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Math Help - Proof by Induction

  1. #1
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    Proof by Induction

    n
    \sum i(2^i) = (n-1) * 2^(n+1) + 2
    i=1
    So far if have the base case 1(2^1) = (1-1) * (1+1) + 2

    Induction part is giving me problems:
    k
    \sum i(2^i) = (k-1) * 2^(k+1) + 2 = ((k+1)-1 * 2^ ((k+1)+1 + 2
    i=1
    Last edited by zaj091000; August 31st 2010 at 08:14 PM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by zaj091000 View Post
    n
    \sum i(2^i) = (n-1) * 2^(n+1) + 2
    i=1
    So far if have the base case 1(2^1) = (1-1) * (1+1) + 2

    Induction part is giving me problems:
    k
    \sum i(2^i) = (k-1) * 2^(k+1) + 2 = ((k+1)-1 * 2^ ((k+1)+1 + 2
    i=1
    Assume that what was to be proved holds for n=k, then:

    \displaystyle \sum_{i=1}^k i 2^i=(k-1)2^{k+1}+2

    Add the next term from the left to both sides:

    \displaystyle (k+1)2^{k+1}+\sum_{i=1}^k i 2^i =\sum_{i=1}^{k+1} i 2^i =(k-1)2^{k+1}+2 + (k+1)2^{k+1}

    Now simplify the right hand side.

    CB
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