. Notice that the summation indices are 
.#2 14. 03. 2015 16:47 — Editoval Pavel (14. 03. 2015 17:00)
Re: Finite K-fold sum
.Backslash je v TeXu tak důležitý jako nekonečno při dělení nulou v tělesech charakteristiky 0.
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#3 14. 03. 2015 19:01
- check_drummer
- Příspěvky: 5577
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Re: Finite K-fold sum
so the sum is .
so the sum is .
) out of the set containing N+K-1 elements - so the sum equals "Máte úhel beta." "No to nemám."
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#4 14. 03. 2015 19:28
Re: Finite K-fold sum
↑ Pavel:
I look forward to seeing your complete proof.
↑ check_drummer:
Thank you. However, the measure of my combinatorics knowledge is allmost zero.
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#5 15. 03. 2015 10:47
Re: Finite K-fold sum
, 
be sequences of real numbers. Let us define:
, if 
and 
for 
.
![$
\sum_{N\ge n_1\geq n_2\geq\dots\geq n_K\geq 1}1&=\sum_{n_1=1}^N\sum_{n_2=1}^{n_1}\cdots\sum_{n_{K-2}=1}^{n_{K-3}}\sum_{n_{K-1}=1}^{n_{K-2}}\sum_{n_K=1}^{n_{K-1}}n_K^{\overline 0}
=\sum_{n_1=1}^N\sum_{n_2=1}^{n_1}\cdots\sum_{n_{K-2}=1}^{n_{K-3}}\sum_{n_{K-1}=1}^{n_{K-2}}n_{K-1}^{\overline 1}\\[.5\baselineskip]
&=\sum_{n_1=1}^N\sum_{n_2=1}^{n_1}\cdots\sum_{n_{K-2}=1}^{n_{K-3}}\frac{n_{K-2}^{\overline 2}}{2}\dots
=\sum_{n_1=1}^N\sum_{n_2=1}^{n_1}\frac{n_2^{\overline{K-2}}}{(K-2)!}
=\sum_{n_1=1}^N\frac{n_1^{\overline{K-1}}}{(K-1)!}\\[.5\baselineskip]
&=\frac{N^{\overline K}}{K!}=\frac{N(N+1)(N+2)\cdots(N+K-1)}{K!}=\color{blue}{{N+K-1}\choose{K}}
$](/mathtex/b9/b933f18456bcdefc40f2cbd894f69010.gif "kopírovat do textarea")
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