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Finding the formula for summation.

  1. Sep 14, 2013 #1
    1. The problem statement, all variables and given/known data

    Find a formula for

    [itex]\sum (2i-1)[/itex] =1+3+5+...+(2n-1)

    2. Relevant equations



    3. The attempt at a solution

    [itex]\sum(2i-1)[/itex]=(1+2+3+...+2n)-(2+4+6+...+2n)
    =(1+2+3+...+2n)-n(n+1)

    I'm unsure what to do with 1+2+3+..+2n ?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 14, 2013 #2

    statdad

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    Maybe start with
    [tex]
    \sum_{i=1}^n \left(2i-1\right) = 2\sum_{i=1}^n i - \sum_{i=1}^n 1
    [/tex]

    and simplify?
     
  4. Sep 14, 2013 #3

    arildno

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    Dearly Missed

    statdad's suggestion is one way of doing this. Alternatively, you might spot a pattern by looking at the partial sums:
    1=?
    1+3=?
    1+3+5=?

    Once you spot the pattern, how could you go about proving whether it is actually universally true, or a mere fluke?
     
  5. Sep 14, 2013 #4
    I think I fugured it out lol.

    The pattern is n^2.
    So then,

    (1+2+3+...+2n)-n(n+1) = (2n(2n+1))/2 -n(n+1)=n(2n+1)- n(n+1) =2n^2+n-n^2-n=n^2
    So the general formula is n^2.
     
  6. Sep 14, 2013 #5

    arildno

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    Yup!
    :smile:
     
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