数列{an}满足:a1=1/2,a1+a2+……+an=n^2an，求an

Question

谢谢！

Answer 1

解：∵a1+a2+……+an=n^2an
∴a1+a2+……+a(n-1)=(n-1)^2a(n-1)
==>an=[a1+a2+……+a(n-1)+an]-[a1+a2+……+a(n-1)]
==>an=n^2an-(n-1)^2a(n-1)
==>(n^2-1)an=(n-1)^2a(n-1)
==>an=[(n-1)^2/(n^2-1)]a(n-1)
==>an=[(n-1)/(n+1)]a(n-1)......(1)
∵a1=1/2
∴由(1)式，得a2=(1/3)a1=(1!)/(3!)
a3=(2/4)a2=(2!)/(4!)
a4=(3/5)a3=(3!)/(5!)
..........
an=[(n-1)/(n+1)]a(n-1)=[(n-1)!/(n+1)!]
=1/[n(n+1)]
故an=1/[n(n+1)]

Answer 2

n^2an
这个是什么意思