题目内容
| lim |
| n→∞ |
| ||||||
| 1+3+5+…+(2n-1) |
分析:先求
+
+
,1+3+5+…+(2n-1),把所求的结果代入到所求的极限中可求
| C | 0 n |
| C | 1 n |
| C | 2 n |
解答:解:∵
+
+
=1+n+
n(n-1)=
n2+
n+1
1+3+5+…+(2n-1)=
•n=n2
∴
=
=
(
+
+
)=
故选B
| C | 0 n |
| C | 1 n |
| C | 2 n |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
1+3+5+…+(2n-1)=
| 1+2n-1 |
| 2 |
∴
| lim |
| n→∞ |
| ||||||
| 1+3+5+…+(2n-1) |
| lim |
| n→∞ |
| ||
| n2 |
| lim |
| n→∞ |
| 1 |
| 2 |
| 1 |
| 2n |
| 1 |
| n2 |
| 1 |
| 2 |
故选B
点评:本题主要考查了二项式的基本运算,等差数列的求和公式及数列极限的求解,分子分母同时除以n2是求解极限的关键
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