已知数列{an}满足a1=0,an+1=(an-根号3)/(根号3an+1) (n属于N+),则该数列中a20=____

由a1=0与a（n+1）=(an-sqr(3))/(sqr(3)an+1)得a2=-sqr(3)
由a（n+1）=(an-sqr(3))/(sqr(3)an+1)
得a(n+2)=(a(n+1)-sqr(3))/(sqr(3)a(n+1)+1)
=-(an+sqr(3))/(sqr(3)an-1)=-a(n-1)
把n用n+1去代得到
a(n+3)=-an
于是a(n+6)=-a(n+3)=an
于是an是以6为周期的周期数列
于是a20=a(3*6+2)=a2=-sqr(3)