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.