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Math Help - Iterated sums

  1. #1
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    Iterated sums

    Sorry if this is the wrong forum.

    Is there a "nice" formula for the following? :

    \sum_{k_2=0}^{k_1}\sum_{k_3=0}^{k_2}\sum_{k_4=0}^{  k_3} ... \sum_{k_n=0}^{k_{n-1}}1

    I do the first few iterations and it just seems to get messy :S

    Thanks for any help
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  2. #2
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    Quote Originally Posted by Aileys. View Post
    Sorry if this is the wrong forum.

    Is there a "nice" formula for the following? :

    \sum_{k_2=0}^{k_1}\sum_{k_3=0}^{k_2}\sum_{k_4=0}^{  k_3} ... \sum_{k_n=0}^{k_{n-1}}1
    You are using the same indices more than once.
    Do you mean
    \sum_{j_2=0}^{k_1}\sum_{j_3=0}^{k_2}\sum_{j_4=0}^{  k_3} ... \sum_{j_n=0}^{k_{n-1}}1?

    If not how can the indices change like that?
    What am I missinng?

    Edit
    I will say that \sum_{j_2=0}^{k_1}\sum_{j_3=0}^{k_2}\sum_{j_4=0}^{  k_3} ... \sum_{j_n=0}^{k_{n-1}}1 =\prod\limits_{j = 1}^{k_{n - 1} } {\left( {k_j  + 1} \right)}
    Last edited by Plato; October 14th 2009 at 03:57 PM.
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  3. #3
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    Yes the indices are supposed top be like that, - that's why it's messy!

    For example \sum_{k_2=0}^{k_1}\sum_{k_3=0}^{k_2}1 = \sum_{k_2=0}^{k_1}(k_2+1) = \frac{k_1(k_1+1)}{2}-(k_1+1)

    Perhaps I'm abusing notation... if so, sorry! But hopefully that^ explains what I'm trying to do
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  4. #4
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    Quote Originally Posted by Aileys. View Post
    Sorry if this is the wrong forum.

    Is there a "nice" formula for the following? :

    \sum_{k_2=0}^{k_1}\sum_{k_3=0}^{k_2}\sum_{k_4=0}^{  k_3} ... \sum_{k_n=0}^{k_{n-1}}1

    I do the first few iterations and it just seems to get messy :S

    Thanks for any help
    The sum is the same as the number of sequences
    0 \leq k_n \leq k_{n-1} \leq \dots \leq k_2 \leq k_1 ,

    which is the same as the number of multisets of size n taken from a set of k_1 objects,

    which is
    \binom{k_1 +n -1}{n}.
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