题目内容
| lim |
| n→∞ |
| 1 |
| n+1 |
| 2 |
| n+1 |
| 3 |
| n+1 |
| 2n-1 |
| n+1 |
| 2n |
| n+1 |
| A、-1 | ||
| B、0 | ||
C、
| ||
| D、1 |
分析:利用同分母分式的加法法则,把原式转化为
,进一步简化为
,由此能求出
(
-
+
-…+
-
)的值.
| lim |
| n→∞ |
| [1+3+5+…+(2n-1)]-[2+4+6+…+2n] |
| n+1 |
| lim |
| n→∞ |
| ||||
| n+1 |
| lim |
| n→∞ |
| -n |
| n+1 |
| lim |
| n→∞ |
| 1 |
| n+1 |
| 2 |
| n+1 |
| 3 |
| n+1 |
| 2n-1 |
| n+1 |
| 2n |
| n+1 |
解答:解:
(
-
+
-…+
-
)
=
=
=
=-1.
故选A.
| lim |
| n→∞ |
| 1 |
| n+1 |
| 2 |
| n+1 |
| 3 |
| n+1 |
| 2n-1 |
| n+1 |
| 2n |
| n+1 |
=
| lim |
| n→∞ |
| [1+3+5+…+(2n-1)]-[2+4+6+…+2n] |
| n+1 |
=
| lim |
| n→∞ |
| ||||
| n+1 |
=
| lim |
| n→∞ |
| -n |
| n+1 |
=-1.
故选A.
点评:本题考查数列的极限,解题时要注意合理地进行等价转化.
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