(1/2+1/3…+1/2003)(1+1/2+1/3+…+1/2004)-(1+1/2+1/3+…+1/2003)(1/2+1/3+…+1/2004)=?

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(1/2+1/3…+1/2003)(1+1/2+1/3+…+1/2004)-(1+1/2+1/3+…+1/2003)(1/2+1/3+…+1/2004)=?(1/2+1/3…+1/2003)(1+1/

(1/2+1/3…+1/2003)(1+1/2+1/3+…+1/2004)-(1+1/2+1/3+…+1/2003)(1/2+1/3+…+1/2004)=?
(1/2+1/3…+1/2003)(1+1/2+1/3+…+1/2004)-(1+1/2+1/3+…+1/2003)(1/2+1/3+…+1/2004)=?

(1/2+1/3…+1/2003)(1+1/2+1/3+…+1/2004)-(1+1/2+1/3+…+1/2003)(1/2+1/3+…+1/2004)=?
设a=(1/2+1/3+...+1/2003) 原式=a(1+a+1/2004)-(1+a)(a+1/2004) =a+a^2+a/2004-a-1/2004-a^2-a/2004 =-1/2004