题目内容
设直线kx+(k+1)y-1=0与坐标轴所围成的直角三角形的面积为Sk,则S1+S2+…+S2009=分析:令x=0,得y=
,令y=0,得x=
,则Sk=
•
=
(
-
),根据三角形面积公式求和.
| 1 |
| k+1 |
| 1 |
| k |
| 1 |
| 2 |
| 1 |
| k(k+1) |
| 1 |
| 2 |
| 1 |
| k |
| 1 |
| k+1 |
解答:解:依题意,得直线与y轴交于(0,
),与x轴交于(
,0),则
则Sk=
•
=
(
-
),
S1+S2+…+S2009
=
(1-
+
-
+…+
-
)
=
(1-
)
=
.
故答案为:
.
| 1 |
| k+1 |
| 1 |
| k |
则Sk=
| 1 |
| 2 |
| 1 |
| k(k+1) |
| 1 |
| 2 |
| 1 |
| k |
| 1 |
| k+1 |
S1+S2+…+S2009
=
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2009 |
| 1 |
| 2010 |
=
| 1 |
| 2 |
| 1 |
| 2010 |
=
| 2009 |
| 4020 |
故答案为:
| 2009 |
| 4020 |
点评:本题考查了一次函数的综合运用.关键是求出一次函数图象与坐标轴的交点,得出面积,再拆项求和.
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