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

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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

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
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

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
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

你的问题没阐述明白