观察式子(1-1/2²)(1-1/3²）=(1-1/2)(1+1/2)(1-1/3)(1+1/3)=1/2x3/2x2/3x4/3=2/3试求
试求(1-1/2²）(1-1/3²)x…x(1-1/2009²)(1-1/2010²）的值
第1个回答 2012-12-18
(1-1/2²）(1-1/3²)x…x(1-1/2009²)(1-1/2010²）
=（1+1/2)(1-1/2)(1+1/3)(1-1/3)............(1+1/2009)(1-1/2009)(1+1/2010)(1-1/2010)
=(3/2)(1/2)(4/3)(2/3)................(2010/2009)(2008/2009)(2011/2010)(2009/2010)
=(1/2)(2011/2010)
=2011/4020
=（1+1/2)(1-1/2)(1+1/3)(1-1/3)............(1+1/2009)(1-1/2009)(1+1/2010)(1-1/2010)
=(3/2)(1/2)(4/3)(2/3)................(2010/2009)(2008/2009)(2011/2010)(2009/2010)
=(1/2)(2011/2010)
=2011/4020
第2个回答 2012-12-18
(1-1/2²）(1-1/3²)x…x(1-1/2009²)(1-1/2010²）
=(1-1/2)(1+1/2)(1-1/3)(1+1/3)x……x(1-1/2010)(1+1/2010)
=1/2x3/2x2/3x4/3x……x2011/2010
=1/2x2011/2010
=2011/4020来自：求助得到的回答本回答被提问者和网友采纳
=(1-1/2)(1+1/2)(1-1/3)(1+1/3)x……x(1-1/2010)(1+1/2010)
=1/2x3/2x2/3x4/3x……x2011/2010
=1/2x2011/2010
=2011/4020来自：求助得到的回答本回答被提问者和网友采纳
第2个回答 2012-12-18
1/2*(......)*(1+1/2010)
=2011/4020
=2011/4020
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