题目
如果有理数a,b满足丨ab-2丨加丨1-b丨=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2011)(b+2011)的值
提问时间:2021-04-01
答案
∵丨ab-2丨+丨1-b丨=0
∴ab-2=0,1-b=0
∴a=2,b=1
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2011)(b+2011)
=1/(1*2)+1/(2*3)+1/(3*4)+……+1/(2013*2012)
=1/1-1/2+1/2-1/3+1/3-1/4+……+1/2012-1/2013
=1-1/2013
=2012/2013
∴ab-2=0,1-b=0
∴a=2,b=1
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2011)(b+2011)
=1/(1*2)+1/(2*3)+1/(3*4)+……+1/(2013*2012)
=1/1-1/2+1/2-1/3+1/3-1/4+……+1/2012-1/2013
=1-1/2013
=2012/2013
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