There is a column of numbers A1, A2, A3 If A1 = 2, then a2011 is () A. 2011B. 2C. -1D. 12

∵a1=2，∴a2=1-12=12，a3=1-2=-1，a4=1-（-1）=2，a5=1-12=12，… And so on, every three numbers for a group of circulation, 2011 △ 3 = 670 So the answer is: 2

If A1 = 1 / 3, A2 is the reciprocal of difference of A1 and A3 is the reciprocal of difference of A2, then A2009 is equal to

a2=1/(1-a1)=1/(1+1/3)=3/4
a3=1/(1-a2)=1/(1-3/4)=4
a4=1/(1-a3)=1/(1-4)=-1/3=a1
So A5 = A2
a6=a3
a7=a4=a1
So this is a cycle of three
The remainder of 2009 / 3 is 2
So A2009 = A2 = 3 / 4

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