题目内容
数列{an}的前n项和为Sn,满足Sn=
an-
-
,设bn=log3(an+
),则数列{
}的前19项和为
______.
| 3 |
| 2 |
| n |
| 2 |
| 3 |
| 4 |
| 1 |
| 2 |
| 1 |
| bn•bn+1 |
令n=1,得到a1=s1=
a1-
-
,解得a1=
,
因为sn=
an-
-
①
当n≥2时求出sn-1=
an-1-
-
②
用①-②得:an=3an-1+1,所以代入求得a2=
,a3=
,a4=
,…
所以数列{an+
}为以3为首项,3为公比的等比数列,
所以通项公式为3n,则bn=
=n,
数列{
}的前19项和为:
+
+…+
=
+
+..+
=1-
+
-
+…+
-
+
-
=
故答案为
.
| 3 |
| 2 |
| 1 |
| 2 |
| 3 |
| 4 |
| 5 |
| 2 |
因为sn=
| 3 |
| 2 |
| n |
| 2 |
| 3 |
| 4 |
当n≥2时求出sn-1=
| 3 |
| 2 |
| n-1 |
| 2 |
| 3 |
| 4 |
用①-②得:an=3an-1+1,所以代入求得a2=
| 17 |
| 2 |
| 53 |
| 2 |
| 161 |
| 2 |
所以数列{an+
| 1 |
| 2 |
所以通项公式为3n,则bn=
| log | 3n3 |
数列{
| 1 |
| bn•bn+1 |
| 1 |
| b1b2 |
| 1 |
| b2b3 |
| 1 |
| b19b20 |
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 19×20 |
=1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 18 |
| 1 |
| 19 |
| 1 |
| 19 |
| 1 |
| 20 |
=
| 19 |
| 20 |
故答案为
| 19 |
| 20 |
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