已知(a-1)的平方根+(ab-2)的平方=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2004)(b+2004)的值
问题描述:
已知(a-1)的平方根+(ab-2)的平方=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2004)(b+2004)的值
答
(a-1)的平方根+(ab-2)的平方=0,即A-1=0 AB-2=0 ==>A=1 B=2又1/AB=1/A-1/B=1-1/21/(A+1)(B+1)=1/(A+1)-1/(B+1)=1/2-1/3...1/(A+2004)(B+2004)=1/(A+2004)-1/(B+2004)=1/2005-1/2006所以1/ab+1/(a+1)(b+1)+1/(a+2)(...