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
OA0
=
a
OA2013
=
b
,用
a
b
表示
OA0
+
OA1
+
OA2
+…+
OA2013
,其结果为______.
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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
OA0
+
OA2013
=2
OA
=
a
+
b

同理可得
OA1
+
OA2012
=
OA2
+
OA2011
=…=
OA1006
+
OA1007

OA0
+
OA1
+
OA2
+…+
OA2013
=1007×2
OA
=1007(
a
+
b

故答案为:1007(
a
+
b
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