12.已知數(shù)列{an}的前n項(xiàng)和為Sn.滿足log2(1+Sn)=n+1.求數(shù)列的通項(xiàng)公式. 解:Sn滿足log2(1+Sn)=n+1.∴1+Sn=2n+1. ∴Sn=2n+1-1. ∴a1=3.an=Sn-Sn-1=(2n+1-1)-(2n-1)=2n(n≥2). ∴{an}的通項(xiàng)公式為an= 查看更多

 

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已知數(shù)列{an}的前n項(xiàng)和為Sn,滿足an+Sn=2n.
(Ⅰ)證明:數(shù)列{an-2}為等比數(shù)列,并求出an;
(Ⅱ)設(shè)bn=(2-n)(an-2),求{bn}的最大項(xiàng).

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已知數(shù)列{an}的前n項(xiàng)和為Sn,滿足Sn=2an+n2-4n(n=1,2,3,…).
(Ⅰ)寫出數(shù)列{an}的前三項(xiàng)a1,a2,a3;
(Ⅱ)求證:數(shù)列{an-2n+1}為等比數(shù)列;
(Ⅲ)求Sn

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已知數(shù)列{an}的前n項(xiàng)和為sn,滿足Sn=2an-2n(n∈N+),
(1)求數(shù)列{an}的通項(xiàng)公式an
(2)若數(shù)列bn滿足bn=log2(an+2),Tn為數(shù)列{
bn
an+2
}的前n項(xiàng)和,求Tn
(3)(只理科作)接(2)中的Tn,求證:Tn
1
2

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已知數(shù)列{an}的前n項(xiàng)和為Sn,滿足關(guān)系式(2+t)Sn+1-tSn=2t+4(t≠-2,t≠0,n=1,2,3,…).
(Ⅰ)當(dāng)a1為何值時(shí),數(shù)列{an}是等比數(shù)列;
(Ⅱ)在(Ⅰ)的條件下,設(shè)數(shù)列{an}的公比為f(t),作數(shù)列{bn}使b1=1,bn=f(bn-1)(n=2,3,4,…),求bn;
(Ⅲ)在(Ⅱ)條件下,如果對(duì)一切n∈N+,不等式bn+bn+1
c2n+1
恒成立,求實(shí)數(shù)c的取值范圍.

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已知數(shù)列{an}的前n項(xiàng)和為Sn,滿足Sn=2an-2n(n∈N*),
(1)求證數(shù)列{an+2}為等比數(shù)列;
(2)若數(shù)列{bn}滿足bn=log2(an+2),Tn為數(shù)列{
bn
an+2
}的前n項(xiàng)和,求證:Tn
3
2

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