若/AB-2/+(A-2)平方=0,求1/AB+1/(A+1)(B+1)+1/(A+2)(B+2)+……+1/(A+2010)(B+2010)的值.

问题描述:

若/AB-2/+(A-2)平方=0,求1/AB+1/(A+1)(B+1)+1/(A+2)(B+2)+……+1/(A+2010)(B+2010)的值.

解得=2011/2012

由/AB-2/+(A-2)平方=0得:AB=2且A-2=0
所以A=2,B=1
代入1/AB+1/(A+1)(B+1)+1/(A+2)(B+2)+……+1/(A+2010)(B+2010)
得:1/1*2+1/2*3+1/3*4+……+1/2011*2012
=1/1-1/2+1/2-1/3+1/3-1/4……+1/2011-1/2012
=1/1-1/2012
=2011/2012

ab-2=0a-2=0∴a=2b=11/AB+1/(A+1)(B+1)+1/(A+2)(B+2)+……+1/(A+2010)(B+2010)=1/1×2+1/2×3+1/3×4+1/4×5+……+1/2011×2012=1-1/2+1/2-1/3+1/3-1/4+1/4-……-1/2011+1/2011-1/2012=1-1/2012=2011/20...