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Help with Fibonacci Identity

  1. Jan 30, 2007 #1
    Can someone guide me on how to prove that
    [tex]F_{4n+3} + F_{4n+6} = F_{2n+1}^2 + F_{2n+4}^2[/tex]

    either side of the above is the difference

    [tex](F_{2n+2}*F_{2n+3} + F_{2n+4}^2) - (F_{2n}*F_{2n+1} + F_{2n+2}^2)[/tex]

    I intend to post this sequence [tex]F_{2n}*F_{2n+1} + F_{2n+2}^2[/tex], with a comment re a few properties thereof, on Sloane's online encyclopedia of integer sequences but would like to verify the above identity first.
     
  2. jcsd
  3. Feb 1, 2007 #2
    First thing springs to mind is to try use general formula for nth
    Fibbonacci number:

    [tex]F_{n}=\frac{\phi^n-(1-\phi)^n}{\sqrt{5}}[/tex]

    In order to proceed with math induction.But I'm unsure will it work or not.
    I'm sure there are better methods ,though.
     
    Last edited: Feb 1, 2007
  4. Feb 4, 2007 #3
    Thanks
    I think there is an identity for the following that works:

    [tex]F_{i}*F_{j} + F_{i+1}*F_{j+1} = F_{?}[/tex]

    Let j = i = 2n+1 then

    [tex]F_{2n+1}^{2} + F_{2n+2}^{2} = F_{4n+3} [/tex]
    [Tex]F_{2n+2}^{2} + F_{2n+3}^{2} = F_{4n+5}[/tex]
    [Tex]F_{2n+2}^{3} + F_{2n+4)^{2} = F_{4n+7}[/tex]
    \\
    [Tex]F_{4n+3} +F_{4n+6} = F_{4n+3} + F_{4n+7} - F_{4n+5}[/tex]
    [Tex] =F_{2n+1}^{2} + F_{2n+4}^{2}[/tex]
     
    Last edited: Feb 4, 2007
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