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Math Help - Fibonaci sequence terms.

  1. #1
    joi
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    Fibonaci sequence terms.

    How to prove that (f(n+1)^2=f(n)f(n+2)+(-1)^n where f(n), f(n+1),f(n+2) are fibonaci sequence terms. I have no idea where to start.
    Last edited by mr fantastic; May 6th 2011 at 01:59 PM. Reason: Re-titled.
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    Quote Originally Posted by joi View Post
    (f(n+1)^2=f(n)f(n+2)+(-1)^n where f(n), f(n+1),f(n+2) are fibonaci sequence terms. I have no idea where to start.
    If this is for n > then something, I would try induction.
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  3. #3
    joi
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    Fibonacci terms

    Thank you dwsmith for answering. You are right: The condition is n>3 or n=3. I forgot to mention that. I will try induction.
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  4. #4
    joi
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    Induction does not work. I find f(k+2)^2=f(k)f(k+4)+(-1)^k. It should be f(k+2)^2=f(k+1)f(k+3)+(-1)^k. I am not sure if I am making any mistake.
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  5. #5
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    Quote Originally Posted by joi View Post
    Induction does not work. I find f(k+2)^2=f(k)f(k+4)+(-1)^k. It should be f(k+2)^2=f(k+1)f(k+3)+(-1)^k.
    No, it should be f(k+2)^2=f(k+1)f(k+3)+(-1)^{k+1}.

    You have to use a bit of ingenuity in the induction in order to get the (-1)^k term to change sign. One way is to start by writing

    f(k+2)^2 = f(k+2)f(k) + [f(k+2) – f(k)]f(k+2).

    Then for the first term on the right-hand side, use the inductive hypothesis in the form f(k+2)f(k) = f(k+1)^2 – (–1)^k. For the second term, use the fact that f(k+2) – f(k) = f(k+1) to write it as f(k+1)f(k+2). You should then be able to complete the inductive step.
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    I might add that what you're trying to prove is know as Cassini's Identity. A google search will give you lots of places where this identity is proven (and with details :-))
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  7. #7
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    Here's how I would try the Induction.

    For notational simplicity, let

    F_k=a,\;\;\;F_{k+1}=b,\;\;\;F_{k+2}=c,\;\;\;F_{k+3  }=d

    a+b=c
    b+c=d


    P(k)

    b^2=ac+(-1)^k


    P(k+1)

    c^2=bd+(-1)^{k+1}

    Proof

    c^2=b(b+c)+(-1)^k(-1)

    c^2=b^2+bc-(-1)^k

    b^2=c^2-bc+(-1)^k=c(c-b)+(-1)^k

    b^2=c(a)+(-1)^k

    which shows that P(k+1) is certainly true if P(k) is true.

    Test for the base case...
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  8. #8
    joi
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    Thank you Opalg, thank you danny, thank you Archie meade for getting involved. I am going to verify your suggestions but I have not done it yet. WHAT ABOUT IF I rewrite the equality as:
    (f(n+1))^2-f(n)f(n+2)=(-1)^n and then substituting the terms f(n+1)=f(n)+f(n-1) and f(n+2)=f(n)+f(n+1) into the equality they get reduced. It seems that if I use the property of infinite decent, combined with well ordering property the whole left hand can be reduced to(-1)^n Can someone tell me if this will be valid? I will review the suggestions but in case that induction does not work.
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  9. #9
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    Well you could start from

    \left(f_{n+1}\right)^2-f_nf_{n+2}=(-1)^n

    and prove

    \left(f_n\right)^2-f_{n-1}f_{n+1}=-(-1)^n

    showing that the sum inevitably alternatives sign as n increases or decreases by 1.


    \left(f_{n+1}\right)^2-\left(f_{n+1}-f_{n-1}\right)\left(f_n+f_{n+1}\right)=(-1)^n

    \left(f_{n+1}\right)^2-f_nf_{n+1}-\left(f_{n+1}\right)^2+f_nf_{n-1}+f_{n-1}f_{n+1}=(-1)^n

    f_n\left(f_{n-1}-f_{n+1}\right)+f_{n-1}f_{n+1}=(-1)^n

    -f_n\left(f_{n+1}-f_{n-1}\right)+f_{n-1}f_{n+1}=(-1)^n

    \left(f_n\right)^2-f_{n-1}f_{n+1}=-(-1)^n

    Now, you only need a reference point.
    Last edited by Archie Meade; May 14th 2011 at 11:22 AM. Reason: typo
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  10. #10
    Junior Member Renji Rodrigo's Avatar
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    Other solution

    Define
    F= \left(\begin{array}{cc}1 & 1 \\ 1 & 0 \\\end{array}\right)


    so
     F^n=\left(\begin{array}{cc}F(n+1) & F(n) \\F(n) & F(n-1) \\ \end{array}\right).
    (by induction)

    take the determinant of both sides, then

    Det( F^n)=(DetF)^n=(-1)^n=F(n+1)F(n-1)-[F(n)]^{2}
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