题目内容
若S=1+
+
+…+
,则S的整数部分是
| 1 | ||
|
| 1 | ||
|
| 1 | ||
|
1998
1998
.分析:S=1+
+
+
+…+
=1+
+
+…+
<1+2×(
+
+…+
)=1999;即S<1999;同理有S=1+
+
+
+…+
=1+
+
+…+
>1+2×(
+
+…+
)
=1+2×[(
-
)+(
-
)…+(
-
)+(
-
)]
=1+2×(
-
)>1998.即1998<S<1999;进而可得答案.
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=1+2×[(
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| 4 |
| 3 |
| 1000000 |
| 999999 |
| 1000001 |
| 1000000 |
=1+2×(
| 1000001 |
| 2 |
解答:解:S=1+
+
+
+…+
=1+
+
+…+
<1+2×(
+
+…+
)
=1+2×[(
-1)+(
-
)+…+(
-
)]
=1+2×(-1+
)
=1+2×(999)
=1999.
即S<1999,
S=1+
+
+
+…+
=1+
+
+…+
>1+2×(
+
+…+
)
=1+2×[(
-
)+(
-
)…+(
-
)+(
-
)]
=1+2×(
-
)>1998.
则1998<S<1999
所以不超过S的最大整数为1998.
故答案为:1998.
| 1 | ||
|
| 1 | ||
|
| 1 | ||
|
| 1 | ||
|
=1+
| 2 | ||||
|
| 2 | ||||
|
| 2 | ||||
|
<1+2×(
| 1 | ||
|
| 1 | ||||
|
| 1 | ||||
|
=1+2×[(
| 2 |
| 3 |
| 2 |
| 1000000 |
| 999999 |
=1+2×(-1+
| 1000000 |
=1+2×(999)
=1999.
即S<1999,
S=1+
| 1 | ||
|
| 1 | ||
|
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|
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|
=1+
| 2 | ||||
|
| 2 | ||||
|
| 2 | ||||
|
>1+2×(
| 1 | ||||
|
| 1 | ||||
|
| 1 | ||||
|
=1+2×[(
| 3 |
| 2 |
| 4 |
| 3 |
| 1000000 |
| 999999 |
| 1000001 |
| 1000000 |
=1+2×(
| 1000001 |
| 2 |
则1998<S<1999
所以不超过S的最大整数为1998.
故答案为:1998.
点评:本题考查有理数域的解法,解题时要认真审题,注意挖掘题设中的隐含条件,合理地运用放缩法进行求解.
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