数列{an}中,a1=1/2,前项的和Sn=n^2an,求an+1

Question

数列{an}中,a1=1/2,前项的和Sn=n^2an,求an+1

Answer 1

n≥2时,
Sn=n²an
Sn-1=(n-1)²a(n-1)
Sn-Sn-1=an=n²an-(n-1)²a(n-1)
(n²-1)an=(n-1)²a(n-1)
(n+1)(n-1)an=(n-1)²a(n-1)
(n+1)an=(n-1)a(n-1)
an/a(n-1)=(n-1)/(n+1)
a(n-1)/a(n-2)=(n-2)/n
…………
a2/a1=1/3
连乘
an/a1=(1/3)(2/4)...(n-1)/(n+1)=[1×2×...×(n-1)]/[3×4×...×(n-1)×n×(n+1)]
=(1×2)/[n(n+1)]=2/[n(n+1)]=2/n - 2/(n+1)
an=a1[2/n - 2/(n+1)]=(1/2)[2/n-2/(n+1)]=1/n - 1/(n+1)
n=1时,a1=1-1/2=1/2,同样满足.
数列{an}的通项公式为an=1/n - 1/(n+1).
a(n+1)=1/(n+1)-1/(n+2).