(1)证明：因为2a(n+1)=(1+1/n)^2*an=[(n+1)/n]^2*an=(n+1)^2/n^2*an
所以2a(n+1)/(n+1)^2=an/n^2,即a(n+1)/(n+1)^2=(1/2)*an/n^2.
因为a1/1^2=1≠0,所以数列{an/n^2}是以1为首项,1/2为公比的等比数列.
所以an/n^2=a1/1^2*q^(n-1)=1*(1/2)^(n-1)=(1/2)^(n-1).
所以an=n^2*(1/2)^(n-1).
(2)因为bn=a(n+1)-（1/2）an=(n+1)^2*(1/2)^n-（1/2）*n^2*(1/2)^(n-1)=(2n+1)*(1/2)^n
所以Sn=3*(1/2)+5*(1/2)^2+7*(1/2)^3+……+(2n+1)*(1/2)^n
1/2Sn= 3*(1/2)^2+5*(1/2)^3+……+(2n-1)*(1/2)^n+(2n+1)*(1/2)^(n+1)
上式-下式,得：
1/2Sn=3/2+2*[(1/2)^2+(1/2)^3+……+(1/2)^n]-(2n+1)*(1/2)^(n+1)
=3/2+2*(1/2)^2*[1-(1/2)^(n-1)]/(1-1/2)]-(2n+1)*(1/2)^(n+1)
=3/2+[1-(1/2)^(n-1)]-(2n+1)*(1/2)^(n+1)
所以Sn=5-(1/2)^(n-2)-(2n+1)*(1/2)^n.