已知:Ia+1I+Iab+2I=0,求1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/（a-2002）(b+2002)
问题描述：
已知:Ia+1I+Iab+2I=0,求1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/（a-2002）(b+2002)
如题
答
你这题有问题,应该
Ia-1I+Iab-2I=0
满足上式的条件是
a-1=0
ab-2=0
解得 a=1 b=2
所以1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/（a+2002）(b+2002)
=1/2*3+1/3*4+...+1/2003*2004
=(1/2-1/3)+(1/3-1/4)+(1/2003-1/2004)
=1/2-1/2004
=1001/2004