数列{an}满足a1=3,an-anan+1=1,An表示{an}前n项之积,则A2013=

由已知用递推公式推出后面的项：
a1=3………………①
由a1－a1a2=1
3－3a2=1
3a2=2
a2=2/3……②
由a2－a2a3=1
2/3－2/3a3=1
2/3a3=-1/3
a3=-1/2……③
由a3－a3a4=1
-1/2－(-1/2)a4=1
1/2a4=3/2
a4=3……④
可见a4=3与a1相等,以后的项也会出现对应相同的数值,即每三项一重复
即：3,2/3,-1/2；3,2/3,-1/2；……循环出现,所以A2010即是2010项的积.
所以先看2010里包含多少个三项,
2013÷3=671
所以
A2010=[3×2/3×(-1/2)]^671
=[-1]^671
=-1

an+1= (an -1)/an = 1- 1/an
an+2= 1- 1/an+1 = 1 - 1/(1- 1/an) = 1- an/(an -1) = -1/(an - 1)
an+3= 1- 1/an+2 = 1- 1/[-1/(an - 1)] = 1+ (an - 1) = an

由于an+3=an，这是一个循环数列，
A2013 = (a1*a2*a3)^671
= [3*(2/3)*(-1/2)]^671
= (-1)671
= -1