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Induction Proof with combination

  1. Nov 7, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove:

    [itex]\sum[/itex][itex]^{n}_{r=0}[/itex]2r([itex]^{n}_{r}[/itex]) = 3n


    2. Relevant equations



    3. The attempt at a solution

    I proceeded by induction:

    Testing the base case for n=0 is correct.


    Moving right along to try to show:

    [itex]\sum[/itex][itex]^{n+1}_{r=0}[/itex]2r([itex]^{n}_{r}[/itex]) = 3n+1



    This is where I'm getting stuck. I can obtain:

    32+2n+1([itex]^{n+1}_{n}[/itex])

    Which I think equals: 32+2n+1([itex]^{n}_{n-1}[/itex])

    Which equals: 32+2n+1*n




    I am not sure how to proceed after this. Any help would be greatly appreciated.

    also, this was a problem on a test I took yesterday.
     
  2. jcsd
  3. Nov 7, 2013 #2

    tiny-tim

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    hi srfriggen! :smile:

    much easier would be a proof using the binomial expansion ((a + b)n) :wink:
     
  4. Nov 7, 2013 #3
    So I can say, by the binomial theorem:

    3n=(2+1)n=[itex]\sum[/itex][itex]^{n}_{r}[/itex]([itex]^{n}_{r}[/itex])2r*1n-r
     
  5. Nov 7, 2013 #4

    tiny-tim

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    yup! :biggrin:

    quicker? o:)
     
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