设数列{an}的首项a1=t,前n项和为Sn,满足5Sn-3Sn-1=3,(n≥2,n∈N*),是否存在常数t,使得数列

假设存在t满足题意,则Sn≠S(n-1) t≠0
n≥2时,
5Sn-3S(n-1)=3
5Sn -15/2=3S(n-1)-9/2
5(Sn -3/2)=3[S(n-1)-3/2]
Sn≠S(n-1),因此Sn≠3/2 a1≠3/2 t≠3/2
(Sn -3/2)/[S(n-1) -3/2]=3/5,为定值.
S1-3/2=a1-3/2=t-3/2
Sn -3/2=(t- 3/2)×(3/5)^(n-1)
Sn=(t-3/2)×(3/5)^(n-1) +3/2
n≥2时,
an=Sn-S(n-1)
=(t-3/2)×(3/5)^(n-1)+3/2-(t-3/2)×(3/5)^(n-2)+3/2
=(-2/5)×(t-3/2)×(3/5)^(n-2)
a(n+1)/an=3/5,为定值.
a1=t=(-2/5)×(t-3/2)×(3/5)^(-1)=(-2/3)×(t-3/2)
5t=3
t=3/5
即存在满足题意的t,t=3/5