已知Iab-2I与Ib-1I互为相反数,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2008)(b+2008)
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已知Iab-2I与Ib-1I互为相反数,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2008)(b+2008)
已知Iab-2I与Ib-1I互为相反数,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2008)(b+2008)
已知Iab-2I与Ib-1I互为相反数,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2008)(b+2008)
Iab-2I+Ib-1I=0
所以ab-2=0,b-1=0
所以b=1
a=2/b=2
所以原式=1/1×2+1/2×3+……+1/2009×2010
=1-1/2+1/2-1/3+……+1/2009-1/2010
=1-1/2010
=2009/2010
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