Chứng minh rằng $\sqrt{2\sqrt{3\sqrt{4\sqrt{5...\sqrt{2000}}}}}< 3$
Chứng minh rằng $\sqrt{2\sqrt{3\sqrt{4\sqrt{5...\sqrt{2000}}}}}< 3$
Started By shinichikudo201, 16-11-2014 - 15:10
#1
Posted 16-11-2014 - 15:10
shinichikudo201
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It is the quality of one's convictions that determines success, not the number of followers
#2
Posted 16-11-2014 - 18:29
Nguyentiendung9372
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Ta có
$\sqrt {2\sqrt {3\sqrt {4\sqrt {...\sqrt {1999\sqrt {2000} } } } } } < \sqrt {2\sqrt {3\sqrt {4\sqrt {...\sqrt {1999.2000} } } } } < \sqrt {2\sqrt {3\sqrt {4\sqrt {...\sqrt {1999.2001} } } } } < \sqrt {2\sqrt {3\sqrt {4\sqrt {...\sqrt {{{2000}^2}} } } } } = \sqrt {2\sqrt {3\sqrt {4\sqrt {...\sqrt {1998.2000} } } } } < \sqrt {2\sqrt {3\sqrt {4\sqrt {...\sqrt {1997\sqrt {{{1999}^2}} } } } } } ... < \sqrt {2.4} < 3$
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