题目内容
数列0.5,0.55,0.555,0.5555,…的前n项之和为
+
-
+
-
.
| 5n |
| 9 |
| 1 |
| 9×10n |
| 1 |
| 9 |
| 5n |
| 9 |
| 1 |
| 9×10n |
| 1 |
| 9 |
分析:先表示出数列0.5,0.55,0.555,0.5555,…的通项公式,然后再根据等比数列前n项和公式计算.
解答:解:将数列0.5,0.55,0.555,0.5555,…记为an,
则an=
(1-10-n),
所以sn=a1+a2+…+an=
(n-
)=
+
-
,
故答案为:
+
-
.
则an=
| 5 |
| 9 |
所以sn=a1+a2+…+an=
| 5 |
| 9 |
| ||||
1-
|
| 5n |
| 9 |
| 1 |
| 9×10n |
| 1 |
| 9 |
故答案为:
| 5n |
| 9 |
| 1 |
| 9×10n |
| 1 |
| 9 |
点评:本题主要考查了通过观察求数列0.5,0.55,0.555,0.5555,…的通项公式,然后根据等比数列前n项公式计算.考查学生的计算能力.
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