题目内容
分析:由题意,s=840(
+
+
+…+
),利用裂项法,可求数列的和.
| 1 |
| 1×3 |
| 1 |
| 2×4 |
| 1 |
| 3×5 |
| 1 |
| 19×21 |
解答:解:由题意,s=840(
+
+
+…+
)
=840×
[(1-
)+(
-
)+(
-
)+…+(
-
)]
=420(1+
-
-
)=589
故选B.
| 1 |
| 1×3 |
| 1 |
| 2×4 |
| 1 |
| 3×5 |
| 1 |
| 19×21 |
=840×
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 19 |
| 1 |
| 21 |
=420(1+
| 1 |
| 2 |
| 1 |
| 20 |
| 1 |
| 21 |
故选B.
点评:本题考查循环结构,考查裂项法求数列的和,确定程序功能是关键.
练习册系列答案
相关题目
| A、798 | ||
B、
| ||
C、
| ||
| D、379 |