To easy设 b_n=(a_n-2)/(a_n+2)则 b_1=1/9, (b_n)^2=b_{n+1}因此 b_n=(b_1)^{2^{n-1}} = (1/9)^{2^{n-1}} =(1/3)^{2^n}故 a_n=2(1+b_n)/(1-b_n), 代入b_n.我以前对递推数列很有研究的,你有啥问题尽管找我吧....