Question

已知数列{an}满足a1=2,an+1=1+an/1-an(n属于N*),则a1×a2×a3×......a2013

Answer 1

解:
a2=(1+a1)/(1-a1)=(1+2)/(1-2)=-3
a3=(1+a2)/(1-a2)=(1-3)/(1+3)=-1/2
a4=(1+a3)/(1-a3)=(1-1/2)/(1+1/2)=1/3
a5=(1+a4)/(1-a4)=(1-1/3)/(1+1/3)=1/2
a6=(1+a5)/(1-a5)=(1+1/2)/(1-1/2)=3
a7=(1+a6)/(1-a6)=(1+3)/(1-3)=-2
a8=(1+a7)/(1-a7)=(1-2)/(1+2)=-1/3
a9=(1+a8)/(1-a8)=(1-1/3)/(1+1/3)=1/2
规律:数列前4项依2-3-1/21/3
2×(-3)×(-1/2)×(1/3)=1
第5项始按1/23-2-1/3循环每4循环
(1/2)×3×(-2)×(-1/3)=1
2013/4=503余1
a2013=1/2
a1×a2×a3×...×a2013
=(a1×a2×a3×a4)×(a5×a6×a7×a8)×...×(a2009×a2010×a2011×a2012)×a2013
=1×1×...×1×a2013
=a2013
=1/2