设数列{an}的前n项和为Sn,满足Sn-tSn-1=n(n大于等于2,n属于N),t为常数,且a1=1.(1)当t=2时,求a2和a3;(2)若{an +1}是等比数列,求t的值;

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设数列{an}的前n项和为Sn,满足Sn-tSn-1=n(n大于等于2,n属于N),t为常数,且a1=1.(1)当t=2时,求a2和a3;(2)若{an+1}是等比数列,求t的值;设数列{an}的前n

设数列{an}的前n项和为Sn,满足Sn-tSn-1=n(n大于等于2,n属于N),t为常数,且a1=1.(1)当t=2时,求a2和a3;(2)若{an +1}是等比数列,求t的值;
设数列{an}的前n项和为Sn,满足Sn-tSn-1=n(n大于等于2,n属于N),t为常数,且a1
=1.(1)当t=2时,求a2和a3;(2)若{an +1}是等比数列,求t的值;

设数列{an}的前n项和为Sn,满足Sn-tSn-1=n(n大于等于2,n属于N),t为常数,且a1=1.(1)当t=2时,求a2和a3;(2)若{an +1}是等比数列,求t的值;
1.当t=2时,取n=2有S2-2S1=2即a1+a2-2a1=2,又a1=1得a2=3；取n=3时有a1+a2+a3-2a1-2a2=3带入得a3=7
2.由Sn-tSn-1=n,a1=1得a2=1+t,a3=1+2t+t^2,因为an +1是等比,带入(a2 +1)^2=(a1+1)*(a3+1)得t^2=0,即t=0,an恒为1

1）当t=2时
Sn-2Sn-1=n,整理得：Sn=-1-n
因S2=a1+a2
所a1+a2=-1-2=-3
因a1=1
所a2=-4
同理a3=-1

(1)因为Sn-2Sn-1=n a1=1
所以 a2=1+t a3=t^2+t+1
所以当t=2时 a2=3 a3=7
(2)因为a2=1+t a3=t^2+t+1
所以当{an +1}是等比数列时
则有(1+t+1)^2=(1+1)(t^2+t+1+1)
t=2 t=0(舍去)

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