1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

(Easy) Maths Challenge

  1. Jan 1, 2010 #1

    FeDeX_LaTeX

    User Avatar
    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

    User Avatar
    Gold Member

    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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook