题目内容
设Sk=| 1 |
| k+1 |
| 1 |
| k+2 |
| 1 |
| 2k |
分析:运用Sk的表达式,得出Sk+1表达式,再将两式作差即可得出
解答:解:由Sk=
+
+…+
,用K+1代替K
可得SK+1=
+
+…+
=
+
+…+
+
+
∴SK+1-SK=
+
-
=
-
∴SK+1=SK+
-
故答案为
-
| 1 |
| k+1 |
| 1 |
| k+2 |
| 1 |
| 2k |
可得SK+1=
| 1 |
| (k+1)+1 |
| 1 |
| (K+1)+2 |
| 1 |
| 2(K+1) |
=
| 1 |
| k+2 |
| 1 |
| k+3 |
| 1 |
| 2k |
| 1 |
| 2K+1 |
| 1 |
| 2K+2 |
∴SK+1-SK=
| 1 |
| 2K+1 |
| 1 |
| 2K+2 |
| 1 |
| K+1 |
| 1 |
| 2k+1 |
| 1 |
| 2k+2 |
∴SK+1=SK+
| 1 |
| 2K+1 |
| 1 |
| 2K+2 |
故答案为
| 1 |
| 2k+1 |
| 1 |
| 2k+2 |
点评:处理数列的递推关系式,从n=k 到n=k+1时,不仅要注意首项和末项的形式,还要注意表达式项数发生的变化,这样才不会出错.
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相关题目
设Sk=
+
+
+…+
,则Sk+1为( )
| 1 |
| k+1 |
| 1 |
| k+2 |
| 1 |
| k+3 |
| 1 |
| 2k |
A、Sk+
| ||||
B、Sk+
| ||||
C、Sk+
| ||||
D、Sk+
|