题目内容
已知数列{an}的首项a1=1且满足 n≥2时,an=
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试题答案
C
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已知数列{an}的首项a1=1,前n项之和Sn满足关系式:3tSn+1-(2t+3)Sn=3t(t>0,n∈N*).
(1)求证:数列{an}是等比数列;
(2)设数列{an}的公比为f(t),数列{bn}满足bn+1=f(
),(n∈N*),且b1=1.
(i)求数列{bn}的通项bn;
(ii)设Tn=b1b2-b2b3+b3b4-b4b5+…+b2n-1b2n-b2nb2n+1,求Tn.
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(1)求证:数列{an}是等比数列;
(2)设数列{an}的公比为f(t),数列{bn}满足bn+1=f(
| 1 | bn |
(i)求数列{bn}的通项bn;
(ii)设Tn=b1b2-b2b3+b3b4-b4b5+…+b2n-1b2n-b2nb2n+1,求Tn.
已知数列{an}的首项a1=1,前n项之和Sn满足关系式:3tSn+1-(2t+3)Sn=3t(t>0,n∈N*).
(1)求证:数列{an}是等比数列;
(2)设数列{an}的公比为f(t),数列{bn}满足
,且b1=1.
(i)求数列{bn}的通项bn;
(ii)设Tn=b1b2-b2b3+b3b4-b4b5+…+b2n-1b2n-b2nb2n+1,求Tn.
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(1)求证:数列{an}是等比数列;
(2)设数列{an}的公比为f(t),数列{bn}满足
,且b1=1.(i)求数列{bn}的通项bn;
(ii)设Tn=b1b2-b2b3+b3b4-b4b5+…+b2n-1b2n-b2nb2n+1,求Tn.
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已知数列{an}的首项a1=a,Sn是数列{an}的前n项和,且满足:
=3n2an+
,an≠0,n≥2,n∈N*.
(1)若数列{an}是等差数列,求a的值;
(2)确定a的取值集合M,使a∈M时,数列{an}是递增数列.
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| S | 2 n |
| S | 2 n-1 |
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已知数列{an}的首项a1=4,且当n≥2时,an-1an-4an-1+4=0,数列{bn}满足bn=
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(Ⅰ)求证:数列{bn}是等差数列,并求{bn}的通项公式;
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| 1 |
| 2-an |
(Ⅰ)求证:数列{bn}是等差数列,并求{bn}的通项公式;
(Ⅱ)若cn=4bn•(nan-6)(n=1,2,3…),如果对任意n∈N*,都有cn+
| 1 |
| 2 |
=3n2an+
,an≠0,n≥2,n∈N*.