Let the area of triangle OAn Bn be Sn, the area of triangle OA 1B 1 be S 1, and the area of all trapezoid An Bn+ 1an+ 1 be S.

The area of triangle OA(n- 1)B(n- 1) is S(n- 1).

s(n- 1)/sn =(oa(n- 1)/oan)^2=(a(n- 1)/an)^2....( 1)

s 1/(s 1+s)=(oa 1/oa2)^2= 1/4

4s 1 = S 1+S S = 3s 1

S(n- 1)= S 1+(n-2)S = S 1+(n-2)* 3s 1 =(3n-5)S 1

sn = S 1+(n- 1)S = S 1+(n- 1)* 3s 1 =(3n-2)S 1

So s (n-1)/sn = (3n-5)/(3n-2) = (a (n-1)/an) 2.

Get a(n- 1)/an = radical sign ((3n-5)/(3n-2)).

An/a(n- 1) = radical sign ((3n-2)/(3n-5)) ...

Can be obtained from formula (2)