Chứng minh:
$\sum_{k=1}^{\frac{p-1}{2}}\frac{\left ( p-1 \right )!}{k}=\frac{2^{p}-2}{p}\left( mod p \right )$
(p là số nguyên tố lẻ)
Edited by NTMFlashNo1, 28-11-2016 - 00:49.
$\sum_{k=1}^{\frac{p-1}{2}}\frac{\left ( p-1 \right )!}{k}$
Started By NTMFlashNo1, 28-11-2016 - 00:47
#1
Posted 28-11-2016 - 00:47
NTMFlashNo1
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$\boxed{\text{Nguyễn Trực-TT-Kim Bài secondary school}}$
#2
Posted 12-08-2017 - 20:46
dogsteven
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Đại úy
Ta có: $p!\sum\limits_{k=1}^{\frac{p-1}{2}}\dfrac{1}{k}\equiv -p!\sum\limits_{k=1}^{p-1}\dfrac{(-1)^{k+1}}{k}\equiv -(p-1)!(2^p-2)\equiv 2^p-2\pmod{p^2}$
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Quyết tâm off dài dài cày hình, số, tổ, rời rạc.
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