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What does 2n on top of the summation expression do diferently than just n?

  1. Sep 20, 2012 #1
    1. The problem statement, all variables and given/known data

    2n
    Ʃ (k)
    k=1

    3. The attempt at a solution

    2n
    Ʃ (k) = 2n(2n+1)/2 (This is just a shot in the dark.)
    k=1
     
  2. jcsd
  3. Sep 20, 2012 #2

    Dick

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    Sure. If you've shown the sum from 1 to n of k is n*(n+1)/2, then the sum to 2n is just 2n*(2n+1)/2. It's just substitution.
     
  4. Sep 20, 2012 #3
    So would this work too?

    2n+1
    Ʃ (k) = (2n+1)((2n+1)+1)/2
    k=1
     
  5. Sep 20, 2012 #4

    Dick

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    Sure. Same thing.
     
  6. Sep 21, 2012 #5

    Ray Vickson

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    [tex] \sum_{k=1}^N k = \frac{N(N+1)}{2},[/tex]
    so if you put N = 2n you get the stated result. If you set N = 2n+1 you get the other result you stated.

    RGV
     
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