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Math Help - Double Summation question

  1. #1
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    Double Summation question

    For all n\in{N} find a formula for \sum{\sum{a_{ij}}}<br />
(first sum between j=1 and n, second between i=1 and j) in each of the following cases:

    i) a_{ij} = 1
    ii) a_{ij} = ij


    How would I go about this? Very confused.
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  2. #2
    Senior Member Peritus's Avatar
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    all you have to remember is the expression for a sum of an arithmetic progression (or use google):

    Here's the second case:

    \sum\limits_{j = 1}^n {\sum\limits_{i = 1}^n {ij}  = } \sum\limits_{j = 1}^n {j\sum\limits_{i = 1}^n i  = } \left[ {\left( {1 + n} \right)\frac{n}<br />
{2}} \right]^2
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  3. #3
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    Sorry, I think you misunderstood the limits. My fault for not knowing the code for it.

    It's this:

     \sum\limits_{j = 1}^n {\sum\limits_{i = 1}^j} {a_{ij}}

    for 1 and ij.
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  4. #4
    Senior Member Peritus's Avatar
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    <br />
\sum\limits_{j = 1}^n {\sum\limits_{i = 1}^j 1  = } \sum\limits_{j = 1}^n {j = \left( {1 + n} \right)\frac{n}<br />
{2}} <br />

    \sum\limits_{j = 1}^n {\sum\limits_{i = 1}^j {ij}  = \sum\limits_{j = 1}^n {j\left( {1 + j} \right)\frac{j}<br />
{2} = } } \frac{1}<br />
{2}\left[ {\sum\limits_{j = 1}^n {j^2  + } \sum\limits_{j = 1}^n {j^3 } } \right]

    these are well known sums Power Sum -- from Wolfram MathWorld
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  5. #5
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    Quote Originally Posted by Peritus View Post
    <br />
\sum\limits_{j = 1}^n {\sum\limits_{i = 1}^j 1  = } \sum\limits_{j = 1}^n {j = \left( {1 + n} \right)\frac{n}<br />
{2}} <br />

    \sum\limits_{j = 1}^n {\sum\limits_{i = 1}^j {ij}  = \sum\limits_{j = 1}^n {j\left( {1 + j} \right)\frac{j}<br />
{2} = } } \frac{1}<br />
{2}\left[ {\sum\limits_{j = 1}^n {j^2  + } \sum\limits_{j = 1}^n {j^3 } } \right]

    these are well known sums Power Sum -- from Wolfram MathWorld
    How did you get to this

    \sum\limits_{j = 1}^n {j\left( {1 + j} \right)\frac{j}<br />
{2} }
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