设数列an的前n项和为Sn,Sn-tSn-1=n,a1=1(1)t=2,求a2,a3(2){an+1}是等比数列,求t的值(3)求sn

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设数列an的前n项和为Sn,Sn-tSn-1=n,a1=1(1)t=2,求a2,a3(2){an+1}是等比数列,求t的值(3)求sn

设数列an的前n项和为Sn,Sn-tSn-1=n,a1=1(1)t=2,求a2,a3(2){an+1}是等比数列,求t的值(3)求sn
设数列an的前n项和为Sn,Sn-tSn-1=n,a1=1
(1)t=2,求a2,a3
(2){an+1}是等比数列,求t的值
(3)求sn

设数列an的前n项和为Sn,Sn-tSn-1=n,a1=1(1)t=2,求a2,a3(2){an+1}是等比数列,求t的值(3)求sn
(1)A1=1,S1=1.根据递推式,S2=4,S3=11,所以A2=3,A3=7
(2)S2=t+2,S3=t^2+2t+3,所以A2=t+1,A3=t^2+t+1,根据题意,2,t+2,t^2+t+2成等比数列,解得t=0或2
(3)对于Sn-t(Sn-1)=n,两边同除t^n,得Sn/(t^n)-S(n-1)/t^(n-1)=n/(t^n)
可用累加法得到Sn/(t^n)=S1/t+2/(t^2)+3/(t^3)+...+n/(t^n)
右边等式是典型的错位相减法,
t=1时,Sn=n(n+1)/2
t=0时,Sn=n
其他情况下,求得Sn/(t^n)后即可求得Sn=[t^(n+1)-t]/(t-1)-n

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