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
+O
=O
+O
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
+O
+O
+…O
=
(O
+O
).
P |
Q |
A |
B |
A1 |
A2 |
A3 |
AN-1 |
n-1 |
2 |
n-1 |
2 |
A |
B |
分析:受:“命题:设P、Q是线段AB的三等分点,则有O
+O
=O
+O
”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
+O
+O
+…O
=
(O
+O
).再利用向量的运算证明结论.
P |
Q |
A |
B |
A1 |
A2 |
A3 |
AN-1 |
n-1 |
2 |
A |
B |
解答:解:设
=a,
=b,能得到的结论是:
+
+…+
═
+
.
证明:
∵
=
-
=
-
(
-
)=
+
,
=
-
=
-
(
-
)=
+
,
…
=
-
=
-
(
-
)=
+
,
∴
+
+…+
=(
+
+…+
)
+(
+
+…+
)
=
+
.
故答案为:
.
OA |
OB |
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.
练习册系列答案
相关题目