A(n+1)-1=(An²-2An+1)/(2An+1)=(An-1)²/(2An+1)
A(n+1)+2=(An²+4An+4)/(2An+1)=(An+2)²/(2An+1)
两式相除 [A(n+1)-1]/[A(n+1)+2]=[(An-1)/(An+2)]²
所以(An-1)/(An+2)={[A(n-1)-1]/[A(n-1)+2]}²
={[A(n-2)-1]/[A(n-2)+2}^(2^2)
.
=[(A1-1)/(A1+2)]^(2^n)
告诉A1,即可求得通项公式