2, (A(n+ 1), 0)=(4/5)(A(n), 0), so {A(n)} is a geometric series and a (n) = 5 * (4/5) (n- 1).

|OB(n+ 1)|=B(n)√2, so {B(n)} is arithmetic progression, and B(n)=(√2)*n,

3、s(n)=( 1/2)*a(n)*b(n)=(5√2/2)*n*(4/5)^(n- 1)，

Assuming that there is a maximum value, there are s (n- 1) < =S(n) and s (n+ 1) < =S(n).

= & gtn- 1 & lt； =(4/5)n and (n+ 1) (4/5)

=>n < = 5 and n & gt=4= >n take 4 or 5.

=> substitute S(n), S(4)=S(5)= 128√2/25.