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Thread: 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:

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

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

    I think I evaluate down to :

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

    Thanks in advance.
    Last edited by mc72; Jul 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 $\displaystyle n1$ ?
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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 $\displaystyle 3n.$
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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 $\displaystyle \sum_{j=1}^n1=n,$ if the upper index of the sum were $\displaystyle 3n,$ 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 $\displaystyle \sum_{j=1}^n1=n,$ if the upper index of the sum were $\displaystyle 3n,$ what's the new value of the sum?

    Ok, so it should evaluate next to


    $\displaystyle \sum_{k=2}^n3n$

    then

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

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

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