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I have a nested sum like so: $$\underbrace{\sum_{k_1=k_0}^{k^*} \ ... \sum_{k_n=k_{n-1}}^{k^*}}_{\text{n times}} 1\quad\ \text{with}\ \ n, k_0, k^* \in \mathbb{N},\ k^*\geq k_0$$

Is there a general, shorter representation that spares me calculating the actual sums?

trion
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    A lot of authors choose to represent such a multiple sum using a chain of inequalities under the summation symbol $$\sum_{k_0\le k_1\le \cdots \le k_n\le k^*} 1$$ – achille hui Feb 22 '14 at 03:23
  • Try induction. See if you can express the sum for n in terms of n-1. – marty cohen Feb 22 '14 at 14:36
  • $\displaystyle\sum_{k_0\le k_1\le\cdots\le k_n\le k^}1=\sum_{1\le k_1-k_0+1\le k_2-k_0+2\le\cdots\le k_n-k_0+n\le k^-k_0+n}1={k^*-k_0+n\choose n}$ – hxthanh Apr 08 '14 at 06:04

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$\displaystyle\sum_{k_0\le k_1\le\cdots\le k_n\le k^*}1=\sum_{1\le k_1-k_0+1< k_2-k_0+2<\cdots< k_n-k_0+n\le k^*-k_0+n}1={k^*-k_0+n\choose n}$

hxthanh
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