题目内容
（2006•广州二模）若Sn=
+
+
+…+
（n∈N*），则
Sn=
．
| 1 |
| 12+2 |
| 1 |
| 22+4 |
| 1 |
| 32+6 |
| 1 |
| n2+2n |
| lim |
| n→∞ |
| 3 |
| 4 |
| 3 |
| 4 |
分析：化简数列的通项，利用裂项法，求出数列的和，然后通过数列极限的运算法则，求出极限．
解答：解：因为
=
(
-
)，
所以Sn=
+
+
+…+
=
(
-
+
-
+
-
+…+
-
)
=
(1+
-
-
)．
所以
Sn=
(1+
-
-
)=
(1+
)=
．
故答案为：
．
| 1 |
| n2+2n |
| 1 |
| 2 |
| 1 |
| n |
| 1 |
| n+2 |
所以Sn=
| 1 |
| 12+2 |
| 1 |
| 22+4 |
| 1 |
| 32+6 |
| 1 |
| n2+2n |
=
| 1 |
| 2 |
| 1 |
| 1 |
| 1 |
| 3 |
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| n |
| 1 |
| n+2 |
=
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| n+1 |
| 1 |
| n+2 |
所以
| lim |
| n→∞ |
| lim |
| n→∞ |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| n+1 |
| 1 |
| n+2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 3 |
| 4 |
故答案为：
| 3 |
| 4 |
点评：本题考查数列通项公式的应用，裂项法求法数列的和，数列极限的应用，考查计算能力．
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