题目内容
若数列{an}通项公式an=
(n∈N+),Sn为其前n项和,
(1)试计算S1,S2,S3的值;
(2)猜测出Sn的公式.
| 1 | n(n+1) |
(1)试计算S1,S2,S3的值;
(2)猜测出Sn的公式.
分析:(1)由an=
-
,利用s1=a1,s2=a1+a2,s3=a1+a2+a3可求
(2)根据(1)中所求式子的规律可作出猜想
| 1 |
| n |
| 1 |
| n+1 |
(2)根据(1)中所求式子的规律可作出猜想
解答:解:(1)∵an=
(n∈N+),
=
-
∴s1=a1=1-
=
s2=a1+a2=1-
+
-
=
s3=1-
+
-
+
-
=
(2)猜想Sn=
| 1 |
| n(n+1) |
=
| 1 |
| n |
| 1 |
| n+1 |
∴s1=a1=1-
| 1 |
| 2 |
| 1 |
| 2 |
s2=a1+a2=1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 2 |
| 3 |
s3=1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 3 |
| 4 |
(2)猜想Sn=
| n |
| n+1 |
点评:本题主要考查了利用数列的通项求解数列的和,属于基础试题
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