=1,a=2
所以1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2011)(b+2011)
=1/(1×2)+1/(2×3)+1/(3×4)+.+1/(2011×2010)
=1-1/2+1/2-1/3+1/3-1/4+ .+1/2010-1/2011
=2010/2011
使用方法为裂项相消
=1,a=2
所以1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2011)(b+2011)
=1/(1×2)+1/(2×3)+1/(3×4)+.+1/(2011×2010)
=1-1/2+1/2-1/3+1/3-1/4+ .+1/2010-1/2011
=2010/2011
使用方法为裂项相消