Question

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

Answer 1

sn=n^2an-n^2+n-----A
sn+1=(n^2+2n+1)an+1-(n+1)^2+n+1-----B
B-A an+1=(n+1)^2an+1-n^2an-2n
0=(n^2+2n)an+1-2n-n^2an
除n
0=(n+2)an+1-2-nan
nan=(n+2)an+1-2
设k
nan+nk=(n+2)an+1+(n+2)k
解得k=-1
即nan-n=(n+2)an+1-(n+2)
移项
n/n+2=an+1-1/an-1
设bn=an-1
b2/b1=1/3 b3/b2=2/4 b4/b3=3/5.bn/bn-1=n-1/n+1
累商叠乘
bn/b1=2/n(n+1)
b1=-1/2
bn=-1/n(n+1)
an=n^2+n-1/n^2+n
带回
Sn=n^2/（n+1)