已知(ab-2)²+|b-1|=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…1/(a
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已知(ab-2)²+|b-1|=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…1/(a+2011)(b+2011)=
![已知(ab-2)²+|b-1|=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…1/(a](/uploads/image/z/3204759-39-9.jpg?t=%E5%B7%B2%E7%9F%A5%EF%BC%88ab-2%EF%BC%89%26%23178%3B%2B%7Cb-1%7C%3D0%2C%E6%B1%821%2Fab%2B1%2F%28a%2B1%29%28b%2B1%29%2B1%2F%28a%2B2%29%28b%2B2%29%2B%E2%80%A61%2F%28a)
(ab-2)²+|b-1|=0可知,b=1,a=2
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…1/(a+2011)(b+2011)
=1/2+1/(2*3)+……+1/(2013*2012)
=1-1/2+1/2-1/3+1/3-1/4+……+1/1012-1/2013
=2012/1013
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…1/(a+2011)(b+2011)
=1/2+1/(2*3)+……+1/(2013*2012)
=1-1/2+1/2-1/3+1/3-1/4+……+1/1012-1/2013
=2012/1013
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