已知(a-1)的平方+ab-2的绝对值=0,求1/ab+1/(a+1)*(b+1)+1/(a+2)*(b+2)+…+1/(a+2009)*(b+2009)+<接下>
<接上>1/(a+2010)*(b+2010)的值
人气:350 ℃ 时间:2019-10-10 05:56:17
解答
由题意a-1=0,ab-2=0
则a=1,b=2
则原式=1/2+1/2*(1/3)+1/3*(1/4)+……+1/2011*(1/2012)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+……+(1/2011-1/2012)
=1-1/2012
=2011/2012
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