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How to Prove Fibonacci Squared Recurrence Relation

  1. Jan 16, 2008 #1
    1. The problem statement, all variables and given/known data
    We were asked in a class discussion if we could try to figure out the recurrence relation of the Fibonacci squared sequence. I correctly figured that the equation was
    F[tex]^{2}_{n}[/tex]=2F[tex]^{2}_{n-1}[/tex] + 2F[tex]^{2}_{n-2}[/tex] - F[tex]^{2}_{n-3}[/tex]

    with F[tex]^{2}_{1}[/tex] = 0 and F[tex]^{2}_{2}[/tex] = 1 and F[tex]^{2}_{3}[/tex] = 1.

    Now, our class homework is to prove it.

    3. The attempt at a solution

    My first instinct is proof by induction. So I start with:

    Test the hypothesis with n=4 (since n=1,2,3 are given by definition).
    F[tex]^{2}_{4}[/tex] = 2F[tex]^{2}_{3}[/tex] + 2F[tex]^{2}_{2}[/tex] - F[tex]^{2}_{1}[/tex]
    = 2 + 2 - 0
    = 4, which agrees with original definition(because F[tex]^{1}_{4}[/tex] = 2, and thus F[tex]^{2}_{4}[/tex] = 4).

    So, the formula holds for some arbitrary but fixed n (n being a member of the natural numbers).

    Test n+1.
    F[tex]^{2}_{n+1}[/tex] = (F[tex]^{1}_{n}[/tex] + F[tex]^{1}_{n-1}[/tex])^{2} (by definition. of Fib. Sequence)

    =F[tex]^{2}_{n}[/tex] + 2 F[tex]^{1}_{n}[/tex] * F[tex]^{1}_{n-1}[/tex] + F[tex]^{2}_{n-1}[/tex] (by distribution)

    = 2F[tex]^{2}_{n-1}[/tex] + 2F[tex]^{2}_{n-2}[/tex] - F[tex]^{2}_{n-3}[/tex] + 2 F[tex]^{1}_{n}[/tex] * F[tex]^{1}_{n-1}[/tex] + 2F[tex]^{2}_{n-2}[/tex] + 2F[tex]^{2}_{n-3}[/tex] - F[tex]^{2}_{n-4}[/tex] (by the induction hypothesis)

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

    =2F[tex]^{1}_{n-1}[/tex] * F[tex]^{1}_{n+1}[/tex] + 4F[tex]^{2}_{n-2}[/tex] + F[tex]^{2}_{n-3}[/tex] - F[tex]^{2}_{n-4}[/tex]

    That's where I'm stuck. I know (or think I know) that I'm supposed to end up with :
    F[tex]^{2}_{n+1}[/tex] = 2F[tex]^{2}_{n}[/tex] + 2F[tex]^{2}_{n-1}[/tex] - F[tex]^{2}_{n-2}[/tex] and so for every n for which the formula is valid, the formula is also valid for n+1. By induction, the formula works for all of the natural numbers.

    Could anyone help me figure out the part I'm stuck at?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jan 16, 2008 #2


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    F2n + 2(Fn)(Fn-1) + F2n-1
    F2n + 2(Fn)(Fn - fn-2) + F2n-1
    F2n +(2F2n) -(2fn-2*fn) + F2n-1
    Use the induction hypothesis on the lone F2n
    2F2n + (2F2n-1) + (2F2n-2) - (F2n-3) - 2(Fn*Fn-2) + F2n-1

    Notice that the first two terms are the first two terms we are looking for. I'm going to leave those off and manipulate the last 4 terms into -Fn-2. The number of steps I used can probably be reduced under scrutiny, but I just did a rough sketch of this.

    F2n-1 + 2F2n-2 - F2n-3 - 2( Fn-1 + Fn-2)*(Fn-2)
    Notice that F2n-2 cancels after expanding the last term.
    F2n-1 - 2Fn-1*Fn-2 - F2n-3
    Write Fn-1 in the middle term as Fn-2 + Fn-3 then expand.

    F2n-1 -2F2n-2 - 2Fn-2Fn-3 - F2n-3
    Now write the first term as (Fn-2 + Fn-3)^2 and expand.
    F2n-2 + 2Fn-2*Fn-3 +f2n-3 -2F2n-2 - 2Fn-2Fn-3 - F2n-3
    Combining and canceling leaves just the term we are looking for -Fn-2
    Combining that with the terms we dropped earlier gives us:
    2F2n+ 2F2n-1 - F2n-2. QED

    Sorry for being really messy, but I really didn't want to use a lot of energy typing this out.
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