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Thread: Prove F_2n = FnLn

  1. #1
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    Prove F_2n = FnLn

    Prove F_2n = FnLn, where Fn are fibonacci numbers and Ln are Lucas numbers.
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  2. #2
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    Quote Originally Posted by yc6489 View Post
    Prove F_2n = FnLn, where Fn are fibonacci numbers and Ln are Lucas numbers.
    That is not true consider $\displaystyle n=2$.
    Then, $\displaystyle F_{2n}=3$.
    $\displaystyle F_n=1$ and $\displaystyle L_n=1$.
    And, $\displaystyle 1\cdot 1 \not = 2$.
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  3. #3
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    Hello, yc6489!

    Prove: .$\displaystyle F_{2n} \:= \:F_n\!\cdot\!L_n$,
    where $\displaystyle F_n$ are Fibonacci numbers and $\displaystyle L_n$ are Lucas numbers.

    Are we allowed to use the closed forms for $\displaystyle F_n$ and $\displaystyle L_n$ ?

    . . $\displaystyle F_n \;=\;\frac{(1 + \sqrt{5})^n - (1 - \sqrt{5})^n}{2^n\sqrt{5}} \qquad L_n \;=\;\frac{(1 + \sqrt{5})^n + (1 - \sqrt{5})^n}{2^n} $



    Then: .$\displaystyle F_n\!\cdot\!L_n\;=\;\frac{(1 + \sqrt{5})^n - (1 - \sqrt{5})^n}{2^n\sqrt{5}}\cdot\frac{(1 + \sqrt{5})^n + (1 - \sqrt{5})^n}{2^n} $

    . . $\displaystyle = \;\frac{(1 + \sqrt{5})^{2n} - (1 - \sqrt{5})^{2n}}{2^{2n}\sqrt{5}} \;=\;F_{2n} $

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    Thanks Soroban for your reply. I think the intention of the exercise is to prove the statement from the definition of Ln = F_n-1 + F_n+1 for n>=2. (The text does not mention your particular definitions of Fn and Ln.)
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    Quote Originally Posted by yc6489 View Post
    Thanks Soroban for your reply. I think the intention of the exercise is to prove the statement from the definition of Ln = F_n-1 + F_n+1 for n>=2. (The text does not mention your particular definitions of Fn and Ln.)
    In that case do it with induction.
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