a1=1/2其前n项和Sn=n2an(n大于等于1)求数列an的通项公式
问题描述：
a1=1/2其前n项和Sn=n2an(n大于等于1)求数列an的通项公式
答
Sn =n^2.an
for n>=2
an = Sn -S(n-1)
an =n^2.an - (n-1)^2.a(n-1)
(n+1)an = (n-1)a(n-1)
an/a(n-1) = (n-1)/(n+1)
an/a1 = 2/[n(n+1)]
an = 1/[n(n+1)]