(1+1/2+1/3+···+1/2003)(1/2+1/3+···+1/2004)-(1+1/2+1/3+···+1/
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(1+1/2+1/3+···+1/2003)(1/2+1/3+···+1/2004)-(1+1/2+1/3+···+1/2004)(1/2+1/3+···+1/2003)
答:
设a=1+1/2+1/3+···+1/2003
(1+1/2+1/3+···+1/2003)(1/2+1/3+···+1/2004)-(1+1/2+1/3+···+1/2004)(1/2+1/3+···+1/2003)
=a(a-1+1/2004)-(a+1/2004)(a-1)
=a(a-1)+a/2004-a(a-1)-(a-1)/2004
=a/2004-a/2004+1/2004
=1/2004
设a=1+1/2+1/3+···+1/2003
(1+1/2+1/3+···+1/2003)(1/2+1/3+···+1/2004)-(1+1/2+1/3+···+1/2004)(1/2+1/3+···+1/2003)
=a(a-1+1/2004)-(a+1/2004)(a-1)
=a(a-1)+a/2004-a(a-1)-(a-1)/2004
=a/2004-a/2004+1/2004
=1/2004
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