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Math Help - proof by induction with sigma

  1. #1
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    proof by induction with sigma

    \sum_{i=1}^{n+1}i\cdot2^i=2^{n+2}+2 for all n\geq 0.
    Last edited by Chris L T521; October 3rd 2009 at 10:28 PM. Reason: reformatted the question.
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  2. #2
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    Hello gohangoten1
    Quote Originally Posted by gohangoten1 View Post
    \sum_{i=1}^{n+1}i\cdot2^i=2^{n+2}+2 for all n\geq 0.
    This is incorrect. n = 2 gives

    \sum_{i=1}^{3}i\cdot2^i=2^{4}+2

    The LHS is 1.2 + 2.4 + 3.8 = 34

    RHS = 16+2=18

    Grandad
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  3. #3
    Junior Member Renji Rodrigo's Avatar
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    I think the formula is \sum^{n}_{k=0} k2^{k} = 2^{n+1}(n-1)+2
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  4. #4
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    Hello everyone

    Thanks to
    Renji Rodrigo for this:
    Quote Originally Posted by Renji Rodrigo View Post
    I think the formula is \sum^{n}_{k=0} k2^{k} = 2^{n+1}(n-1)+2
    Adapting this formula to look more like the original, we get

    \sum^{n+1}_{i=1} i\cdot2^{i} = 2^{n+2}n+2

    So let P(n) be the propositional function S_n=\sum^{n+1}_{i=1} i\cdot2^{i} = 2^{n+2}n+2

    P(1) is S_1=\sum^{2}_{i=1} i\cdot2^{i} = 2^{3}\cdot1+2

    i.e. S_1=1\cdot2 + 2\cdot2^2 = 10 = 2^{3}\cdot1+2

    So P(1) is true.

    Now P(n) \Rightarrow S_{n+1} = S_n+(n+2)\cdot2^{n+2}

    =2^{n+2}n+2+(n+2)\cdot2^{n+2}

    =2^{n+2}(n+n+2)+2

    =2^{n+2}(2n+2)+2

    =2^{n+2}\cdot2(n+1)+2

    =2^{(n+1)+2}\cdot(n+1)+2

    i.e. P(n) \Rightarrow P(n+1)

    So P(n) is true for all n \ge 1

    Grandad
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