設(shè)為實(shí)數(shù).成等比數(shù)列.且成等差數(shù)列.則的值是 . 查看更多

 

題目列表(包括答案和解析)

設(shè)等比數(shù)列{an}的首項(xiàng)為a1=2,公比為q(q為正整數(shù)),且滿足3a3是8a1與a5的等差中項(xiàng);等差數(shù)列{bn}滿足2n2-(t+bn)n+
32
bn
=0(t∈R,n∈N*).
(Ⅰ)求數(shù)列{an},{bn}的通項(xiàng)公式;
(Ⅱ) 若對(duì)任意n∈N*,有anbn+1+λanan+1≥bnan+1成立,求實(shí)數(shù)λ的取值范圍;
(Ⅲ)對(duì)每個(gè)正整數(shù)k,在ak和a k+1之間插入bk個(gè)2,得到一個(gè)新數(shù)列{cn}.設(shè)Tn是數(shù)列{cn}的前n項(xiàng)和,試求滿足Tm=2cm+1的所有正整數(shù)m.

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等差數(shù)列{an}的各項(xiàng)均為正整數(shù),a1=3,前n項(xiàng)和為Sn,等比數(shù)列{bn}中,b1=1,且b2•S2=16,{ban}是公比為4的等比數(shù)列
(1)求an與bn
(2)設(shè)Cn=
1
S1
+
1
S2
+
1
S2
+…+
1
Sn
,若對(duì)任意正整數(shù)n,當(dāng)m∈[-1,1]時(shí),不等式t2-2mt+
3
4
>Cn恒成立,求實(shí)數(shù)t的取值范圍.

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數(shù)列{an}是首項(xiàng)a1=4的等比數(shù)列,且S3,S2,S4成等差數(shù)列,
(1)求數(shù)列{an}的通項(xiàng)公式;
(2)若bn=log2|an|,設(shè)Tn為數(shù)列{
1bnbn+1
}
的前n項(xiàng)和,若Tn≤λbn+1對(duì)一切n∈N*恒成立,求實(shí)數(shù)λ的最小值.

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等差數(shù)列{a}是遞增數(shù)列,前n項(xiàng)和為Sn,且a1,a2,a5成等比數(shù)列,S5=a32
(1)求通項(xiàng)an;
(2)令bn=
1
2
(
an+1
an
+
an
an+1
)
,設(shè)Tn=b1+b2+…+bn-n,若M>Tn>m對(duì)一切正整數(shù)n恒成立,求實(shí)數(shù)M、m的取值范圍;
(3)試構(gòu)造一個(gè)函數(shù)g(x),使f(n)=a1g(1)+a2g(2)+…+ang(n)<
1
3
(n∈N+)
恒成立,且對(duì)任意的m∈(
1
4
,
1
3
)
,均存在正整數(shù)N,使得當(dāng)n>N時(shí),f(n)>m.

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等差數(shù)列{an}的各項(xiàng)均為正整數(shù),a1=3,前n項(xiàng)和為Sn,等比數(shù)列{bn}中,b1=1,且b2•S2=16,{ban}是公比為4的等比數(shù)列
(1)求an與bn
(2)設(shè)Cn=
1
S1
+
1
S2
+
1
S2
+…+
1
Sn
,若對(duì)任意正整數(shù)n,當(dāng)m∈[-1,1]時(shí),不等式t2-2mt+
3
4
>Cn恒成立,求實(shí)數(shù)t的取值范圍.

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