已知(a-1)的平方根+(ab-2)的平方=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2004)(b+2004)的值

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已知(a-1)的平方根+(ab-2)的平方=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2004)(b+2004)的值
已知(a-1)的平方根+(ab-2)的平方=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2004)(b+2004)的值

已知(a-1)的平方根+(ab-2)的平方=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2004)(b+2004)的值
(a-1)的平方根+(ab-2)的平方=0,
即A-1=0 AB-2=0 ==>A=1 B=2
又1/AB=1/A-1/B=1-1/2
1/(A+1)(B+1)=1/(A+1)-1/(B+1)=1/2-1/3
...
1/(A+2004)(B+2004)=1/(A+2004)-1/(B+2004)=1/2005-1/2006
所以1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2004)(b+2004)
=1-1/2+1/2-1/3.+1/2005-1/2006
=1-1/2006=2005/2006

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