题目内容
（2007•嘉定区一模）已知f(n)=1+
+
+…+
(n∈N)，则f（n+1）-f（n）=（ ）
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3n-1 |
分析：由f（n）=1+
+
+…+
+
+
，知f（n+1）=1+
+
+…+
+
+
+
+
+
，由此能求出f（n+1）-f（n）．
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3n-3 |
| 1 |
| 3n-2 |
| 1 |
| 3n-1 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3n-3 |
| 1 |
| 3n-2 |
| 1 |
| 3n-1 |
| 1 |
| 3n |
| 1 |
| 3n+1 |
| 1 |
| 3n+2 |
解答：解：∵f（n）=1+
+
+…+
+
+
，
∴f（n+1）=1+
+
+…+
+
+
+
+
+
，
∴f（n+1）-f（n）=
+
+
．
故选D．
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3n-3 |
| 1 |
| 3n-2 |
| 1 |
| 3n-1 |
∴f（n+1）=1+
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3n-3 |
| 1 |
| 3n-2 |
| 1 |
| 3n-1 |
| 1 |
| 3n |
| 1 |
| 3n+1 |
| 1 |
| 3n+2 |
∴f（n+1）-f（n）=
| 1 |
| 3n |
| 1 |
| 3n+1 |
| 1 |
| 3n+2 |
故选D．
点评：本题考查数列的函数性质，解题时要认真审题，注意总结规律，合理地进行等价转化．
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