数列an中,已知a1=2,设Sn是数列的前n的和,若Sn=(n^2)*an,求an通项公式?

Sn=n^2an①
S(n-1)=(n-1)^2a(n-1)②
①-②,an=n^2an-(n-1)^2a(n-1)
整理得an/a(n-1)=(n-1)/(n+1)
所以an=a1*(a2/a1)(a3/a2)*……*（an/an-1)
=1*(1/3)*(2/4)*(3/5)*……*[(n-2)/n]*[(n-1)/(n+1)]
=2/[n(n+1)]

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