题目内容
设Sn=
+
+
+…+
,且Sn•Sn+1 =
,则n=
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 12 |
| 1 |
| n2+n |
| 3 |
| 4 |
6
6
.分析:由于
=
=
-
,可考虑利用裂项求和求Sn,代入SnSn+1可求n
| 1 |
| n2+n |
| 1 |
| n(n+1) |
| 1 |
| n |
| 1 |
| n+1 |
解答:解:∵
=
=
-
∴Sn=
+
+
+…+
=1-
+
-
+…+
-
=1-
=
∴SnSn+1=
•
=
=
∴n=6
故答案为:6
| 1 |
| n2+n |
| 1 |
| n(n+1) |
| 1 |
| n |
| 1 |
| n+1 |
∴Sn=
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 12 |
| 1 |
| n2+n |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| n |
| 1 |
| n+1 |
=1-
| 1 |
| n+1 |
| n |
| n+1 |
∴SnSn+1=
| n |
| n+1 |
| n+1 |
| n+2 |
| n |
| n+2 |
| 3 |
| 4 |
∴n=6
故答案为:6
点评:本题主要考查利用裂项求和的方法的应用,这是数列求和的一个重要方法,要注意掌握.
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