Question

已知数列｛an｝的前n项和为Sn，Sn=1/3（an-1）（n∈N*)（1）求a1，a2（2）求Sn

Answer 1

1)
a1=S1=1/3(a1-1)
a1=-1/2

a2=S2-S1=1/3(a2-1)+1/2
3a2=a2-1+3/2
2a2=1/2
a2=1/4

2）
∵Sn=1/3(An-1)
∴Sn-1=1/3(An-1 -1)
∴An=Sn-Sn-1=1/3(An-1)-1/3(An-1 -1)
化简得2/3An=-1/3An-1
即An/An-1=-1/2=q
∴数列为等比数列
a1=1/3(a1-1)
a1=-1/2.
所以an=(-1/2)*(-1/2)^(n-1)=（-1/2）^n

Answer 2

1、n=1,则：s1=(a1-1)/3
s1=a1
解得：a1=-1/2
n=2, s2=(a2-1)/3 s2=a1+a2 解得：a2=1/4
2、 3sn=an-1=sn-s(n-1)-1
2sn=-s(n-1)-1
2(sn+1/3)=-[s(n-1)+1/3]
(sn+1/3)/[s(n-1)+1/3]=-1/2
sn+1/3是等比数列，首项为s1+1/3=a1+1/3=-1/6，公比为-1/2
sn+1/3=(-1/6)*(-1/2)^(n-1)
sn=(1/3)(-1/2)^n-1/3=[(-1/2)^n-1]/3

Answer 3

S1=a1,解得a1=-0.5
S(n+1)-Sn=1/3[a(n+1)-an]=a(n+1)(n∈N*)，移项化简发现an是等比数列，q=-0.5
故a2=0.25 an=(-0.5)^n(n∈N*)
Sn=1/3[(-0.5)^n-1](n∈N*)