Question

已知根号a-1+(ab-2)平方=0.求1/ab+1/(a+2)(b+2)+...+1/(a+2011)(b+2011)的值

Answer 1

a-1=0
ab-2=0
∴a=1
b=2
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+……+1/2012×2013
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/2012+1/2012-1/2013
=2012/2013

Answer 2

解：由题意可知
a-1=0，ab-2=0
所以a=1，b=2
原式=1/(1+0)(1+1)+1/(1+1)(1+2)+...+1/(1+2011)(2+2011)
因为1/a(a+1)=1/a-1/(a+1)（通分计算一下就可以了）
所以原式=1-1/2+1/2-1/3+...+1/2012-1/2013
=2012/2013

Answer 3

a-1=0
ab-2=0
∴a=1
b=2
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+……+1/2012×2013
=1-1/2+1/2-1/3+1/3-1/4+1/4......1/2012+1/2012-1/2013
=2012/2013