已知a-1的绝对值+ab-2的绝对值=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2005)(b+2005)的值
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已知a-1的绝对值+ab-2的绝对值=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2005)(b+2005)的值
已知a-1的绝对值+ab-2的绝对值=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2005)(b+2005)的值
已知a-1的绝对值+ab-2的绝对值=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2005)(b+2005)的值
由已知a-1的绝对值+ab-2的绝对值=0
得a-1=0且ab-2=0
所以a=1 ab=2 b=2
设所求式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2005)(b+2005)的第n项为An,可以看出,An=1/(a+n-1)(b+n-1)
=1/[ab+(n-1)(a+b)+(n-1)^2]
将a=1 ab=2 b=2代入上式
An=1/[2+3(n-1)+(n-1)^2]
=1/(n-1+2)(n-1+1)
=1/n(n+1)
=1/n-1/(n+1)
所以1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2005)(b+2005)
=1-1/2+1/2-1/3+……+1/2005-1/2006+1/2006-1/2007
=1-1/2007
=2006/2007
由已知a-1的绝对值+ab-2的绝对值=0
得a-1=0且ab-2=0
所以a=1 ab=2 b=2
设所求式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2005)(b+2005)的第n项为An,可以看出,An=1/(a+n-1)(b+n-1)
=1/[ab+(n-1)(a+b)+(n-1)^2]
将a=1 ab=2 b=...
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由已知a-1的绝对值+ab-2的绝对值=0
得a-1=0且ab-2=0
所以a=1 ab=2 b=2
设所求式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2005)(b+2005)的第n项为An,可以看出,An=1/(a+n-1)(b+n-1)
=1/[ab+(n-1)(a+b)+(n-1)^2]
将a=1 ab=2 b=2代入上式
An=1/[2+3(n-1)+(n-1)^2]
=1/(n-1+2)(n-1+1)
=1/n(n+1)
=1/n-1/(n+1)
所以1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2005)(b+2005)
=1-1/2+1/2-1/3+……+1/2005-1/2006+1/2006-1/2007
=1-1/2007
=2006/2007懂了吗?
收起
由已知a-1的绝对值+ab-2的绝对值=0
得a-1=0且ab-2=0
所以a=1 ab=2 b=2
设所求式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2005)(b+2005)的第n项为An,可以看出,An=1/(a+n-1)(b+n-1)
=1/[ab+(n-1)(a+b)+(n-1)^2]
将a=1 ab=2 b=...
全部展开
由已知a-1的绝对值+ab-2的绝对值=0
得a-1=0且ab-2=0
所以a=1 ab=2 b=2
设所求式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2005)(b+2005)的第n项为An,可以看出,An=1/(a+n-1)(b+n-1)
=1/[ab+(n-1)(a+b)+(n-1)^2]
将a=1 ab=2 b=2代入上式
An=1/[2+3(n-1)+(n-1)^2]
=1/(n-1+2)(n-1+1)
=1/n(n+1)
=1/n-1/(n+1)
所以1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2005)(b+2005)
=1-1/2+1/2-1/3+……+1/2005-1/2006+1/2006-1/2007
=1-1/2007
=2006/2007
收起