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Math Help - Induction proof :D

  1. #1
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    Induction proof :D

    Can you help me with this one? Take it slow please....

    Show by induction: 2+2^2+2^3+.....+2^n=2*(2^n-1)

    Thanks for all of your answers !
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  2. #2
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    Hello, Neffets!

    LaTeX isn't back yet . . . Hope you can read my notation . . .


    Show by induction: .2 + 2^2 + 2^3 + . . . + 2^n .= .2(2^1 - 1}

    Verify S(1): .2 .= .2(2^1 - 1) . . . yes!

    Assume S(k): .2 + 2^2 + 2^3 + . . . + 2^k .= .2(2^k - 1)


    Add 2^{k+1} to both sides:

    . . 2 + 2^2 + 2^3 + . . . + 2^k + 2^{k+1} .= .2(2^k - 1) + 2^{k+1}

    The right side is: .2^{k+1} - 2 + 2^{k+1} .= .22^{k+1} - 2 .= .2(2^{k+1} - 1)


    And we have: .2 + 2^2 + 2^3 + . . . + 2^{k+1} .= .2(2^{k+1} - 1)

    . . There!

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  3. #3
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    It's a bit tricky without the LaTex, but i did understand what you wrote.. Although where did you get this from:
    The right side is: .2^{k+1} - 2 + 2^{k+1}
    Thanks for all of your help...
    Last edited by Neffets...:P; September 16th 2006 at 05:25 AM.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Neffets...:P View Post
    It's a bit tricky without the LaTex, but i did understand what you wrote.. Although where did you get this from:

    Thanks for all of your help...
    Look at the previous line, on the right hand side of the "=" sign you
    have:

    2(2^k - 1) + 2^{k+1},

    Now multiply the first bracket out to get:

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

    which is what you asked about, which Soroban then simplifies a bit
    and puts back equal to the left hand side back at the earlier equation
    where he had picked up the right hand side.

    RonL
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  5. #5
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    Of course...

    Thnx Soroban and Captain Black!
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  6. #6
    Senior Member OReilly's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Fast cars, fast women, fast algorithms... what more could a man want?
    -- Joe Mattis
    RonL
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