Oqn (2n- 1, 0), and the distance from PN to x axis is (2/3) (2n- 1).

s△opnqn=(2n- 1)*(2/3)^(2n- 1)/2

s'=(2/3)^(2n- 1)+(2/3)^(2n- 1)*(ln2/3)=(2/3)^(2n- 1)[ 1+(2n- 1)ln(2/3)]=0

2n- 1 =- 1/(LN2/3)= 1/(ln3-LN2)

n = 1/(2ln 3-2ln 2)+ 1/2

1/2 & lt； n & lt 1

When n= 1, s is the largest, and the maximum value = 1/3.