lab-2l+(b-1)^2=0
lab-2l>=0且(b-1)^2>=0
所以ab-2=0且b-1=0
即a=2,b=1
又有1/n*(n+1)=1/n-1/(n+1)
所以1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...1/(a+99)(b+99)
=1/2+1/2*3+1/3*4+……+1/100*101
=1-1/2+1/2-1/3+1/3-1/4+……+1/100-1/101
=1-1/101
=100/101
如果lab-2l+(b-1)的2次幂=0,试求：1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...1/
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