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Double summation,

  1. Jun 16, 2013 #1
    1. The problem statement, all variables and given/known data
    How can I compute the sum
    An example to calculate
    [tex]\sum_{i=1}^n\sum_{j=i+1}^n(i+2j)[/tex]?? I only have an example where n=1 and it gives a sum of 0 (why?)

    Maybe with n=3, what would the expanded form look like?


    2. Relevant equations
    I know how to do double sums, but when I have indexes ranging from a constant to a constant. But I haven't encountered one where the index is dependent on the other sum's index.


    3. The attempt at a solution
    I tried computing the inner sum first:

    [tex]\sum_{i=1}^3 \left( \sum_{j=i+1}^{3}i +2\sum_{j=i+1}^{3}j \right)[/tex]
    [tex]=\sum_{i=1}^3 \left(i(3-(i+1)+1) +2\sum_{j=i+1}^{3}j \right)[/tex]
    [tex]=\sum_{i=1}^3 \left(3i-i^2 +2\sum_{j=i+1}^{3}j \right)[/tex]
    [tex]=3(1+2+3)-1^2-2^2-3^2+2\sum_{i=1}^3\sum_{j=i+1}^{3}j[/tex]

    I don't know what to do about the summation of j, since it's indexed at j but it starts at i+1?

    I only know that the answer is 20 because of Wolfram for when n = 3
    http://www.wolframalpha.com/input/?i=Sum[i+2j,+{i,+1,+3},+{j,+i+1,+3}]
     
    Last edited: Jun 16, 2013
  2. jcsd
  3. Jun 16, 2013 #2

    lurflurf

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    Just do the inner sum first
    $$\sum_{i=1}^n\sum_{j=i+1}^n(i+2j)=\sum_{i=1}^n((n-i)i+n(n+1)-i(i+1))=\sum_{i=1}^n(n(n+1)+(n-1)i-2i^2)$$
     
  4. Jun 16, 2013 #3

    SteamKing

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    When the starting value of the summation index is greater than the final value of the summation index, then by implication, the sum = 0, since there are no terms to be summed.
     
  5. Jun 16, 2013 #4
    May I ask how you are computing the "2j" sum? I'm guessing it's the 2 last terms n(n+1) - i(i+1) since (n-i)i sums the "i" term.
     
    Last edited: Jun 16, 2013
  6. Jun 16, 2013 #5
    I tried going back to this and elaborated more, by thinking that the sum of j is just like the sum of integers formula

    [tex]\sum_{i=1}^n \left( \sum_{j=i+1}^{n}i +2\sum_{j=i+1}^{n}j \right)[/tex]
    [tex]\sum_{i=1}^n \left(i(n-(i+1)+1) +2\sum_{j=1}^{n-i}(j+i) \right)[/tex] by index shifting
    [tex]\sum_{i=1}^n \left(i(n-i) +2\sum_{j=1}^{n-i}j+2\sum_{j=1}^{n-i}i \right)[/tex]
    [tex]\sum_{i=1}^n \left(i(n-i) +2*\frac{(n-i)(n-i+1)}{2}+2*(i(n-i-1+1)) \right)[/tex]using the sum of consectuive integers formula on j
    [tex]\sum_{i=1}^n \left(i(n-i) +(n-i)(n-i+1)+2i(n-i) \right)[/tex]

    Though this doesn't give the same answer..
     
  7. Jun 16, 2013 #6

    haruspex

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    Doesn't give the same answer as what? Looks right, and it produces 20 for n=3. How about finishing the summation into closed form?
     
  8. Jun 16, 2013 #7

    lurflurf

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    There are many ways to manipulate this. One way it to note

    $$\sum_{j=i+1}^n=\sum_{j=1}^n-\sum_{j=i}^n$$
    Remember the outside index is treated as constant in the inside sum
    use these formula you should know
    $$\sum_{k=1}^n 1=n \\
    \sum_{k=1}^n k=n(n+1)/2 \\
    \sum_{k=1}^n k^2=n(n+1)(2n+1)/6 $$

    here is how to expand n=3 maybe that will help
    $$\require{cancel} \sum_{i=1}^3 \sum_{j=i+1}^3 (i+2j)=\sum_{j=1+1}^3 (1+2j)+ \sum_{j=2+1}^3 (2+2j)+ \cancel{\sum_{j=3+1}^3 (3+2j)} \\
    =(1+2\cdot2)+(1+2\cdot 3)+(2+2\cdot 3)=5+7+8=20$$

    The crossed out sum is 0
     
  9. Jun 16, 2013 #8
    Oh it did turn out to be the same answer. I think I confused one of my "i"s with a "1" on paper. Thanks a lot though for the supplemental info above!! :)
     
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