已知S(x)=a1x+a2x^2+...+anx^n,且a1,a2,...,an组成等差数列,n为正偶数

（1）因a1,a2,。。。,an组成等差数列,所以设公差为d,又因S（1)=n^2,则a1+a2+。。。+an=n^2,则由等差数列求和公式Sn=[2na1+n(n-1)d]/2可得[2na1+n(n-1)d]/2=n^2,化简得a1+d(n-1)/2=n,即a1=n(1-d/2)+d/2,因为a1是等差数列的首项,所以和n无关,故1-d/2=0,即d=2所以a1=1,得an=a1+(n-1)d,带入an=1+(n-1)*2,得{an}的通项公式为 an=2n-1；（2）由（1）可得S(x)=x+3x^2+5x^3+。。。+(2n-1)x^n,则S(1/2)=(1/2)+3*(1/2)^2+5*(1/2)^3+。。。+(2n-1)*(1/2)^n,(1/2)S(1/2)=(1/2)^2+3*(1/2)^3+5*(1/2)^4+。。。+(2n-1)*(1/2)^(n+1),则S(1/2)-(1/2)S(1/2)=(1/2)+2*(1/2)^2+2*(1/2)^3+。。。+2*(1/2)^n-(2n-1)*(1/2)^(n+1),化简得(1/2)S(1/2)=-(1/2)+(1-(1/2)^n)/(1-(1/2))-(2n-1)*(1/2)^(n+1),S(1/2)=-1+4-(1/2)^(n-2)-(2n-1)*(1/2)^n,S(1/2)=3-(1/2)^(n-2)-(2n-1)*(1/2)^n,所以S(1/2)