题目内容

S=1+
1
2
+
1
3
+…+
1
1000000
,则S的整数部分是
1998
1998
分析:S=1+
1
2
+
1
3
+
1
4
+…+
1
1000000
=1+
2
2
+
2
+
2
3
+
3
+…+
2
1000000
+
1000000
<1+2×(
1
2
+1
+
1
3
+
2
+…+
1
1000000
+
999999
)=1999;即S<1999;同理有S=1+
1
2
+
1
3
+
1
4
+…+
1
1000000
=1+
2
2
+
2
+
2
3
+
3
+…+
2
1000000
+
1000000
>1+2×(
1
3
+
2
+
1
4
+
3
+…+
1
1000001+
1000000

=1+2×[(
3
-
2
)+(
4
-
3
)…+(
1000000
-
999999
)+(
1000001
-
1000000
)]
=1+2×(
1000001
-
2
)>1998.即1998<S<1999;进而可得答案.
解答:解:S=1+
1
2
+
1
3
+
1
4
+…+
1
1000000

=1+
2
2
+
2
+
2
3
+
3
+…+
2
1000000
+
1000000

<1+2×(
1
2
+1
+
1
3
+
2
+…+
1
1000000
+
999999

=1+2×[(
2
-1)+(
3
-
2
)+…+(
1000000
-
999999
)]
=1+2×(-1+
1000000

=1+2×(999)
=1999.
即S<1999,
S=1+
1
2
+
1
3
+
1
4
+…+
1
1000000

=1+
2
2
+
2
+
2
3
+
3
+…+
2
1000000
+
1000000

>1+2×(
1
3
+
2
+
1
4
+
3
+…+
1
1000001+
1000000

=1+2×[(
3
-
2
)+(
4
-
3
)…+(
1000000
-
999999
)+(
1000001
-
1000000
)]
=1+2×(
1000001
-
2
)>1998.
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.
练习册系列答案
相关题目

违法和不良信息举报电话:027-86699610 举报邮箱:58377363@163.com

精英家教网