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