Question

已知数列{an}的前n项和为Sn，满足Sn=n2an-n2(n-1)，且a1=1/2，求{an}的通项

Answer 1

S1=a1=1/2。n>=2时，有：
Sn=n^2*an-n^2*(n-1)=n^2*(Sn-S(n-1))-n^2*(n-1)
(n^2-1)Sn=n^2*S(n-1)+n^2*(n-1) 等式两边同除n(n-1)
[(n+1)/n]Sn=[n/(n-1)]S(n-1)+n
[n/(n-1)]S(n-1)=[(n-1)/(n-2)]S(n-2)+(n-1)
…………
(4/3)S3=(3/2)S2+3
(3/2)S2=2S1+2=1+2
以上各式相加：[(n+1)/n]Sn=n+(n-1)+…+3+2+1=n(n+1)/2
Sn=n^2/2，代入Sn=n^2*an-n^2*(n-1)可得：an=n-1/2(n=1,2,3,……,)