题目内容

如图,有以下命题:设P、Q是线段AB的三等分点,则有O
P
+O
Q
=O
A
+O
B
把此命题推广,设点A1,A2,A3…An-1是线段AB的n等分点(n≥3,n∈N*),则有O
A1
+O
A2
+O
A3
+
O
AN-1
=
n-1
2
n-1
2
O
A
+O
B
).
分析:受:“命题:设P、Q是线段AB的三等分点,则有O
P
+O
Q
=O
A
+O
B
”的启示,如果点A1,A2,A3,…,An-1是AB的n(n≥3)等分点,归纳得出猜想O
A1
+O
A2
+O
A3
+
O
AN-1
=
n-1
2
O
A
+O
B
).再利用向量的运算证明结论.
解答:解:设
OA
=a
OB
=b
,能得到的结论是:
OA1
+
OA2
+…+
OAn-1
n-1
2
a
+
n-1
2
b

证明:
OA1
=
OA
-
A1A
=
a
-
1
n
a
-
b
)=
n-1
n
a
+
1
n
b

OA2
=
OA
-
A2A
=
a
-
2
n
a
-
b
)=
n-2
n
a
+
2
n
b


OAn-1
=
OA
-
An-1A
=
a
-
n-1
n
a
-
b
)=
1
n
a
+
n-1
n
b

OA1
+
OA2
+…+
OAn-1
=(
n-1
n
+
n-2
n
+…+
1
n
a
+(
n-1
n
+
n-2
n
+…+
1
n
b
 
=
n-1
2
a
+
n-1
2
b

故答案为:
n-1
2
点评:本题考查类比推理、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.
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