(1/2+1/3+…+1/2003)(1+1/2+…+1/2002)-(1+1/2+…+1/2003)(1/2+1/3+…+1/2002)=?
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![(1/2+1/3+…+1/2003)(1+1/2+…+1/2002)-(1+1/2+…+1/2003)(1/2+1/3+…+1/2002)=?](/uploads/image/z/11514141-45-1.jpg?t=%281%2F2%2B1%2F3%2B%E2%80%A6%2B1%2F2003%29%281%2B1%2F2%2B%E2%80%A6%2B1%2F2002%29-%281%2B1%2F2%2B%E2%80%A6%2B1%2F2003%29%281%2F2%2B1%2F3%2B%E2%80%A6%2B1%2F2002%29%3D%3F)
(1/2+1/3+…+1/2003)(1+1/2+…+1/2002)-(1+1/2+…+1/2003)(1/2+1/3+…+1/2002)=?
(1/2+1/3+…+1/2003)(1+1/2+…+1/2002)-(1+1/2+…+1/2003)(1/2+1/3+…+1/2002)=?
(1/2+1/3+…+1/2003)(1+1/2+…+1/2002)-(1+1/2+…+1/2003)(1/2+1/3+…+1/2002)=?
令1/2+1/3+…+1/2002=x
则原式
=(x+1/2003)(1+x)-(1+x+1/2003)x
=x+x²+x/2003+1/2003-x-x²-x/2003
=1/2003