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Math Help - Proving Lucas Numbers and Fibonacci Numbers

  1. #1
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    Proving Lucas Numbers and Fibonacci Numbers

    The Lucas numbers l_{0}, l_{1}, l_{2},...,l_{n}... are defined on the same recurrence relation defining the Fibonacci numbers, but the Lucas numbers posses different initial conditions.

    l_{n}=l_{n-1} + l_{n-2}, (n≥2),  l_{0}=2, l_{1} = 1

    (a) l_{n} = f_{n-1} + f_{n+1} for n≥1.

    My hardest part is getting started. I don't really know how I can start off relating these two functions.

    EDIT:
    I don't know why I didn't think of this (mind isn't all there today), but I can use induction, correct?
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  2. #2
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    So Let n=1
    l_{1}=f_{0}+f_{2}
    1 = 0 +1

    Check

    Assume l_{n}=f_{n-1}+f_{n+1}

    l_{n+1}= l_{n} + l_{n-1} stated in our first assumption

    l_{n}= f_{n-1}+f_{n+1} and * l_{n-1}=f_{n-1}+f_{n+1}*

    l_{n+1}=f_{n-1}+f_{n+1} +f_{n-1}+f_{n+1}

    then by simplifying with Fibonacci Identities I get

    l_{n+1} = f_{n} + f_{n+2} for n≥1.

    My question now is with the part between the **, can I make that assumption based on my induction assumption since that is n-1 and my induction assumption is n.
    Last edited by hockey777; March 18th 2008 at 08:29 AM.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by hockey777 View Post
    So Let n=1
    l_{1}=f_{0}+f_{2}
    1 = 0 +1

    Check

    Assume l_{n}=f_{n-1}+f_{n+1}

    l_{n+1}= l_{n} + l_{n-1} stated in our first assumption

    l_{n}= f_{n-1}+f_{n+1} and * l_{n-1}=f_{n-1}+f_{n+1}*

    l_{n+1}=f_{n-1}+f_{n+1} +f_{n-1}+f_{n+1}

    then by simplifying with Fibonacci Identities I get

    l_{n+1} = f_{n} + f_{n+2} for n≥1.

    My question now is with the part between the **, can I make that assumption based on my induction assumption since that is n-1 and my induction assumption is n.

    Use strong induction, then the assumption is that:

    l_{k}=f_{k-1}+f_{k+1} ~\forall k \in [1,n] for some n > 1


    RonL
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