(1-1\2²)(1-1\3²)(1-1\4²)……(1-1\1004²)过程
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(1-1\2²)(1-1\3²)(1-1\4²)……(1-1\1004²)过程
(1-1\2²)(1-1\3²)(1-1\4²)……(1-1\1004²)过程
(1-1\2²)(1-1\3²)(1-1\4²)……(1-1\1004²)过程
解:原式=
(1-1/2²)(1-1/3²)(1-1/4²)(1-1/5²)……(1-1/2004²)(1-1/2005²)
=(1+1/2)(1-1/2)(1+1/3)(1-1/3)……(1+1/2004)(1-1/2004)(1+1/2005)(1-1/2005)
=(3/2)(1/2)(4/3)(2/3)(...
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解:原式=
(1-1/2²)(1-1/3²)(1-1/4²)(1-1/5²)……(1-1/2004²)(1-1/2005²)
=(1+1/2)(1-1/2)(1+1/3)(1-1/3)……(1+1/2004)(1-1/2004)(1+1/2005)(1-1/2005)
=(3/2)(1/2)(4/3)(2/3)(5/4)(3/4)(6/5)(4/5)…(2005/2004)(2003/2004)(2006/2005)(2004/2005)
注意到上式中第一项(3/2)与(2/3)的积为1;第三项(4/3)与第六项(3/4)的积为1,以后的各项按照该方式均为1.
但其中的第二项1/2和2006/2005保留
所以:
上式=1/2*2006/2005
=2003/2005
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