Question

已知数列{an}的前n项和为Sn,a1=1/2，且Sn=n^2An-n(n-1),求an

Answer 1

n≥2时，
Sn=n²an-n(n-1)
Sn-1=(n-1)²a(n-1)-(n-1)(n-2)
an=Sn-Sn-1=n²an-n(n-1)-(n-1)²a(n-1)+(n-1)(n-2)
an=n²an-(n-1)²a(n-1)-2n+2
(n+1)(n-1)an-(n-1)²a(n-1)-2(n-1)=0
(n+1)an-(n-1)a(n-1)=2
an/(n-1)-a(n-1)/(n+1)=2/[(n+1)(n-1)]=1/(n-1)-1/(n+1)
(an-1)/(n-1)=[a(n-1)-1]/(n+1)
(an-1)/[a(n-1)-1]=(n-1)/(n+1)
[a(n-1)-1]/[a(n-2)-1]=(n-2)/n
…………
(a2-1)/(a1-1)=1/3
连乘
(an-1)/(a1-1)=[1×2×...×(n-1)]/[3×4×...×(n-1)×n×(n+1)]=(1×2)/[n(n+1)]
an-1=2(1/2-1)/[n(n+1)]=-1/[n(n+1)]
an=1-1/[n(n+1)]=(n²+n-1)/[n(n+1)]
n=1时，a1=(1+1-1)/(1×2)=1/2，同样满足。
数列{an}的通项公式为an=(n²+n-1)/[n(n+1)]

Answer 2

因为Sn=n^2An-n(n-1)
所以Sn-S(n-1)=An
即An=n^2An-n(n-1)-[(n-1)^2A(n-1)^-(n-1)（n-2)]
整理得 (n+1)An-(n-1)A(n-1)=2
An=2/(n+1)+{(n-1)/(n+1)}A(n-1)
A1=1/2
A2=5/6
A3=11/12
1/2=1/(1*2) 5/6=5/(2*3) 11/12=11/(3*4)
所以 An=[n(n+1)-1]/n(n+1)