(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

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(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

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(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 = p2ann ∈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.
题型:解答题难度:偏易来源:不详

答案

(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 =an = 2n – 2
Tn =                          ①
                            ②
① – ②得

=" 4" – 2 ×
=" 4" – 2 ×
Tn =…………8分
TnTn – 1 =
n > 2时,TnTn – 1< 0
所以,当n > 2时,0 < TnT3 = 3
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
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
n > M时,an > 1恒成立.…………14分

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