已知(a-1)²+|ab-2|=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2009)(b+2009)+1/(a+2010)(b+2010)

已知(a-1)²+|ab-2|=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2009)(b+2009)+1/(a+2010)(b+2010)
已知(a-1)²+|ab-2|=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2009)(b+2009)+1/(a+2010)(b+2010)

已知(a-1)²+|ab-2|=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2009)(b+2009)+1/(a+2010)(b+2010)
(a-1)²+|ab-2|=0,则a=1,b=2
所以原式=1/2+1/2*3+1/3*4+……+1/2010*2011+1/2011*2012
=1/2+1/2-1/3+1/3-1/4+1/4-1/5+……+1/2010-1/2011+1/2011-1/2012
=1-1/2012
=2011/2012

(a-1)²+|ab-2|=0
因为平方和绝对值非负
所以a-1=0,ab-2=0
a=1,b=2,代入所求式
得1/1*2+1/2*3+1/3*4+........+1/2011*2012
利用公式1/(n*(n+1))=1/n-1/(n+1)
所求=(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+........+(1/2011-1/2012)
=1-1/2012
=2011/2012

a=1 b=2 代入变成一个简单的式子拆分为1-1/2+1/2-1/3+...+1/2011-1/2012=2011/2012