數學知識交流---求證

2011-07-22 3:38 am
(1) 求證:1/64 < (1/2) × (3/4) × (5/6) × ... × (2007/2008) × (2009/2010) < 1/44

(2) x - 0 = 0 , 求 x .

回答 (3)

2011-07-22 6:58 am
✔ 最佳答案
1)(1/2) × (2/3) × (4/5) × ... × (2006/2007) × (2008/2009)
<
(1/2) × (3/4) × (5/6) × ... × (2007/2008) × (2009/2010)
==>
(1/2) × (2/3) × (4/5) × ... × (2006/2007) × (2008/2009)
x (1/2) × (3/4) × (5/6) × ... × (2007/2008) × (2009/2010)
<
[ (1/2) × (3/4) × (5/6) × ... × (2007/2008) × (2009/2010) ]²
==>
(1/2) x (1/2)(2/3)(3/4)(4/5)(5/6) x ... x (2007/2008)(2008/2009)(2009/2010)
<
[ (1/2) × (3/4) × (5/6) × ... × (2007/2008) × (2009/2010) ]²
==>
(1/2) x (1/2010) < [ (1/2) × (3/4) × (5/6) × ... × (2007/2008) × (2009/2010) ]²
==>
1/4020 < [ (1/2) × (3/4) × (5/6) × ... × (2007/2008) × (2009/2010) ]²
1/4096 < [ (1/2) × (3/4) × (5/6) × ... × (2007/2008) × (2009/2010) ]²
1/64 < (1/2) × (3/4) × (5/6) × ... × (2007/2008) × (2009/2010) .....(1);(1/2) × (3/4) × (5/6) × ... × (2007/2008) × (2009/2010)
<
(2/3) × (4/5) × (6/7) × ... × (2008/2009) × (2010/2011)
==>
[ (1/2) × (3/4) × (5/6) × ... × (2007/2008) × (2009/2010) ]²
<
(2/3) × (4/5) × (6/7) × ... × (2008/2009) × (2010/2011)
x (1/2) × (3/4) × (5/6) × ... × (2007/2008) × (2009/2010)
==>
[ (1/2) × (3/4) × (5/6) × ... × (2007/2008) × (2009/2010) ]² < 1/2011
[ (1/2) × (3/4) × (5/6) × ... × (2007/2008) × (2009/2010) ]² < 1/1936
(1/2) × (3/4) × (5/6) × ... × (2007/2008) × (2009/2010) < 1/44 .....(2)
(1) & (2) :
1/64 < (1/2) × (3/4) × (5/6) × ... × (2007/2008) × (2009/2010) < 1/44

2)x - 0 = 0x = 0
2011-07-22 5:36 am
(2) x - 0 = 0 , 求 x .
x-0=0
x=0+0=0

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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110721000051KK01009

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