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Thread: Fibonacci number

  1. #1
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    Fibonacci number

    For $\displaystyle n \in N $ let $\displaystyle f_n $ be the $\displaystyle n^{th}$ Fibonacci number :
    $\displaystyle f_1 = 1 \ , f_2 =1 , f_n= f_{n-1}+ f_{n-2} , n \geq 3 .$
    Prove that :
    $\displaystyle f_n \ is \ even \iff 3 \ divides \ n $
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  2. #2
    MHF Contributor chisigma's Avatar
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    According to...

    Fibonacci number - Wikipedia, the free encyclopedia

    ... a basic property of the fibonacci's numbers is that 'every 3rd number of the sequence is even and more generally, every k-th number of the sequence is a multiple of $\displaystyle f_{k}$'. In case of $\displaystyle k=3$ is $\displaystyle f_{k}=3$...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by flower3 View Post
    For $\displaystyle n \in N $ let $\displaystyle f_n $ be the $\displaystyle n^{th}$ Fibonacci number :
    $\displaystyle f_1 = 1 \ , f_2 =1 , f_n= f_{n-1}+ f_{n-2} , n \geq 3 .$
    Prove that :
    $\displaystyle f_n \ is \ even \iff 3 \ divides \ n $
    Proof by induction:

    Let the statement be P(n).

    $\displaystyle f_3=1+1=2$

    Thus P(3) is true.

    --------------------------------------------------

    Let P(m) be true.

    $\displaystyle m=3a$

    $\displaystyle f_m=2k$

    --------------------------------------------------

    Consider P(m+3).

    $\displaystyle m+3=3a+3=3(a+3)$

    Thus 3 divides m+3.

    $\displaystyle f_{m+3}=f_{m+2}+f_{m+1}$

    $\displaystyle f_{m+3}=f_{m+1}+f_m+f_{m+1}$

    $\displaystyle f_{m+3}=2f_{m+1}+f_m$

    $\displaystyle f_{m+3}=2f_{m+1}+2k$

    $\displaystyle f_{m+3}=2(f_{m+1}+k)$

    Thus $\displaystyle f_{m+3}$ is even.

    --------------------------------------------------

    Hence, by the principle of induction, the statement is true.
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