F

为AM的中点，又 E为AA1的中点，∴EF//A1M，

在三棱柱ABC A1B1C1中，D,M分别为A1B1,AB的中点，

A1D//BM

，且A1D BM，

则四边形A1DBM为平行四边形， A1M//BD，

EF//BD，又 BD 平面BC1D，EF 平面BC1D，

··························································································· 6 EF//平面BC1D． ·

分

（Ⅱ）设AC上存在一点G，使得平面EFG将三棱柱分割成两部分的体积之比为1︰31，则VE AFG:VABC ABC 1:32．

111

1

VE AFGVABC A1B1C1

1

1AF AG AE

AB AC A1A

2

111AG1AG

，

342AC24AC1AG1AG3

，即··································· 12分 ，所以符合要求的点G存在．·

24AC32AC4

19．解析 （Ⅰ）由2an 1 3Sn 3n 4，得2an 3Sn 1 3n 1（n 2）， 两式相减得2an 1 2an 3(Sn Sn 1) 3，即2an 1 an 3， ·································· 2分 ∴an 1 an

21

32

，则an 1 1 (an 1)（n 2）， ············································ 4分

2

1

1

1 ， 由a1 2，又2a2 3S1 7，得a2 ，则

a1 12 122

1

1

a2 1

故数列{an 1}是以a1 1 1为首项， 为公比的等比数列．

2

n 1

则an 1 (a1 1) ( )n 1 ( )n 1，∴······································· 6分 an ( ) 1， ·

1

111

222

（Ⅱ）由（Ⅰ）得，bn [( )n 1 1] n2 ( )n 1 n2，

2

2

11

由题意得b2n 1 b2n，则有 ( )2n 2 (2n 1)2 ( )2n 1 (2n)2，

2

2

11

即 ( )

2

1

2n 2

[1 ( )] (2n 1) (2n)

2

n

*

1

22

，∴

(4n 1) 4

6

n

n

， ······························· 10分

(4 1) 4

6

2，

而

(4n 1) 4

6

对于n N时单调递减，则

(4n 1) 4

6

的最大值为

故 2．·········································································································· 12分 20．解析（Ⅰ）将(1,1)

与代入椭圆C的方程，

1 1

1, 3 a2b2

得 解得a2 3，b2

2 3 3 1,

22

4b 2a

．