数列{an}的前n项和记为Sn,a1=t,点(Sn,an+1)在直线y=3x+1上,n∈自然数.当实数t为何值时,数列{an}是等比数列?

Question

数列{an}的前n项和记为Sn,a1=t,点(Sn,an+1)在直线y=3x+1上,n∈自然数.当实数t为何值时,数列{an}是等比数列?

Answer 1

x=Sn y=a(n+1)代入直线方程
a(n+1)=3Sn +1
S(n+1)-Sn=3Sn +1
S(n+1)=4Sn +1
S(n+1) +1/3=4Sn +4/3=4(Sn +1/3)
t=-1/3时,S1+1/3=-1/3+1/3=0
Sn +1/3=0 Sn=-1/3,为定值,数列是第一项为-1/3,以后各项均为0的数列,不满足题意,因此
t≠-1/3
[S(n+1)+1/3]/(Sn +1/3)=4,为定值
S1+1/3=t+ 1/3
数列{Sn +1/3}是以t+ 1/3为首项,4为公比的等比数列
Sn +1/3=(t+ 1/3)×4^(n-1)
Sn=(t+1/3)×4^(n-1) -1/3
n≥2时,an=Sn-S(n-1)=(3t+1)×4^(n-2)
a(n+1)/an=[(3t+1)×4^(n-1)]/[(3t+1)×4^(n-2)]=4,为定值,数列{an}是以4为公比的等比数列.
x=S1=a1=t,y=a2代入直线方程
a2=3t+1,要a1是等比数列中的项,则
a2/a1=4
(3t+1)/t=4,解得t=1,即t=1时,数列{an}是以1为首项,4为公比的等比数列.