题目内容
（1992•云南）
[
+
+
+…+
]=
．
| lim |
| n→∞ |
| 1 |
| 1•4 |
| 1 |
| 4•7 |
| 1 |
| 7•10 |
| 1 |
| (3n-2)(3n+1) |
| 1 |
| 3 |
| 1 |
| 3 |
分析：首先利用列项相消法求出数列的和，然后取极限即可得到答案．
解答：解：
[
+
+
+…+
]
=
[(1-
)+(
-
)+…+(
-
)
=
(1-
)
=
•
=
．
故答案为
．
| lim |
| n→∞ |
| 1 |
| 1•4 |
| 1 |
| 4•7 |
| 1 |
| 7•10 |
| 1 |
| (3n-2)(3n+1) |
=
| lim |
| n→∞ |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 4 |
| 1 |
| 7 |
| 1 |
| 3n-2 |
| 1 |
| 3n+1 |
=
| lim |
| n→∞ |
| 1 |
| 3 |
| 1 |
| 3n+1 |
=
| lim |
| n→∞ |
| 1 |
| 3 |
| 3n |
| 3n+1 |
| 1 |
| 3 |
故答案为
| 1 |
| 3 |
点评：本题考查了列项相消法求数列的前n项和，考查了数列极限的求法，是基础的运算题．
练习册系列答案
相关题目