题目内容
若an=| 1 |
| n+1 |
| 1 |
| n+2 |
| 1 |
| 2n |
分析：由题意知，an+1-an=(
+
+…+
+
+
)-（
+
+…+
），由此能够推陈出新出此结果．
| 1 |
| n+2 |
| 1 |
| n+3 |
| 1 |
| 2n |
| 1 |
| 2n+1 |
| 1 |
| 2n+2 |
| 1 |
| n+1 |
| 1 |
| n+2 |
| 1 |
| 2n |
解答：解：an+1-an=(
+
+…+
+
+
)-（
+
+…+
）
=
+
-
=
-
．
故答案为：
-
．
| 1 |
| n+2 |
| 1 |
| n+3 |
| 1 |
| 2n |
| 1 |
| 2n+1 |
| 1 |
| 2n+2 |
| 1 |
| n+1 |
| 1 |
| n+2 |
| 1 |
| 2n |
=
| 1 |
| 2n+1 |
| 1 |
| 2n+2 |
| 1 |
| n+1 |
=
| 1 |
| 2n+1 |
| 1 |
| 2n+2 |
故答案为：
| 1 |
| 2n+1 |
| 1 |
| 2n+2 |
点评：本题考查数列的运算，解题的关键是an+1=(
+
+…+
+
+
)，别丢掉
这一项．
| 1 |
| n+2 |
| 1 |
| n+3 |
| 1 |
| 2n |
| 1 |
| 2n+1 |
| 1 |
| 2n+2 |
| 1 |
| 2n+1 |
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若an=
+
+…+
（n是正整数），则an+1=an+（ ）
| 1 |
| n+1 |
| 1 |
| n+2 |
| 1 |
| 2n |
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