题目内容
设f(n)=1+
+
+…+
，那么f（2k+1）-f（2k）=______．
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| n |
∵f(n)=1+
+
+…+
，
∴f（2k+1）-f（2k）=1+
+
+…+
+
+
+…+
-(1+
+
+…+
)
=
+
+…+
．
故答案为：
+
+…+
．
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| n |
∴f（2k+1）-f（2k）=1+
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2k |
| 1 |
| 2k+1 |
| 1 |
| 2k+2 |
| 1 |
| 2k+ 2k |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2k |
=
| 1 |
| 2k+1 |
| 1 |
| 2k+2 |
| 1 |
| 2k+1 |
故答案为：
| 1 |
| 2k+1 |
| 1 |
| 2k+2 |
| 1 |
| 2k+1 |
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