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Math Help - another fibonacci proof

  1. #1
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    another fibonacci proof

    Prove by induction that for all n>=1, u(n+1)^2 - u(n+2)u(n)= (-1)^n. Any help would be greatly appreciated. I have no idea how to show the right side will end up (-1)^n+1
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  2. #2
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    Hello, padsinseven!

    This is the Fibonacci sequence: . u_n\:=\:1,1,2,3,5,8,\cdots


    Prove by induction that for all n \geq 1\!:\;\;\left(u_{n+1}\right)^2 - \left(u_{n+2}\right)\left(u_n\right)\;= \;(-1)^n
    Verify S(1)\!:\;\;(u_2)^2 - (u_3)(u_1) \;=\;1^2 - (2)(1) \;=\;-1\;=\;(-1)^1 . . . True!


    Assume S(k) is true: . \left(u_{k+1}\right)^2 - \left(u_{k+2}\right)\left(u_k\right)  \;=\;(-1)^k

    And we want prove S(k+1)\!:\;\;\left(u_{k+2}\right)^2 - \left(u_{k+3}\right)\left(u_{k+1}\right) \;=\;(-1)^{k+1}


    The left side of S(k+1) is:

    . . . . . . \left(u_{k+2}\right)^2 - \underbrace{\left(u_{k+3}\right)}\left(u_{k+1}\rig  ht)

    . . = \;\left(u_{k+2}\right)^2 - \overbrace{\left[u_{k+2} + u_{k+1}\right]}\left(u_{k+1}\right)

    . . = \;\left(u_{k+2}\right)^2 - \left(u_{k+2}\right)\left(u_{k+1}\right) - \left(u_{k+1}\right)^2

    . . = \;\left(u_{k+2}\right)\cdot\underbrace{\left[u_{k+2} - u_{k+1}\right]} - \left(u_{k+1}\right)^2

    . . = \qquad\:\left(u_{k+2}\right)\cdot\overbrace{\left(  u_k\right)} - \left(u_{k+1}\right)^2

    . . = \;\qquad\; - \underbrace{\bigg[\left(u_{k+1}\right)^2 - \left(u_{k+2}\right)\left(u_k\right)\bigg]}_{\text{This is }S(k)}

    . . = \;\qquad\qquad\qquad - \overbrace{(-1)^k}

    . . = \qquad\qquad\qquad\quad (-1)^{k+1} . . . . ta-DAA!

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