题目内容
已知正项等比数列{an}满足a2=
,a4=
,n∈N*
(1)求数列{an}通项公式;
(2)设数列{bn}满足bn=log3an•log3an+1,求数列{
}的前n和Tn.
| 1 |
| 9 |
| 1 |
| 81 |
(1)求数列{an}通项公式;
(2)设数列{bn}满足bn=log3an•log3an+1,求数列{
| 1 |
| bn |
考点:数列的求和,等比数列的性质
专题:等差数列与等比数列
分析:(Ⅰ)设公比为q,由已知条件得
=
=q2,从而得到q=
,或q=-
.由此利用分类思想能求出数列{an}通项公式.
(Ⅱ)由bn=log3|an|•log3|an+1|=n(n+1),得
=
=
-
,由此利用裂项求和法能求出数列{
}的前n和Tn.
| a4 |
| a2 |
| 1 |
| 9 |
| 1 |
| 3 |
| 1 |
| 3 |
(Ⅱ)由bn=log3|an|•log3|an+1|=n(n+1),得
| 1 |
| bn |
| 1 |
| n(n+1) |
| 1 |
| n |
| 1 |
| n+1 |
| 1 |
| bn |
解答:
角:(Ⅰ)设公比为q,∵a2=
,a4=
,n∈N*,∴
=
=q2,
∴q=
,或q=-
.
①当q=
时,a1=
=
,
an=(
)n,n∈N*.
②q=-
时,a1=
=-
,
an=(-
)n,n∈N*.
(Ⅱ)∵bn=log3|an|•log3|an+1|=n(n+1),
∴
=
=
-
,
∴Tn=1-
+
-
+…+
-
=1-
=
,n∈N*.
| 1 |
| 9 |
| 1 |
| 81 |
| a4 |
| a2 |
| 1 |
| 9 |
∴q=
| 1 |
| 3 |
| 1 |
| 3 |
①当q=
| 1 |
| 3 |
| ||
|
| 1 |
| 3 |
an=(
| 1 |
| 3 |
②q=-
| 1 |
| 3 |
| ||
-
|
| 1 |
| 3 |
an=(-
| 1 |
| 3 |
(Ⅱ)∵bn=log3|an|•log3|an+1|=n(n+1),
∴
| 1 |
| bn |
| 1 |
| n(n+1) |
| 1 |
| n |
| 1 |
| n+1 |
∴Tn=1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| n |
| 1 |
| n+1 |
=1-
| 1 |
| n+1 |
=
| n |
| n+1 |
点评:本题考查数列的通项公式的求法,考查数列的前n项和的求法,解题时要认真审题,注意裂项求和法的合理运用.
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