摘要:设等比数列{an}(n∈N*)的首项.公比.且a1+a3+-+a2n-1=.则n= .
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设等比数列{an}的前n项和Sn,首项a1=1,公比q=f(λ)=
(λ≠-1,0).
(Ⅰ)证明:Sn=(1+λ)-λan;
(Ⅱ)若数列{bn}满足b1=
,bn=f(bn-1)(n∈N*,n≥2),求数列{bn}的通项公式;
(Ⅲ)若λ=1,记cn=an(
-1),数列{cn}的前项和为Tn,求证:当n≥2时,2≤Tn<4.
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| λ |
| 1+λ |
(Ⅰ)证明:Sn=(1+λ)-λan;
(Ⅱ)若数列{bn}满足b1=
| 1 |
| 2 |
(Ⅲ)若λ=1,记cn=an(
| 1 |
| bn |