Matematické Fórum

stuart clark · 01. 02. 2017 15:26

finding $\displaystyle \sum\sum_{0 \leq i < \leq n}(i+j)\binom{n}{i}\binom{n}{j}$

Brano · 09. 02. 2017 02:28

$\sum_{i<j}(i+j){n\choose i}{n\choose j}=\sum_{i<j}i{n\choose i}{n\choose j}+\sum_{i<j}j{n\choose i}{n\choose j}=\sum_{i\not=j}i{n\choose i}{n\choose j}=n\sum_{i\not=j}{n-1\choose i-1}{n\choose j}=$
$=n\left[\sum_{i,j}{n-1\choose i-1}{n\choose j}-\sum_{i}{n-1\choose i-1}{n\choose i}\right]=n\left[2^{n-1}2^n-\sum_{i}{n-1\choose n-i}{n\choose i}\right]=n\left[2^{2n-1}-{2n-1\choose n}\right]$

stuart clark · 23. 02. 2017 14:02

Thanks ↑ Brano:

Matematické Fórum

#1 01. 02. 2017 15:26

double summation

#2 09. 02. 2017 02:28

Re: double summation

#3 23. 02. 2017 14:02

Re: double summation

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