f(1)=a1+a2+.....+an=n²
an=Sn -S(n-1) =n²-(n-1)² =2n-1 (n≥2)
因为a1=s1=1 =2-1 也成立
所以通项公式为 an =2n-1 (n∈N+)

2)f(1/3) =1*(1/3)+3*(1/3)²+....+(2n-1)(1/3)^n .....(1)
(1/3)f(1/3) =1*(1/3)²+....+(2n-1)(1/3)^(n+1)........(2)

(1)-(2)
(2/3)f(1/3)
=1/3+2[(1/3)² +(1/3)³+...+(1/3)^n] -(2n-1)(1/3)^(n+1)
=1/3+2*{(1/3)²[1 -(1/3)^(n-1)]/[1 -(1/3)]}-(2n-1)(1/3)^(n+1)
=1/3+1/3 -(1/3)^n -(2n-1)(1/3)^(n+1)
<2/3
==>(2/3)f(1/3) <2/3
所以 ,f(1/3)<1