摘要:(3)bn=Sn+1-Sn=an+12=.由bn<.得m>.
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设数列{an}的前n项和为Sn,且Sn=(1+λ)-λan,其中λ为常数,且λ≠-1,0,n∈N+
(1)证明:数列{an}是等比数列.
(2)设数列{an}的公比q=f(λ),数列{bn}满足b1=
,bn=f(bn-1)(n∈N+,n≥2),求数列{bn}的通项公式.
(3)设λ=1,Cn=an(
-1),数列{Cn}的前n项和为Tn,求证:当n≥2时,2≤Tn<4.
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(1)证明:数列{an}是等比数列.
(2)设数列{an}的公比q=f(λ),数列{bn}满足b1=
| 1 |
| 2 |
(3)设λ=1,Cn=an(
| 1 |
| bn |
已知各项均为正数的数列{an},设其前n项和为Sn,且满足:Sn=(
)2.
(1)求a1,a2,a3;
(2)求出数列{an}的通项公式;
(3)设bn=
,求数列{bn}的前n项和.
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| an+1 |
| 2 |
(1)求a1,a2,a3;
(2)求出数列{an}的通项公式;
(3)设bn=
| 1 |
| anan+1 |
已知等差数列数﹛an﹜的前n项和为Sn,等比数列﹛bn﹜的各项均为正数,公比是q,且满足:a1=3,b1=1,b2+S2=12,S2=b2q.
(Ⅰ)求an与bn;
(Ⅱ)设cn=3bn-λ•2
(λ∈R),若﹛cn﹜满足:cn+1>cn对任意的n∈N°恒成立,求λ的取值范围.
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(Ⅰ)求an与bn;
(Ⅱ)设cn=3bn-λ•2
| an | 3 |