設(shè)數(shù)列=

A.                             B.3                        C.-2                       D.

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科目:高中數(shù)學(xué) 來(lái)源: 題型:

已知數(shù)列{an}和{bn}滿足:a1=λ,an+1=
23
an+n-4,bn=(-1)n(an-3n+21)
,其中λ為實(shí)數(shù),n為正整數(shù).
(Ⅰ)對(duì)任意實(shí)數(shù)λ,證明數(shù)列{an}不是等比數(shù)列;
(Ⅱ)試判斷數(shù)列{bn}是否為等比數(shù)列,并證明你的結(jié)論;
(Ⅲ)設(shè)0<a<b,Sn為數(shù)列{bn}的前n項(xiàng)和.是否存在實(shí)數(shù)λ,使得對(duì)任意正整數(shù)n,都有a<Sn<b?若存在,求λ的取值范圍;若不存在,說(shuō)明理由.

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科目:高中數(shù)學(xué) 來(lái)源: 題型:

(2008•閘北區(qū)一模)已知數(shù)列{an}和{bn}滿足:a1=λ,an+1=
23
an+n-4,bn=(-1)n(an
-3n+21),其中λ為實(shí)數(shù),n為正整數(shù).Sn為數(shù)列{bn}的前n項(xiàng)和.
(1)對(duì)任意實(shí)數(shù)λ,證明:數(shù)列{an}不是等比數(shù)列;
(2)對(duì)于給定的實(shí)數(shù)λ,試求數(shù)列{bn}的通項(xiàng)公式,并求Sn
(3)設(shè)0<a<b(a,b為給定的實(shí)常數(shù)),是否存在實(shí)數(shù)λ,使得對(duì)任意正整數(shù)n,都有a<Sn<b?若存在,求λ的取值范圍;若不存在,說(shuō)明理由.

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科目:高中數(shù)學(xué) 來(lái)源: 題型:

已知數(shù)列{an}和{bn}滿足:a1=λ,an+1=
23
an+n
,bn=(-1)n(an-3n+9),其中λ為實(shí)數(shù),n為正整數(shù).
(1)若數(shù)列{an}前三項(xiàng)成等差數(shù)列,求λ的值;
(2)試判斷數(shù)列{bn}是否為等比數(shù)列,并證明你的結(jié)論;
(3)設(shè)0<a<b,Sn為數(shù)列{bn}的前n項(xiàng)和.是否存在實(shí)數(shù)λ,使得對(duì)任意正整數(shù)n,都有a<Sn<b?若存在,求λ的取值范圍;若不存在,說(shuō)明理由.

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科目:高中數(shù)學(xué) 來(lái)源: 題型:

A已知數(shù)列{an}是首項(xiàng)為a1=
1
4
,公比q=
1
4
的等比數(shù)列,設(shè)bn+2=3log
1
4
an  (n∈N*)
,數(shù)列{cn}滿足cn=an•bn
(1)求證:{bn}是等差數(shù)列;
(2)求數(shù)列{cn}的前n項(xiàng)和Sn;
(3)若cn
1
4
m2+m-1
對(duì)一切正整數(shù)n恒成立,求實(shí)數(shù)m的取值范圍.
B已知數(shù)列{an}和{bn}滿足:a1=λ,an+1=
2
3
an+n-4
,bn=(-1)n(an-3n+21),其中λ為實(shí)數(shù),n為正整數(shù).
(Ⅰ)對(duì)任意實(shí)數(shù)λ,證明:數(shù)列{an}不是等比數(shù)列;
(Ⅱ)證明:當(dāng)λ≠-18時(shí),數(shù)列{bn}是等比數(shù)列;
(Ⅲ)設(shè)0<a<b(a,b為實(shí)常數(shù)),Sn為數(shù)列{bn}的前n項(xiàng)和.是否存在實(shí)數(shù)λ,使得對(duì)任意正整數(shù)n,都有a<Sn<b?若存在,求λ的取值范圍;若不存在,說(shuō)明理由.

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科目:高中數(shù)學(xué) 來(lái)源: 題型:

已知數(shù)列{an}和{bn}滿足:a1=λ,an+1=
23
an+n-4,bn=(-1)n(an-3n+21)其中λ為實(shí)數(shù),且λ≠-18,n為正整數(shù).
(Ⅰ)求證:{bn}是等比數(shù)列;
(Ⅱ)設(shè)0<a<b,Sn為數(shù)列{bn}的前n項(xiàng)和.是否存在實(shí)數(shù)λ,使得對(duì)任意正整數(shù)n,都有a<Sn<b?若存在,求λ的取值范圍;若不存在,說(shuō)明理由.

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