如果la-1l+lab-2l=0.求1/ab+1/(a+1)(b+1)+...+1/(a+2004)(b+2004)

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如果la-1l+lab-2l=0.求1/ab+1/(a+1)(b+1)+...+1/(a+2004)(b+2004)
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如果la-1l+lab-2l=0.求1/ab+1/(a+1)(b+1)+...+1/(a+2004)(b+2004)
如果la-1l+lab-2l=0.求1/ab+1/(a+1)(b+1)+...+1/(a+2004)(b+2004)

如果la-1l+lab-2l=0.求1/ab+1/(a+1)(b+1)+...+1/(a+2004)(b+2004)
∵la-1l+lab-2l=0
∴a-1=0 ab-2=0
∴a=1 b=2
∴原式=1/(1×2)+1/(2×3)+1/(2005×2006)
=1-1/2+1/2-1/3+1/2005-1/2006
=1-1/2006
=2005/2006