已知(a-1)^2+│ab-2│=0求1/ab+1/(a+1)(b+2)+1/(a+2)(b+2)+...+1/(a+2005)(b+2005)的值
问题描述：
已知(a-1)^2+│ab-2│=0求1/ab+1/(a+1)(b+2)+1/(a+2)(b+2)+...+1/(a+2005)(b+2005)的值
答
因为平方和绝对值都是非负的,已知中两者相加为0,只能是两者都为0于是 a=1,b=2同时1/ab=1/a-1/b=1-1/21/(a+1)(b+1)=1/(2*3)=1/2-1/3...1/(a+2005)(b+2005)=1/(2006*2007)=1/2006-1/2007累加可得等式右边的被减数与下...