已知数列{an}的前n项和为Sn,满足Sn=n2an-n2(n-1),且a1=1/2,求{an}的通项

S1=a1=1/2.n>=2时,有：Sn=n^2*an-n^2*(n-1)=n^2*(Sn-S(n-1))-n^2*(n-1) (n^2-1)Sn=n^2*S(n-1)+n^2*(n-1) 等式两边同除n(n-1)[(n+1)/n]Sn=[n/(n-1)]S(n-1)+n[n/(n-1)]S(n-1)=[(n-1)/(n-2)]S(n-2)+(n-1)…………(4/3)S...