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Math Help - Fibonacci proof

  1. #1
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    Fibonacci proof

    Here is my latest problem. Little help please. NOTE: u(n) is read u "sub" n

    Prove by induction that u(n) is even if 3 divides n. In addition I also have a logic question just in reading this problem.

    Is (3 divides n implies u(n) is even) the same thing as
    u(n) is even if 3 divides n.
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  2. #2
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    Hello, padsinseven!

    Your interpretation of that implication is correct.


    Prove by induction that u_n is even if 3 divides n.

    We wish to prove that: .If n = 3k, then u_n is even.


    Verify S(1)\!:\;k = 1,\:n = 3\quad\Rightarrow\quad u_3 \:=\:2 . . . True!


    Assume S(k) is true: . u_{3k} \:=\:2a for some integer a.

    . . and we wish to prove that: . u_{3(k+1)} is even.


    We have: . u_{3(k+1)} \;=\;u_{3k+3}

    . . . . . . . . . . . . = \;\underbrace{u_{3k+2}} +\, u_{3k+1}

    . . . . . . . . . . =\;\overbrace{\left(u_{3k+1} + u_{3k}\right)} +\, u_{3k+1}

    . . . . . . . . . . =\;2\!\cdot\!u_{3k+1} + \underbrace{u_{3k}}_{\text{This is }2a}

    . . . . . . . . . . = \;2\!\cdot\!u_{3k+1} + 2a

    . . . . . . . . . . = \;2\left(u_{3k+1} + a\right)


    Therefore: . u_{3(k+1)} is even.

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