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用[x]表示不大于x的最大整数,如[3]=3,[3.1]=3.设S=1[(10×11−1)210×11]+1[(11×12−1)211×12]+…+1[(49×50−1)249×50],则[20S]=()A.0B.1C.2D.3
题目详情
用[x]表示不大于x的最大整数,如[3]=3,[3.1]=3.设S=
+
+…+
,则[20S]=( )
A.0
B.1
C.2
D.3
| 1 | ||
[
|
| 1 | ||
[
|
| 1 | ||
[
|
A.0
B.1
C.2
D.3
▼优质解答
答案和解析
[
]=[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.
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