题目内容
(本小题满分14分)
已知各项均为正数的数列{an}前n项和为Sn,(p – 1)Sn = p2 – an,n ∈N*,p > 0且p≠1,数列{bn}满足bn = 2logpan.
(Ⅰ)若p =
,设数列
的前n项和为Tn,求证:0 < Tn≤4;
(Ⅱ)是否存在自然数M,使得当n > M时,an > 1恒成立?若存在,求出相应的M;若不存在,请说明理由.
已知各项均为正数的数列{an}前n项和为Sn,(p – 1)Sn = p2 – an,n ∈N*,p > 0且p≠1,数列{bn}满足bn = 2logpan.
(Ⅰ)若p =


(Ⅱ)是否存在自然数M,使得当n > M时,an > 1恒成立?若存在,求出相应的M;若不存在,请说明理由.
(Ⅰ)解:由(p – 1)Sn = p2 – an(n∈N*) ①
由(p – 1)Sn – 1 = p2 – an – 1 ②
① – ②得
(n≥2)
∵an > 0(n∈N*)
又(p – 1)S1 = p2 – a1,∴a1 = p
{an}是以p为首项,
为公比的等比数列
an = p
bn = 2logpan = 2logpp2 – n
∴bn =" 4" – 2n …………4分
证明:由条件p =
得an = 2n – 2
∴Tn =
①
②
① – ②得

=" 4" – 2 ×
=" 4" – 2 ×
∴Tn =
…………8分
Tn – Tn – 1 =
当n > 2时,Tn – Tn – 1< 0
所以,当n > 2时,0 < Tn≤T3 = 3
又T1 = T2 = 4,∴0 < Tn≤4.…………10分
(Ⅱ)解:若要使an > 1恒成立,则需分p > 1和0 < p < 1两种情况讨论
当p > 1时,2 – n > 0,n < 2
当0 < p < 1时,2 – n < 0,n > 2
∴当0 < p < 1时,存在M = 2
当n > M时,an > 1恒成立.…………14分
由(p – 1)Sn – 1 = p2 – an – 1 ②
① – ②得

∵an > 0(n∈N*)
又(p – 1)S1 = p2 – a1,∴a1 = p
{an}是以p为首项,

an = p

bn = 2logpan = 2logpp2 – n
∴bn =" 4" – 2n …………4分
证明:由条件p =

∴Tn =


① – ②得

=" 4" – 2 ×

=" 4" – 2 ×

∴Tn =

Tn – Tn – 1 =

当n > 2时,Tn – Tn – 1< 0
所以,当n > 2时,0 < Tn≤T3 = 3
又T1 = T2 = 4,∴0 < Tn≤4.…………10分
(Ⅱ)解:若要使an > 1恒成立,则需分p > 1和0 < p < 1两种情况讨论
当p > 1时,2 – n > 0,n < 2
当0 < p < 1时,2 – n < 0,n > 2

当n > M时,an > 1恒成立.…………14分
略

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