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Math Help - Evaluating summations

  1. #1
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    Evaluating summations

    I'm still struggling with basic summation evaluation.

    I've managed to get a formula like this:

    \sum_{k=2}^n  \sum_{j=1}^{3n} 1

    But the 3n is confusing me and I'm not sure what my next step should be.

    I think I evaluate down to :

    \sum_{k=2}^n(3n-j)? and then where?

    Thanks in advance.
    Last edited by mc72; July 21st 2009 at 09:41 AM.
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  2. #2
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    Don't know what the summand exactly is.

    What's n1 ?
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  3. #3
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    Oops sorry, that was just a missing bracket in latex. Should be fixed now.
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  4. #4
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    That makes sense, but the inner sum is actually 3n.
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  5. #5
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    Thanks for the help, but I'm still a little fuzzy. I'm not sure what you mean by the inner sum is actually 3n?
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  6. #6
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    You're respected to know that is \sum_{j=1}^n1=n, if the upper index of the sum were 3n, what's the new value of the sum?
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  7. #7
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    Quote Originally Posted by Krizalid View Post
    You're respected to know that is \sum_{j=1}^n1=n, if the upper index of the sum were 3n, what's the new value of the sum?

    Ok, so it should evaluate next to


    \sum_{k=2}^n3n

    then

    3 \sum_{k=2}^nn^2 ?

    Sorry my summation skills are really fuzzy.
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  8. #8
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    That sum is w.r.t. k, then you can pull back that 3n and you'll have \sum_{k=2}^n1=\sum_{k=1}^{n-1}1=n-1.
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  9. #9
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    Quote Originally Posted by Krizalid View Post
    then you can pull back that 3n and you'll have \sum_{k=2}^n1=\sum_{k=1}^{n-1}1=n-1.

    Sorry again, but I'm not sure what you mean by pull back that 3n?
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