在数列{an}中,a1=1/2,前n项的和Sn=n平方an,求an+1?

s(n+1)=(n+1)^2Xa(n+1)=sn+a(n+1)=n^2Xan+a(n+1) 就有下列等式
a(n+1)/an=n/(n+2) 我们可以得出
a(n+1)=a(n+1)/anXan/a(n-1)X.Xa2/a1Xa1=n/(n+2)X(n-1)/(n+1)...1/3Xa1
=2a1/((n+1)(n+2))=1/((n+1)(n+2))