最新人教A版高中数学必修5第二章测评试卷及答案

第二章测评
(时间:120分钟满分:150分)
一、选择题(每小题5分,共60分)
1.已知数列{a n}是等差数列,a1=2,其公差d≠0.若a5是a3和a8的等比中项,则S18=()
A.398
B.388
C.189
D.199
a52=a3·a8,公差d≠0,a1=2,∴(a1+4d)2=(a1+2d)·(a1+7d),代入数据可得
d=189.故选C.
(2+4d)2=(2+2d)·(2+7d),解得d=1,∴S18=18a1+18×17
2
2.已知数列{b n}是等比数列,b9是1和3的等差中项,则b2b16=()
A.16
B.8
C.4
D.2
b9是1和3的等差中项,所以2b9=1+3,即b9=2.由等比数列{b n}的性质可得b2b16=b92=4.
3.已知在递减的等差数列{a n}中,a3=-1,a1,a4,-a6成等比数列,若S n为数列{a n}的前n项和,则S7的值为
() A.-14 B.-9
C.-5
D.-1
{a n}的公差为d,由已知得a3=a1+2d=-1,a42=a1·(-a6),即(a1+3d)2=a1·(-a1-5d),且{a n}为递减
d=7-21=-14.
数列,则d=-1,a1=1.故S7=7a1+7×6
2
4.等差数列{a n}中,S16>0,S17<0,当其前n项和取得最大值时,n=()
A.8
B.9
C.16
D.17
,S16>0,即a1+a16=a8+a9>0,S17<0,即a1+a17=2a9<0,所以a9<0,a8>0,所以等差数列{a n}为递减数列,且前8项为正数,从第9项以后为负数,所以当其前n项和取得最大值时,n=8.故选A.
5.(2020·全国Ⅱ高考,文6)记S n为等比数列{a n}的前n项和.若a5-a3=12,a6-a4=24,则S n
=()
a n
A.2n-1
B.2-21-n
C.2-2n-1
D.21-n-1
{a n}的公比为q.
∵a5-a3=12,a6-a4=24,∴a6-a4
=q=2.
a5-a3
又a 5-a 3=a 1q 4-a 1q 2=12a 1=12,∴a 1=1.
∴a n =a 1·q n-1=2n-1,
S n =a 1
(1-q n )1-q =
1×(1-2n )1-2
=2n
-1. ∴S n
a n
=
2n -12
n -1=2-
1
2
n -1=2-2
1-n
.
故选B .
6.已知数列{a n }满足a n +a n+1=1
2(n ∈N *),a 2=2,S n 是数列{a n }的前n 项和,则S 21为( ) A.5 B.72
C.92
D.132
a n +a n+1=1
2,a 2=2,∴a n ={-3
2,n 为奇数,2,n 为偶数.
∴S 21=11×(-32)+10×2=7
2.故选B .
7.我国古代数学巨著《九章算术》中,有如下问题:“今有女子善织,日自倍,五日织五尺,问日织几何?”这个问题用今天的白话叙述为:“有一位善于织布的女子,每天织的布都是前一天的2倍,已知她5天共织布5尺,问这位女子每天分别织布多少?”根据上面的已知条件,可求得该女子第4天所织布的尺数为( ) A .815
B .1615
C .2031
D .4031
n 天织的布为a n 尺,且数列{a n }为公比q=2的等比数列,
由题意可得a 1(1-25)1-2=5,解得a 1=5
31.
所以该女子第4天所织布的尺数为a 4=a 1q 3=40
31. 故选D .
8.在各项都为正数且不相等的等比数列{a n }中,S n 为其前n 项和,若a m ·a 2m+2=a 72
=642(m ∈N *),且a m =8,
则S 2m =( ) A.127 B.255 C.511
D.1 023
{a n }的公比为q ,则a 1q m-1·a 1q 2m+1=(a 1q 6)2.因为等比数列{a n }的各项都为正数且不相
等,所以m-1+2m+1=12,解得m=4,故a 4=8.又因为a 72=642,所以a 7=64,q 3=a
7a 4
=8,解得q=2,所以
a 1=a 4
2
3=1.故
S 2m =S 8=1-28
1-2
=255.
9.已知在各项均为正数的数列{a n }中,a 1=1,a 2=2,2a n 2
=a n -12+a n+12(n ≥2),b n =1
a n +a
n+1
,记数列{b n }的前n 项和为S n ,若S n =3,则n 的值是( ) A.99
B.33
C.48
D.9
2a n 2
=a n -12+a n+12(n ≥2),
∴数列{a n 2}是首项为1,公差为22-1=3的等差数列,∴a n 2
=1+3(n-1)=3n-2.
又a n >0,∴a n =√3n -2,
∴b n =1
a
n +a n+1
=
√3n -2+√3n+1
=1
3
·(√3n +1−√3n -2), 故数列{b n }的前n 项和S n =1
3[(√4−√1)+(√7−√4)+…+(√3n +1−√3n -2)]=1
3·(√3n +1-1).
由S n =1
3(√3n +1-1)=3,解得n=33.故选B 10.已知数列{a n }满足a 1+3a 2+32a 3+…+3n-1a n =n
3(n ∈N *),则a n =( ) A.1
3n B.
1
3n -1
C.13n
D.
1
3
n+1
a 1+3a 2+32a 3+…+3n-1a n =n 3
,①
a 1+3a 2+32a 3+…+3n-2a n-1=n -1
3
(n ≥2),② ①-②,得3n-1a n =n
3
−n -13
=13
(n ≥2),∴a n =13
n (n ≥2).由①得a 1=13
,经验证也满足上式,∴a n =13
n (n ∈
N *).故选C .
11.对于正项数列{a n },定义:G n =
a 1+2a 2+3a 3+…+na n
n
为数列{a n }的“匀称值”.已知数列{a n }的“匀称值”为G n =n+2,则该数列中的a 10等于( ) A .8
3
B .12
5
C .9
4
D .21
10
G n=a1+2a2+3a3+…+na n
,G n=n+2,∴n·G n=n·(n+2)=a1+2a2+3a3+…+na n,∴
n
.故10×(10+2)=a1+2a2+3a3+…+10a10;9×(9+2)=a1+2a2+3a3+…+9a9,两式相减得10·a10=21,∴a10=21
10
选D.
12.在数列{a n}中,a1=1,a2=2,且a n+2-a n=1+(-1)n(n∈N*),则S100=()
A.0
B.1 300
C.2 600
D.2 602
a n+2-a n=1+(-1)n(n∈N*),当n=1时,得a3-a1=0,即a3=a1;当n=2时,得a4-a2=2.由此可得,当n为
+a2=n.
奇数时,a n=a1;当n为偶数时,a n=2×n-2
2
所以S100=a1+a2+…+a100
=(a1+a3+…+a99)+(a2+a4+…+a100)
=50a1+(2+4+ (100)
=2 600.
=50+50×(100+2)
2
二、填空题(每小题5分,共20分)
13.若数列{a n}的前n项和S n=n2-8n,n=1,2,3,…,则满足a n>0的n的最小值为.
,当n=1时,a1=S1=-7,当n≥2时,a n=S n-S n-1=2n-9.而a1=2×1-9=-7.
综上,a n=2n-9.
,又因为n∈N*.
由2n-9>0,得n>9
2
故满足a n>0的n的最小值为5.
14.已知在公差不为零的正项等差数列{a n}中,S n为其前n项和,lg a1,lg a2,lg a4也成等差数列.若a5=10,则S5=.
{a n}的公差为d,则d>0.
由lg a1,lg a2,lg a4成等差数列,
得2lg a2=lg a1+lg a4,则a22=a1a4,
即(a1+d)2=a1(a1+3d),d2=a1d.
因为d>0,所以d=a1,a5=5a1=10,
解得d=a1=2.故S5=5a1+5×4
×d=30.
2
15.若等差数列{a n}的前n项和为S n,且a2=0,S5=10,数列{b n}满足b1=0,且b n+1=a n+1+b n,则数列{b n}的通项公式为.
{a n }的公差为d ,则{
a 1+d =0,5a 1+10d =10,解得{a 1=-2,
d =2.
于是a n =-2+2(n-1)=2n-4.因此a n+1=2n-2.于
是b n+1-b n =2n-2,b n =b 1+(b 2-b 1)+(b 3-b 2)+…+(b n -b n-1)=0+0+2+…+(2n-4)=n 2-3n+2,故数列{b n }的通项公式为b n =n 2-3n+2.
n =n 2-3n+2
16.(2020·全国Ⅰ高考,文16)数列{a n }满足a n+2+(-1)n a n =3n-1,前16项和为540,则a 1= .
n 为偶数时,有a n+2+a n =3n-1,
则(a 2+a 4)+(a 6+a 8)+(a 10+a 12)+(a 14+a 16)=5+17+29+41=92, 因为前16项和为540,所以a 1+a 3+a 5+a 7+a 9+a 11+a 13+a 15=448.
当n 为奇数时,有a n+2-a n =3n-1,由累加法得a n+2-a 1=3(1+3+5+…+n )-1+n
2=3
4n 2+n+1
4,所以a n+2=3
4
n 2+n+14
+a 1,
所以a 1+3
4×12+1+1
4+a 1+3
4×32+3+1
4+a 1+3
4×52+5+1
4+a 1+3
4×72+7+1
4+a 1+
34×92+9+14
+a 1+34×112+11+14+a 1+34×132+13+1
4+a 1=448,解得a 1=7.
三、解答题(共6小题,共70分)
17.(本小题满分10分)已知数列{a n }是等差数列,前n 项和为S n ,且满足a 2+a 7=23,S 7=10a 3. (1)求数列{a n }的通项公式;
(2)若a 2,a k ,a k+5(k ∈N *)构成等比数列,求k 的值.
设等差数列{a n }的公差是d.
根据题意有{a 1+d +a 1+6d =23,
7a 1+7×6
2d =10(a 1+2d ), 解得{a 1=1,d =3.
所以数列{a n }的通项公式为a n =3n-2. (2)由(1)得a 2=4,a k =3k-2,a k+5=3(k+5)-2, 由于a 2,a k ,a k+5(k ∈N *)构成等比数列, 所以(3k-2)2=4[3(k+5)-2],
整理得3k 2-8k-16=0,解得k=4(舍去k =-4
3). 故k=4.
18.(本小题满分12分)已知各项均为正数的等比数列{a n }的前n 项和为S n ,且2a 2=S 2+1
2,a 3=2. (1)求数列{a n }的通项公式;
(2)若b n =log 2a n +3,数列1b n b n+1的前n 项和为T n ,求满足T n >1
3的正整数n 的最小值.
由题意知,2a 2=S 2+12
,∴2a 2=a 1+a 2+12
,得a 2=a 1+12
.
设等比数列{a n }的公比为q ,
∵a 3=2,∴2q =2q 2+1
2,化简得q 2-4q+4=0,解得q=2, ∴a n =a 3·q n-3=2·2n-3=2n-2.
(2)由(1)知,b n =log 2a n +3=log 22n-2+3=n-2+3=n+1,
∴1
b n b n+1
=1
(n+1)(n+2)=1
n+1−1
n+2, ∴T n =1
b
1b 2
+
1b 2b 3+…+1b n b n+1=12−13+13−14+…+
1n+1−1n+2
=12−
1n+2
=
n
2(n+2)
. 令T n >1
3,得n
2(n+2)>1
3,解得n>4,
∴满足T n >1
3的正整数n 的最小值是5.
19.(本小题满分12分)已知数列{a n }满足2a n+1
=
1a n
+
1a n+2
(n ∈N *),且a 3=15
,a 2=3a 5.
(1)求{a n }的通项公式;
(2)若b n =3a n a n+1(n ∈N *),求数列{b n }的前n 项和S n .
由
2a n+1
=
1a n
+
1a n+2(n ∈N *)可知数列{1a n
}为等差数列.
由已知得1a 3
=5,
1a 2
=13·
1
a 5
, 设其公差为d ,则1
a 1
+2d=5,1
a 1
+d=1
3(1
a 1
+4d),
解得1a 1
=1,d=2,于是1
a n
=1+2(n-1)=2n-1,
整理得a n =1
2n -1.
(2)由(1)得b n =3a n a n+1=3
(2n -1)(2n+1)=3
2(1
2n -1-1
2n+1), 所以S n =3
2(1-1
3+1
3−1
5+…+1
2n -1−1
2n+1)=3n
2n+1. 20.(本小题满分12分)已知数列{a n }的前n 项和S n =2a n -2n . (1)求a 1,a 2.
(2)设c n =a n+1-2a n ,证明数列{c n }是等比数列.
(3)求数列{
n+1
2c n
}的前n 项和T n .
a 1=S 1,2a 1=S 1+2,∴a 1=S 1=2.
由2a n =S n +2n ,知2a n+1=S n+1+2n+1=a n+1+S n +2n+1,∴a n+1=S n +2n+1,①
∴a 2=S 1+22=2+22=6.
①式知a n+1-2a n =(S n +2n+1)-(S n +2n )=2n+1-2n =2n ,即c n =2n ,∴c
n+1c n
=2(常数). ∵c 1=21=2,
∴{c n }是首项为2,公比为2的等比数列.
c n =2n ,∴n+12c n
=
n+12n+1
.
∴数列{n+12c n
}的前n 项和T n =
2
2
2+
32
3+
42
4+…+
n+12n+1,12T n =2
23
+
32
4+…+
n 2
n+1+
n+12n+2
,
两式相减,得12T n =
2
2
2
+
12
3
+
12
4
+
12
5+…+1
2
n+1
−
n+12
n+2
=
12+123×(1-1
2n -1)1-12
−n+12n+2=34−
12
n+1
−
n+12n+2
=
34−n+32n+2.∴T n =32−n+3
2
n+1. 21.(本小题满分12分)已知数列{a n }的前n 项和S n =a n +12n 2+3
2n-2(n ∈N *). (1)求数列{a n }的通项公式; (2)若
b n ={1
(a n -1)(a n +1),n 为奇数,
4·(12
)a n
,n 为偶数,
且数列{b n }的前n 项和为
T n ,求T 2n .
由于S n =a n +12
n 2+32
n-2,所以当n ≥2时,S n-1=a n-1+12
(n-1)2+32
(n-1)-2,两式相减得a n =a n -a n-1+n+1,于是a n-1=n+1,
所以a n =n+2. (2)由(1)得
b n ={1
(n+1)(n+3)
,n 为奇数,(1
2)n ,n 为偶数,
所以T 2n =b 1+b 2+b 3+…+b 2n =(b 1+b 3+…+b 2n-1)+(b 2+b 4+…+b 2n ).
因为b 1+b 3+…+b 2n-1=12×4+14×6+16×8+…+12n×(2n+2)=14[11×2+12×3+…+1n×(n+1)]=1
4(1-12+12-13+…+1n -1n+1
)=n 4(n+1)
,b 2+b 4+…+b 2n =(12)2
+(14)4+…+(12)
2n =14[1-(14
)n ]1-14
=
13
[1-(14)n
],于是T 2n =n
4(n+1)+1
3[1-(14)n
].
22.(本小题满分12分)已知数列{a n }满足3(n+1)a n =na n+1(n ∈N *),且a 1=3. (1)求数列{a n }的通项公式; (2)求数列{a n }的前n 项和; (3)若
a n
b n
=
2n+3n+1,求证:5
6
≤
1
b 1
+
1b 2+…+1
b n
<1.
3(n+1)a n =na n+1,
所以a
n+1
a n
=
3(n+1)
n
(n ∈N *), 则a
2a 1=3×21,
a 3a 2=3×32,a 4a 3=3×43,……a n a n -1=3×n n -1,累乘可得a
n a 1
=3n-1×n. 又因为a 1=3,所以a n =n×3n (n ∈N *).
{a n }的前n 项和为S n ,则S n =1×3+2×32+3×33+…+(n-1)×3n-1+n×3n ,①
3S n =1×32+2×33+3×34+…+(n-1)×3n +n×3n+1,② ①-②,可得
-2S n =3+32+33+…+3n -n×3n+1
=3(1-3n )
1-3-n×3n+1
=3
2(3n -1)-n×3n+1 =(1
2-n)×3n+1-3
2
. 所以S n =(n 2-1
4)×3n+1+3
4.
因为a
n b n
=2n+3n+1, 所以1
b n
=2n+3
n+1×1
n×3n =2n+3
n (n+1)×1
3n
=3(n+1)-n
n (n+1)×1
3n =(3
n -1
n+1)×1
3n =1
n ×1
3n -1−1
n+1×1
3n , 则
1
b 1+
1b 2+…+1b n
=(1×
13
-1
2
×
13
1)
+(12
×
13
1
-13
×
13
2)+…+(1
n
×
13
n -1
-
1n+1×13n )=1-1n+1×13
n .
因为n ∈N *,所以0<1
n+1×13n
≤1
6,
即5
6≤1-1
n+1×
1
3n <1, 于是5
6≤1
b 1+1
b 2
+…+1
b n <1.。