1)
f(1)=a1+a2+...+an=n^2
a1+a2+...an-1=(n-1)^2
两式相减得 an=n^2-(n-1)^2=2n-1
2)
f(x)=x+3x^2+.(2n-1)x^n
f(1/3)=1/3+3(1/3)^2+.(2n-1)(1/3)^n
1/3f(1/3)=(1/3)^2+3(1/3)^3+.(2n-1)(1/3)^(n+1)
上两式相减得
2/3f(1/3)=1/3+2(1/3)^2+2(1/3)^3+.2(1/3)^n-(2n-1)(1/3)^(n+1)
=1/3+2[(1/3)^2-(1/3)^(n+1)]/(1-1/3)-(2n-1)(1/3)^(n+1)