题目内容
已知直线l:y=-n+1 |
n |
1 |
n |
3 |
2 |
1 |
2 |
分析:首先求得S1,S2,Sn的值,然后由规律:
×
=
-
求解即可求得答案.
1 |
n+1 |
1 |
n |
1 |
n |
1 |
n+1 |
解答:解:当n=1时,直线l1:y=-2x+1与x轴和y轴分别交于点A1和B1,
则A1(
,0),B1(0,1),
∴S1=
×
×1,
∵当n=2时,直线l2:y=-
x+
与x轴和y轴分别交于点A2和B2,
则A2(
,0),B2(0,
),
∴S2=
×
×
,
∴直线ln与x轴和y轴分别交于点An和Bn,
△AnOBn的面积为Sn=
×
×
,
∴S1+S2+S3+…+Sn=
×
×1+
×
×
+…+
×
×
,
=
×(1-
+
-
+…+
-
),
=
×(1-
),
=
.
故答案为:
.
则A1(
1 |
2 |
∴S1=
1 |
2 |
1 |
2 |
∵当n=2时,直线l2:y=-
3 |
2 |
1 |
2 |
则A2(
1 |
3 |
1 |
2 |
∴S2=
1 |
2 |
1 |
3 |
1 |
2 |
∴直线ln与x轴和y轴分别交于点An和Bn,
△AnOBn的面积为Sn=
1 |
2 |
1 |
n+1 |
1 |
n |
∴S1+S2+S3+…+Sn=
1 |
2 |
1 |
2 |
1 |
2 |
1 |
3 |
1 |
2 |
1 |
2 |
1 |
n+1 |
1 |
n |
=
1 |
2 |
1 |
2 |
1 |
2 |
1 |
3 |
1 |
n |
1 |
n+1 |
=
1 |
2 |
1 |
n+1 |
=
n |
2n+2 |
故答案为:
n |
2n+2 |
点评:此题考查了一次函数的应用.解题的关键是找到规律:△AnOBn的面积为Sn=
×
×
与
×
=
-
.
1 |
2 |
1 |
n+1 |
1 |
n |
1 |
n+1 |
1 |
n |
1 |
n |
1 |
n+1 |
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