設(shè)O是直線AB外一點(diǎn),
OA
=
a
,
OB
=
b
,點(diǎn)A1,A2,A3…An-1是線段AB的n,(n≥2)等分點(diǎn),則
OA1
+
OA2
+
OA3
+…+
OAn-1
=______(用
a
b
,n表示).
由題意可得
OA1
=
OA
+
AA1
=
OA
+
1
n
AB
=
OA
+
1
n
(
OB
-
OA
)=
a
+
1
n
(
b
-
a
)

OA2
=
OA
+AA2
=
OA
+
2
n
AB
=
OA
+
2
n
(
OB
-
OA
)=
a
+
2
n
(
b
-
a
)


OAn-1
=
OA
+AAn-1
=
OA
+
n-1
n
AB
=
OA
+
n-1
n
(
OB
-
OA
)=
a
+
n-1
n
(
b
-
a
)
,
把以上n-1個(gè)式子相加得
OA1
+
OA2
+
OA3
+…+
OAn-1
=(n-1)
a
+
1+2+3+…+(n-1)
n
(
b
-
a
)

=(n-1)
a
+
n(n-1)
2n
(
b
-
a
)
=
n-1
2
(
a
+
b
)

故答案為
n-1
2
(
a
+
b
)
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