若│a-1│+│ab-2│=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2002)(b+2002)的值,本人有急用,
问题描述：
若│a-1│+│ab-2│=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2002)(b+2002)的值,本人有急用,
答
解得：a=1,b=2
根据分解公式：
1/ab+1/(a+1)(b+1)+...+1/(a+2002)(b+2002)
=1/(1*2)+1/(2*3)+...+1/(2003*2004)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+.+(1/2002-1/2003)+(1/2003-1/2004)
=1-1/2004
=2003/2004