因为 Iab-2I+Ib-1I=0 所以b=1 a=2
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2001)(b+2001)
=1/1*2+1/2*3+1/3*4+1/4*5...+1/2002*2003
=1-1/2+1/2-1/3+1/3-1/4+1/4...+1/2002-1/2003
=1-1/2003
=2002/2003
已知 Iab-2I+Ib-1I=0 求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+200
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