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(Easy) Maths Challenge

  1. Jan 1, 2010 #1

    FeDeX_LaTeX

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    Gold Member

    Prove that:

    [tex]F_{n} = \frac {\varphi^n - (1-\varphi)^n}{\sqrt5}[/tex]

    holds true for any number in the Fibonacci Sequence.

    Who can answer the fastest?
     
  2. jcsd
  3. Jan 1, 2010 #2

    Char. Limit

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    Hmm... well, defining phi as (1+sqrt(5))/2, this means that the second term there is equal to -(phi+1), or -1/phi.

    So... for odd n, the denominator reduces to phi^n+phi^-n. Multiplying top and bottom by phi^n gives us phi^2n+1 in the numerator and sqrt(5)*phi^n in the denominator. Splitting the fraction gives us (phi^n)/sqrt(5) + 1/(sqrt(5)*phi^n). Beyond there... I don't know.
     
  4. Jan 1, 2010 #3
    [tex]F_n+F_{n+1}=\frac{1}{\sqrt{5}}\left(\phi^n-(1-\phi)^n+\phi^{n+1}-(1-\phi)^{n+1}\right)=\frac{1}{\sqrt{5}}\left(\phi^n(1+\phi)-(1-\phi)^{n}(2-\phi)\right)[/tex]
    Since by definition
    [tex]\phi+1=\phi^2[/tex]
    and we also get
    [tex](1-\phi)^2=1-2\phi+\phi^2=2-\phi[/tex]
    then
    [tex]F_n+F_{n+1}=\frac{1}{\sqrt{5}}\left(\phi^n\phi^2-(1-\phi)^{n}(1-\phi)^2\right)=\frac{1}{\sqrt{5}}\left(\phi^{n+2}-(1-\phi)^{n+2}\right)=F_{n+2}[/tex]
    That's all you need.
     
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