问题
解答题
已知
(Ⅰ)求
(Ⅱ)求
(III)求四边形AnAn+1Bn+1Bn面积的最大值. |
答案
(Ⅰ)∵
=3An-1An
⇒AnAn+1
=AnAn+1 1 3
,An-1An
∴
=A4A5 1 3
=(A3A4
)21 3
=(A2A3
)31 3
=A1A2
(1 27
-OA2
)=OA1 1 3
.(3分)J
(II)由(1)知
=AnAn+1 1 3n-1
=A1A2 1 3n-3
,j
=OAn
+OA1
+…A1A2
=An-1An
+j
+…+A1A2 An-1An
=
+9j
+3j
+…+j 1 3n-3
=j
+j 9[1-(
)n-1]1 3 1- 1 3
=j 29-(
)n-41 3 2
.(6分)j
∵|
|=2Bn-1Bn
且Bn-1,Bn均在射线y=x(x≥0)上,2
∴
=2Bn-1Bn
+2i
.∴j
=OBn
+OB1
+B1B2
+…+B2B3
=3i+3Bn-1Bn
+(n-1)(2j
+2i
)j
(III)∵|
|=AnAn+1
,△AnAn+1Bn+1的底面边AnAn+1的高为h1=2n+3.1 3n-3
又|
|=2BnBn+1
,An(0,2
)到直线y=x的距离为h2=29-(
)n-41 3 2
.29-(
)n-41 3 2 2
∴Sn=
•(2n+3)•1 2
+1 3n-3
•21 2
•2
=29-(
)n-41 3 2 2
+29 2
,(10分)n 3n-3
而Sn-Sn-1=
-n 3n-3
=n-1 3n-4
<0,-2n+3 3n-3
∴S1>S2>…>Sn>…
∴Smax=S1=
+29 2
=1 3-2
+9=29 2
.47 2
(12分)