Question

如图，有以下命题：设点P、Q是线段AB的三等分点，则有

OP

+

OQ

=

OA

+

OB

，把此命题推广，设点A₁、A₂、…、A_n-1是AB的n等分点（n≥3），则

OA1

+

OA2

+…+

OAn-1

=

n-1

2

(

OA

+

OB

)．

Answer 1

分析：用向量加法的法则和几何意义知

OAk

=

OA

-

AkA

=

OA

-

k

n

(

OA

-

OB

)化简

OA1

+

OA2

+…+

OAn-1

的解析式，利用等差数列求和公式得到它的结果．

解答：解：∵

OA1

=

OA

-

A1A

=

OA

-

1

n

(

OA

-

OB

)，

OA2

=

OA

-

A2A

=

OA

-

2

n

(

OA

-

OB

)，…

OAk

=

OA

-

AkA

=

OA

-

k

n

(

OA

-

OB

)，∴

OA1

+

OA2

+…+

OAn-1

=

n-1

2

(

OA

+

OB

)，
故答案为

n-1

2

点评：本题考查向量加法、减法的运算法则和几何意义，并且运用等差数列求和公式进行计算化简以及进行合情推理．