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已知丨a-1丨+丨ab-2丨=0,求1/ab+1/（a+1)(b+1)+1/(a+2）(b+2）+...+1/（a+1998)(b+1998)值
题目详情
已知丨a-1丨+丨ab-2丨=0,求1/ab+1/（a+1)(b+1)+1/(a+2）(b+2）+
...+1/（a+1998)(b+1998)值
...+1/（a+1998)(b+1998)值
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答案和解析
根号(a-1)+(ab-2)^2=0
a-1=0
ab-2=0
a=1
b=2
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2004)(b+2004)
=1/1*2+1/2*3+1/3*4+-------+1/2005*2006
=1-1/2+1/2-1/3+1/3-1/4+------+1/2005-1/2006
=1-1/2006
=2005/2006
你把3006那几个改一下 就行了 我懒得算
a-1=0
ab-2=0
a=1
b=2
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2004)(b+2004)
=1/1*2+1/2*3+1/3*4+-------+1/2005*2006
=1-1/2+1/2-1/3+1/3-1/4+------+1/2005-1/2006
=1-1/2006
=2005/2006
你把3006那几个改一下 就行了 我懒得算
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