已知:实数a,b满足条件√a-1+(ab-2)²=0试求1/ab+1/(a+1)(b+1)+..+ 1/（a+

这是一道很古老的题目了.
算术平方根与平方项均恒非负,两非负项之和=0,两非负项分别=0
a-1=0 a=1
ab-2=0 b=2/a=2/1=2
b=a+1
1/(ab)+1/[(a+1)(b+1)]+...+1/[(a+2013)(b+2013)]
=1/[a(a+1)]+1/[(a+1)(a+2)]+...+1/[(a+2013)(a+2014)]
=1/a-1/(a+1)+1/(a+1)-1/(a+2)+...+1/(a+2013)-1/(a+2014)
=1/a -1/(a+2014)
=1/1-1/(1+2014)
=1-1/2015
=2014/2015

√(a-1)+(ab-2)²=0
√(a-1)>=0,(ab-2)²>=0
所以a-1=0,ab-2=0
a=1,
b=2
所以1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+........+1/(a+2013)(b+2013)
=1/2+1/(2*3)+1/(3*4)+.....+1/(2014*2015)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+.....+(1/2014-1/2015)
=1-1/2015
=2014/2015

a-1=0 ab-2=0
a=1, b=2
1/ab+1/(a+1)(b+1)+..+ 1/（a+2013﹚﹙b+2013﹚
=1/1*2+1/2*3+...+1/2014*2015
=1-1/2+1/2-1/3+1/3-1/4+...+1/2013-1/2014+1/2014-1/2015
=1-1/2015
=2014/2015