+F_{2n+1}

    =F_{2n+2}. \ \clubsuit
    Last edited by NonCommAlg; September 16th 2008 at 04:38 PM. Reason: the base of induction are the cases: n = 1, 2, 3.
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