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=____
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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=____
![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=____](/uploads/image/z/2486362-58-2.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D0%2Can%2B1%3D%28an-%E6%A0%B9%E5%8F%B73%29%2F%28%E6%A0%B9%E5%8F%B73an%2B1%29+%28n%E5%B1%9E%E4%BA%8EN%2B%29%2C%E5%88%99%E8%AF%A5%E6%95%B0%E5%88%97%E4%B8%ADa20%3D____)
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)
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)
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=____
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