Question

已知数列{an}满足a1=2,an+1=(2/1-an)-1,(n∈N*),则a1·a2·a3...·a2015的值为 A.2 B.-6

Answer 1

a(n+1)=2/(1-an)
-1=(2-1+an)/(1-an)=(1+an)/(1-an)
a2=(1+2)/(1-2)=-3
a3=(1-3)/(1+3)=-½
a4=(1-½)/(1+½)=⅓
a5=(1+⅓)/(1-⅓)=2=a1
数列{an}是以4为周期的周期数列
a1·a2·a3·a4=2·(-3)·(-½)·⅓=1
a1·a2·a3·...·a2015
=(a1·a2·a3·a4)·(a5·a6·a7·a8)·...·(a2013·a1014·a2015·a2016)/a2016
=1×1×...×1/a4
=1/⅓
=3
选择项没有写全。你看一下，哪个选项是3，就是选那个选项。