题目内容
用[x]表示不大于x的最大整数,如[3]=3,[3.1]=3.设S=
+
+…+
,则[20S]=( )
| 1 | ||
[
|
| 1 | ||
[
|
| 1 | ||
[
|
| A、0 | B、1 | C、2 | D、3 |
分析:首先可得[
]=(n-1)(n+2),即可得3S=
+
+
-
-
-
,则可求得[20S]的值.
| [n(n+1)-1]2 |
| n(n+1) |
| 1 |
| 9 |
| 1 |
| 10 |
| 1 |
| 11 |
| 1 |
| 49 |
| 1 |
| 50 |
| 1 |
| 51 |
解答:解:[
]=[n(n+1)-2+
]=n(n+1)-2=(n-1)(n+2),
∵
=
(
-
),
∴3S=
-
+
-
+
-
+
-
+…+
-
=
+
+
-
-
-
,
∴[20S]=[
×(
+
+
-
-
-
)]≈[1.6]=1.
故选B.
| [n(n+1)-1]2 |
| n(n+1) |
| 1 |
| n(n+1) |
∵
| 1 |
| (n-1)(n+2) |
| 1 |
| 3 |
| 1 |
| n-1 |
| 1 |
| n+2 |
∴3S=
| 1 |
| 9 |
| 1 |
| 12 |
| 1 |
| 10 |
| 1 |
| 13 |
| 1 |
| 11 |
| 1 |
| 14 |
| 1 |
| 12 |
| 1 |
| 15 |
| 1 |
| 48 |
| 1 |
| 51 |
| 1 |
| 9 |
| 1 |
| 10 |
| 1 |
| 11 |
| 1 |
| 49 |
| 1 |
| 50 |
| 1 |
| 51 |
∴[20S]=[
| 20 |
| 3 |
| 1 |
| 9 |
| 1 |
| 10 |
| 1 |
| 11 |
| 1 |
| 49 |
| 1 |
| 50 |
| 1 |
| 51 |
故选B.
点评:此题考查了取整函数的意义.解题的关键是注意将复杂的[
]化简.
| [n(n+1)-1]2 |
| n(n+1) |
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