问题
填空题
设O是直线AB外一点,
|
答案
由题意可得
=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个式子相加得
+OA1
+OA2
+…+OA3
=(n-1)OAn-1
+a
(1+2+3+…+(n-1) n
-b
)a
=(n-1)
+a
(n(n-1) 2n
-b
)=a
(n-1 2
+a
),b
故答案为
(n-1 2
+a
).b