在数列{an}中,a1=1,a2=-3,且在数列{an}中,an+1=an+an+2,则a2012=

问题描述:

在数列{an}中,a1=1,a2=-3,且在数列{an}中,an+1=an+an+2,则a2012=
a(n+1)=an+a(n+2)

a(n+1)=an+a(n+2) (1)a(n+2)=a(n+1)+a(n+3) (2)(1)+(2)a(n+1)+a(n+2)=a(n+1)+a(n+2)+an+a(n+3)a(n+3)+an=0a(n+3)=-ana(n+6)=-a(n+3)=an数列为循环数列,每6个循环一次.2012=2+335×6a2012=a2=-3