急救！数列{an}中,a1=2,前n项和为Sn=n^2*an,求Sn

an=Sn-S(n-1)=n^2*an - (n-1)^2*a(n-1)
a(n-1)*(n-1)^2=an*(n-1)*(n+1)
an= (n-1)/(n+1) *a(n-1)
= (n-1)/[(n-1)+2] *a(n-1)
= (n-1)/[(n-1)+2] *(n-2)/[(n-2)+2] *a(n-2)
= (n-1)/[(n-1)+2] *(n-2)/[(n-2)+2] *(n-3)/[(n-3)+2] *a(n-3)
= (n-1)/[(n-1)+2] *(n-2)/[(n-2)+2] *(n-3)/[(n-3)+2].......2/4 *1/3 *a1
= 2a1/n(n+1)
=4/n(n+1)

sn=n^2 * 4/n(n+1) =4n/(n+1)

an=Sn-S(n-1)=n^2*an-(n-1)^2*a(n-1)
a(n-1)*(n-1)^2=an*(n-1)*(n+1)
a(n-1)*(n-1)=an*(n+1)
an=a(n-1)*(n-1)/(n+1)
=a(n-2)*(n-2)/n*(n-1)/(n+1)
=...=a1*[1*2*...*(n-1)]/[3*4*...*(n+1)]
=4/[n*(n+1)]
Sn=n^2*an
=4n^2/[n*(n+1)]