设[1/2+
1
3+…+
1
2010]=x
则原式=x(1+x+[1/2011])-(1+x)(x+[1/2011])
=x+x2+[1/2011]x-(x+[1/2011]+x2+[1/2011]x)
=x+x2+[1/2011]x-x-[1/2011]-x2-[1/2011]x
=-[1/2011].
设[1/2+
1
3+…+
1
2010]=x
则原式=x(1+x+[1/2011])-(1+x)(x+[1/2011])
=x+x2+[1/2011]x-(x+[1/2011]+x2+[1/2011]x)
=x+x2+[1/2011]x-x-[1/2011]-x2-[1/2011]x
=-[1/2011].