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所以b1+
b2
2+
b3
3+… +
bn-1
n-1=an(n≥2),
两式相减得:
bn
n =an+1-an =2,n≥2,
所以bn =2n,n≥2,
又b1 =6不满足上式,所以bn =
6, n=1,
2n,n≥2
{ .
(2)当 n≥ 2时, 1anbn
= 1
(2n+2)(2n) =
1
4n(n+1)= (14 1n- 1n+ )1 ,
所以Sn =
1
a1b1
+ 1a2b2
+ 1a3b3
+… + 1anbn
= 14×6+ [14 12-( )13 (+ 13- )14
+… (+ 1n- 1n+ ) ]1
= 124+
1
4
1
2-
1
n+( )1
= 124+
n-1
8(n+1)=
2n-1
12(n+1),
又当n=1时,S1=
1
a1b1
= 14×6=
1
24满足上式,
所以数列 1
anb{ }n 的前n项和Sn=
2n-1
12(n+1)(n∈N+).
21.解:(1)当n=1时,a1+S1 =a1+a1 =2,
所以a1 =1.
因为Sn =2-an,即an+Sn =2,
所以an+1+Sn+1 =2.
两式相减得an+1-an+Sn+1-Sn =0,即an+1-an
+an+1 =0,故有2an+1=an,由Sn =2-an,知an≠0,
所以
an+1
an
= 12(n∈N+),
所以{an}是首项为1,公比为
1
2的等比数列,其通
项公式为an =( )12
n-1
,
因为bn+1 =bn+an(n=1,2,3,…),
所以bn+1-bn =( )12
n-1
,
所以bn-bn-1+bn-1-bn-2+… +b2-b1
=bn-b1
=( )12
n-2
+( )12
n-3
+… +( )12
0
=
1-( )12
n-1
1-12
=2-( )12
n-2
(n=2,3,…),
又b1 =1,所以bn =3-( )12
n-2
(n=2,3,…),
当n=1时也满足上式,
所以bn =3-( )12
n-2
(n∈N+).
(2)因为cn =n(3-bn)=2n( )12
n-1
,
所以Tn= [2 ( )12
0
+2×( )12
1
+3×( )12
2
+…
+(n-1)×( )12
n-2
+n×( )12
n- ]1 . ①
1
2Tn = [2 ( )12
1
+2×( )12
2
+3×( )12
3
+…
+(n-1)×( )12
n-1
+n×( )12 ]
n
. ②
① -②得,12Tn = [2 ( )12
0
+( )12
1
+( )12
2
+
… +( )12
n- ]1 -2×n×( )12
n
(n∈N+),
Tn =4×
1-( )12
n
1-12
-4×n×( )12
n
=8-(8+4n)×1
2n
(n∈N+).
22.解:(1)因为a2n+1 =6Sn+9n+1,
所以a2n =6Sn-1+9(n-1)+1,
所以a2n+1-a
2
n =6an+9(n≥2),
所以a2n+1 =(an+3)
2,
而数列{an}各项均为正数,
所以an+1 =an+3(n≥2),
又a2 =4,a
2
2 =4
2 =6a1+9+1,
所以a1 =1,所以a2-a1 =3,
所以{an}为公差为3的等差数列,故an=3n-2.
因为b1 =1,b3 =4,所以bn =2
n-1.
(2)由(1)得cn =(3n-2)·2
n-1,
①Tn =1·2
0+4·21+… +(3n-2)·2n-1,
2Tn =1·2
1+4·22+… +(3n-2)·2n,
所以 -Tn=1+3(2
1+22+… +2n-1)-(3n-2)
·2n =1+6(2n-1-1)-(3n-2)·2n,
所以Tn =(3n-5)·2
n+5.
②由题得(3n-5)·2n·m≥6n2-31n+35恒成立,
即 m≥ 6n
2-31n+35
(3n-5)·2n
=(3n-5)(2n-7)
(3n-5)·2n
=
2n-7
2n
恒成立,设kn =
2n-7
2n
,
则kn+1-kn =
2n-5
2n+1
-2n-7
2n
=9-2n
2n+1
,
当n≤4时,kn+1 >kn,n≥5时,kn+1 <kn,
又k4 =
1
24
<k5 =
3
25
,
所以(kn)max=k5 =
3
32,所以m≥
3
32.
故实数m [的取值范围是 332,+ )∞ .
高考数学信息优化卷(六)
概率与统计参考答案与解题提示
一、选择题
1~6 BBBDDC 7~12 DBBBCD
提示:
1.从黄、白、蓝、红4种颜色中任意选2种颜色的所
有基本事件有{黄白},{黄蓝},{黄红},{白蓝},{白
红},{蓝红},共6种.
其中包含白色