数列{an}满足a1＝1／2,a1＋a2＋……＋an＝n的平方×an,则数列{an}的通项公式?

Sn=n^2*anSn-1=(n-1)^2*an-1Sn-Sn-1=an=n^2*an-(n-1)^2*an-1an(n^2-1)=an-1*(n-1)^2an-1/an=(n+1)/(n-1)a1/a2*a2/a3*a3/a4*.*an-1/an=3/1*4/3*5/4*...(n+1)/(n-1)a1/an=n+1an=1/2(n+1)