(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)

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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)

(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)
(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)

(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t=1/2+1/3+···+1/2003
原式=（1+t）（t+1/2004）-（1+t+1/2004）t
=[t²+（1+1/2004）t+1/2004] - [t²+（1+1/2004）t]
=1/2004

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