摘要:已知正项数列{an}的前n项和为Sn..数列{bn}满足a1b1+a2b2+a3b3+-+anbn= Ⅰ)求数列{an}的通项公式, Ⅱ)求数列{bn}的通项公式
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已知正项数列{an}的前n项和为Sn,且an+
=2Sn,n∈N*.
(Ⅰ)求证:数列{Sn2}是等差数列;
(Ⅱ)求解关于n的不等式an+1(Sn-1+Sn)>4n-8;
(Ⅲ)记数列bn=2Sn3,Tn=
+
+…+
,证明:1-
<Tn<
-
.
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| an |
(Ⅰ)求证:数列{Sn2}是等差数列;
(Ⅱ)求解关于n的不等式an+1(Sn-1+Sn)>4n-8;
(Ⅲ)记数列bn=2Sn3,Tn=
| 1 |
| b1 |
| 1 |
| b2 |
| 1 |
| bn |
| 1 | ||
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| 2 |
| 1 | ||
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已知正项数列{an}的前n项和为Sn,且an和Sn满足:4Sn=(an+1)2(n=1,2,3…),
(1)求{an}的通项公式;
(2)设bn=
,求{bn}的前n项和Tn;
(3)在(2)的条件下,对任意n∈N*,Tn>
都成立,求整数m的最大值.
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(1)求{an}的通项公式;
(2)设bn=
| 1 |
| an•an+1 |
(3)在(2)的条件下,对任意n∈N*,Tn>
| m |
| 23 |