题目内容
如图,O为直线A0A2013外一点,若A0,A1,A2,A3,A4,A5,…,A2013中任意相邻两点的距离相等,设
=
,
=
,用
,
表示
+
+
+…+
,其结果为______.
| OA0 |
| a |
| OA2013 |
| b |
| a |
| b |
| OA0 |
| OA1 |
| OA2 |
| OA2013 |
设A0A2013的中点为A,则A也是A1A2012,…A1006A1007的中点,
由向量的中点公式可得
+
=2
=
+
,
同理可得
+
=
+
=…=
+
,
故
+
+
+…+
=1007×2
=1007(
+
)
故答案为:1007(
+
)
由向量的中点公式可得
| OA0 |
| OA2013 |
| OA |
| a |
| b |
同理可得
| OA1 |
| OA2012 |
| OA2 |
| OA2011 |
| OA1006 |
| OA1007 |
故
| OA0 |
| OA1 |
| OA2 |
| OA2013 |
| OA |
| a |
| b |
故答案为:1007(
| a |
| b |
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