千里之行，始于足下讓知識(shí)帶有溫度。第2頁/共2頁精品文檔推薦十年高考真題分類匯編(2022-2022)數(shù)學(xué)專題08數(shù)列Word版含答案解析版十年高考真題分類匯編（2022—2022）數(shù)學(xué)

專題08數(shù)列

一、挑選題

1.(2022·全國1·理T9)記Sn為等差數(shù)列{an}的前n項(xiàng)和.已知S4=0,a5=5,則()A.an=2n-5B.an=3n-10

C.Sn=2n2

-8n

D.Sn=12

n2

-2n

2.(2022·浙江·T10)設(shè)a,b∈R,數(shù)列{an}滿足a1=a,an+1=an2

+b,n∈N*

,則()

A.當(dāng)b=1

2時(shí),a10>10B.當(dāng)b=1

4時(shí),a10>10C.當(dāng)b=-2時(shí),a10>10

D.當(dāng)b=-4時(shí),a10>10

3.(2022·全國1·理T4)記Sn為等差數(shù)列{an}的前n項(xiàng)和,若3S3=S2+S4,a1=2,則a5=()A.-12B.-10C.10

D.12

4.(2022·浙江·T10)已知a1,a2,a3,a4成等比數(shù)列,且a1+a2+a3+a4=ln(a1+a2+a3).若a1>1,則()A.a1a3,a2a4D.a1>a3,a2>a4

5.(2022·北京·理T4文T5)“十二平均律”是通用的音律體系,明代朱載堉最早用數(shù)學(xué)辦法計(jì)算出半音比例,為這個(gè)理論的進(jìn)展做出了重要貢獻(xiàn).十二平均律將一個(gè)純八度音程分成十二份,依次得到十三個(gè)單音,從其次個(gè)單音起,每一個(gè)單音的頻率與它的前一個(gè)單音的頻率的比都等于√212

.若第一個(gè)單音的頻率為f,則第八個(gè)單音的頻率為()A.√23

f

B.√223

f

C.√2512

f

D.√2712

f

6.(2022·全國1·理T12)幾位高校生響應(yīng)國家的創(chuàng)業(yè)號(hào)召,開發(fā)了一款應(yīng)用軟件.為激發(fā)大家學(xué)習(xí)數(shù)學(xué)的愛好,他們推出了“解數(shù)學(xué)題獵取軟件激活碼”的活動(dòng).這款軟件的激活碼為下面數(shù)知識(shí)題的答案:已知數(shù)列1,1,2,1,2,4,1,2,4,8,1,2,4,8,16,…,其中第一項(xiàng)是20

,接下來的兩項(xiàng)是20

,21

,再接下來的三項(xiàng)是20

,21

,22

,依此類推.求滿足如下條件的最小整數(shù)N:N>100且該數(shù)列的前N項(xiàng)和為2的整數(shù)冪.那么該款軟件的激活碼是()

A.440

B.330

C.220

D.110

7.(2022·全國3·理T9)等差數(shù)列{an}的首項(xiàng)為1,公差不為0.若a2,a3,a6成等比數(shù)列,則{an}前6項(xiàng)的和為()A.-24B.-3

C.3

D.8

8.(2022·全國1·理T3)已知等差數(shù)列{an}前9項(xiàng)的和為27,a10=8,則a100=()

A.100

B.99

C.98

D.97

9.(2022·浙江·理T13)已知{an}是等差數(shù)列,公差d不為零,前n項(xiàng)和是Sn,若a3,a4,a8成等比數(shù)列,則()

A.a1d>0,dS4>0

B.a1d0,dS40

10.(2022·全國2·文T5)設(shè)Sn是等差數(shù)列{an}的前n項(xiàng)和,若a1+a3+a5=3,則S5=()

A.5

B.7

C.9

D.11

11.(2022·全國1·文T7)已知{an}是公差為1的等差數(shù)列,Sn為{an}的前n項(xiàng)和.若S8=4S4,則a10=()

A.17

2B.19

2

C.10

D.12

12.(2022·全國2·理T4)已知等比數(shù)列{an}滿足a1=3,a1+a3+a5=21,則a3+a5+a7=()

A.21

B.42

C.63

D.84

13.(2022·全國2·文T9)已知等比數(shù)列{an}滿足a1=1

4

,a3a5=4(a4-1),則a2=()

A.2

B.1

C.1

D.1

14.(2022·大綱全國·文T8)設(shè)等比數(shù)列{an}的前n項(xiàng)和為Sn.若S2=3,S4=15,則S6=()

A.31

B.32

C.63

D.64

15.(2022·全國2·文T5)等差數(shù)列{an}的公差為2,若a2,a4,a8成等比數(shù)列,則{an}的前n項(xiàng)和Sn=()

A.n(n+1)

B.n(n-1)

C.n(n+1)

2D.n(n-1)

2

16.(2022·全國2·理T3)等比數(shù)列{an}的前n項(xiàng)和為Sn.已知S3=a2+10a1,a5=9,則a1=()

A.1

3B.-1

3

C.1

9

D.-1

9

17.(2022·全國1·文T6)設(shè)首項(xiàng)為1,公比為2

3

的等比數(shù)列{an}的前n項(xiàng)和為Sn,則()

A.Sn=2an-1

B.Sn=3an-2

C.Sn=4-3an

D.Sn=3-2an

18.(2022·全國1·理T12)設(shè)△AnBnCn的三邊長(zhǎng)分離為an,bn,cn,△AnBnCn的面積為Sn,n=1,2,3,….若

b1>c1,b1+c1=2a1,an+1=an,bn+1=cn+an

2,cn+1=bn+an

2

,則()

A.{Sn}為遞減數(shù)列

B.{Sn}為遞增數(shù)列

C.{S2n-1}為遞增數(shù)列,{S2n}為遞減數(shù)列

D.{S2n-1}為遞減數(shù)列,{S2n}為遞增數(shù)列

19.(2022·全國1·理T7)設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn,若Sm-1=-2,Sm=0,Sm+1=3,則m=()A.3B.4C.5D.6

20.(2022·全國·理T5)已知{an}為等比數(shù)列,a4+a7=2,a5a6=-8,則a1+a10=()A.7B.5C.-5

D.-7

21.(2022·全國·文T12)數(shù)列{an}滿足an+1+(-1)n

an=2n-1,則{an}的前60項(xiàng)和為()A.3690B.3660C.1845D.1830

二、填空題

1.(2022·全國3·文T14)記Sn為等差數(shù)列{an}的前n項(xiàng)和.若a3=5,a7=13,則S10=.

2.(2022·全國3·理T14)記Sn為等差數(shù)列{an}的前n項(xiàng)和.若a1≠0,a2=3a1,則S

10S5

=.

3.(2022·江蘇·T8)已知數(shù)列{an}(n∈N*

)是等差數(shù)列,Sn是其前n項(xiàng)和.若a2a5+a8=0,S9=27,則S8的值是.

4.(2022·北京·理T10)設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn.若a2=-3,S5=-10,則a5=,Sn的最小值為.

5.(2022·全國1·文T14)記Sn為等比數(shù)列{an}的前n項(xiàng)和.若a1=1,S3=3

4

,則S4=.

6.(2022·全國1·理T14)記Sn為等比數(shù)列{an}的前n項(xiàng)和.若a1=13

,a42

=a6,則S5=________.

7.(2022·全國1·理T14)記Sn為數(shù)列{an}的前n項(xiàng)和.若Sn=2an+1,則S6=.8.(2022·北京·理T9)設(shè){an}是等差數(shù)列,且a1=3,a2+a5=36,則{an}的通項(xiàng)公式為.

9.(2022·上?！10)設(shè)等比數(shù)列{an}的通項(xiàng)公式為an=qn-1

(n∈N*

),前n項(xiàng)和為Sn,若limn→∞Snan+1

=1

2,則q=.

10.(2022·江蘇·T14)已知集合A={x|x=2n-1,n∈N*},B={x|x=2n,n∈N*

}.將A∪B的全部元素從小到大依次羅列構(gòu)成一個(gè)數(shù)列{an}.記Sn為數(shù)列{an}的前n項(xiàng)和,則使得Sn>12an+1成立的n的最小值為.11.(2022·全國2·理T15)等差數(shù)列{an}的前n項(xiàng)和為Sn,a3=3,S4=10,則∑

k=1n

1

Sk

=____________.

12.(2022·全國3·理T14)設(shè)等比數(shù)列{an}滿足a1+a2=-1,a1-a3=-3,則a4=.

13.(2022·江蘇·理T9文T9)等比數(shù)列{an}的各項(xiàng)均為實(shí)數(shù),其前n項(xiàng)和為Sn.已知S3=74,S6=63

4,則a8=.14.(2022·浙江·理T13文T13)設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,若S2=4,an+1=2Sn+1,n∈N*

,則a1=,S5=.15.(2022·北京·理T12)已知{an}為等差數(shù)列,Sn為其前n項(xiàng)和.若a1=6,a3+a5=0,則S6=.16.(2022·全國1·理T15)設(shè)等比數(shù)列{an}滿足a1+a3=10,a2+a4=5,則a1a2…an的最大值為.17.(2022·全國1·文T13)在數(shù)列{an}中,a1=2,an+1=2an,Sn為{an}的前n項(xiàng)和.若Sn=126,則n=.18.(2022·湖南·理T14)設(shè)Sn為等比數(shù)列{an}的前n項(xiàng)和,若a1=1,且3S1,2S2,S3成等差數(shù)列,則an=.

19.(2022·福建·文T16)若a,b是函數(shù)f(x)=x2

-px+q(p>0,q>0)的兩個(gè)不同的零點(diǎn),且a,b,-2這三個(gè)數(shù)可適當(dāng)排序后成等差數(shù)列,也可適當(dāng)排序后成等比數(shù)列,則p+q的值等于.20.(2022·江蘇·理T11)設(shè)數(shù)列{an}滿足a1=1,且an+1-an=n+1(n∈N*

).則數(shù)列{1

an

}前10項(xiàng)的和為____________.

21.(2022·全國2·理T16)設(shè)Sn是數(shù)列{an}的前n項(xiàng)和,且a1=-1,an+1=SnSn+1,則Sn=.22.(2022·廣東·理T10)在等差數(shù)列{an}中,若a3+a4+a5+a6+a7=25,則a2+a8=.

23.(2022·陜西·文T13)中位數(shù)為1010的一組數(shù)構(gòu)成等差數(shù)列,其末項(xiàng)為2015,則該數(shù)列的首項(xiàng)為.

24.(2022·江蘇·理T7)在各項(xiàng)均為正數(shù)的等比數(shù)列{an}中,若a2=1,a8=a6+2a4,則a6的值是.25.(2022·廣東·文

T13)等比數(shù)列{an}的各項(xiàng)均為正數(shù),且

a1a5=4,則

log2a1+log2a2+log2a3+log2a4+log2a5=.

26.(2022·安徽·理T12)數(shù)列{an}是等差數(shù)列,若a1+1,a3+3,a5+5構(gòu)成公比為q的等比數(shù)列,則q=.27.(2022·全國2·文T16)數(shù)列{an}滿足an+1=

1

1-an

,a8=2,則a1=____________.

28.(2022·北京·理T12)若等差數(shù)列{an}滿足a7+a8+a9>0,a7+a100,m∈N*

,q∈(1,√2m

],證實(shí):存在d∈R,使得|an-bn|≤b1對(duì)n=2,3,…,m+1均成立,并求d的取值范圍(用b1,m,q表示).

10.(2022·天津·文T18)設(shè){an}是等差數(shù)列,其前n項(xiàng)和為Sn(n∈N*

);{bn}是等比數(shù)列,公比大于0,其前n項(xiàng)和為Tn(n∈N*

).已知b1=1,b3=b2+2,b4=a3+a5,b5=a4+2a6.(1)求Sn和Tn;

(2)若Sn+(T1+T2+…+Tn)=an+4bn,求正整數(shù)n的值.

11.(2022·天津·理T18)設(shè){an}是等比數(shù)列,公比大于0,其前n項(xiàng)和為Sn(n∈N*

),{bn}是等差數(shù)列.已知a1=1,a3=a2+2,a4=b3+b5,a5=b4+2b6.(1)求{an}和{bn}的通項(xiàng)公式;

(2)設(shè)數(shù)列{Sn}的前n項(xiàng)和為Tn(n∈N*

),①求Tn;

②證實(shí)∑k=1

n

(Tk+bk+2)bk

(k+1

)(k+2)=

2n+2-2(n∈N*

).12.(2022·全國2·理T17文T17)記Sn為等差數(shù)列{an}的前n項(xiàng)和,已知a1=-7,S3=-15.(1)求{an}的通項(xiàng)公式;(2)求Sn,并求Sn的最小值.

13.(2022·全國1·文T17)已知數(shù)列{an}滿足a1=1,nan+1=2(n+1)an.設(shè)bn=a

n

n.

(1)求b1,b2,b3;

(2)推斷數(shù)列{bn}是否為等比數(shù)列,并說明理由;(3)求{an}的通項(xiàng)公式.

14.(2022·全國3·理T17文T17)等比數(shù)列{an}中,a1=1,a5=4a3.(1)求{an}的通項(xiàng)公式;

(2)記Sn為{an}的前n項(xiàng)和,若Sm=63,求m.

15.(2022·全國1·文T17)設(shè)Sn為等比數(shù)列{an}的前n項(xiàng)和,已知S2=2,S3=-6.(1)求{an}的通項(xiàng)公式;

(2)求Sn,并推斷Sn+1,Sn,Sn+2是否成等差數(shù)列.

16.(2022·全國2·文T17)已知等差數(shù)列{an}的前n項(xiàng)和為Sn,等比數(shù)列{bn}的前n項(xiàng)和為Tn,a1=-1,b1=1,a2+b2=2.

(1)若a3+b3=5,求{bn}的通項(xiàng)公式;

(2)若T3=21,求S3.

17.(2022·全國3·文T17)設(shè)數(shù)列{an}滿足a1+3a2+…+(2n-1)an=2n.

(1)求{an}的通項(xiàng)公式;

}的前n項(xiàng)和.

(2)求數(shù)列{an

2n+1

18.(2022·天津·理T18)已知{an}為等差數(shù)列,前n項(xiàng)和為Sn(n∈N*),{bn}是首項(xiàng)為2的等比數(shù)列,且公比大于0,b2+b3=12,b3=a4-2a1,S11=11b4.

(1)求{an}和{bn}的通項(xiàng)公式;

(2)求數(shù)列{a2nb2n-1}的前n項(xiàng)和(n∈N*).

19.(2022·山東·理T19)已知{xn}是各項(xiàng)均為正數(shù)的等比數(shù)列,且x1+x2=3,x3-x2=2.

(1)求數(shù)列{xn}的通項(xiàng)公式;

(2)如圖,在平面直角坐標(biāo)系xOy中,依次銜接點(diǎn)P1(x1,1),P2(x2,2)…Pn+1(xn+1,n+1)得到折線P1P2…Pn+1,求由該折線與直線y=0,x=x1,x=xn+1所圍成的區(qū)域的面積Tn.

20.(2022·山東·文T19)已知{an}是各項(xiàng)均為正數(shù)的等比數(shù)列,且a1+a2=6,a1a2=a3.

1)求數(shù)列{an}的通項(xiàng)公式;

}的前n項(xiàng)和Tn.

(2){bn}為各項(xiàng)非零的等差數(shù)列,其前n項(xiàng)和為Sn.已知S2n+1=bnbn+1,求數(shù)列{bn

an

21.(2022·天津·文T18)已知{an}為等差數(shù)列,前n項(xiàng)和為Sn(n∈N*),{bn}是首項(xiàng)為2的等比數(shù)列,且公比大于0,b2+b3=12,b3=a4-2a1,S11=11b4.

(1)求{an}和{bn}的通項(xiàng)公式;

(2)求數(shù)列{a2nbn}的前n項(xiàng)和(n∈N*).

22.(2022·全國2·理T17)Sn為等差數(shù)列{an}的前n項(xiàng)和,且a1=1,S7=28.記bn=[lgan],其中[x]表示不超過x的最大整數(shù),如[0.9]=0,[lg99]=1.

(1)求b1,b11,b101;

(2)求數(shù)列{bn}的前1000項(xiàng)和.

23.(2022·全國2·文T17)等差數(shù)列{an}中,a3+a4=4,a5+a7=6.(1)求{an}的通項(xiàng)公式;

(2)設(shè)bn=[an],求數(shù)列{bn}的前10項(xiàng)和,其中[x]表示不超過x的最大整數(shù),如[0.9]=0,[2.6]=2.24.(2022·浙江·文T17)設(shè)數(shù)列{an}的前n項(xiàng)和為Sn.已知S2=4,an+1=2Sn+1,n∈N*

.(1)求通項(xiàng)公式an;

(2)求數(shù)列{|an-n-2|}的前n項(xiàng)和.

25.(2022·北京·文T15)已知{an}是等差數(shù)列,{bn}是等比數(shù)列,且b2=3,b3=9,a1=b1,a14=b4.(1)求{an}的通項(xiàng)公式;

(2)設(shè)cn=an+bn,求數(shù)列{cn}的前n項(xiàng)和.

26.(2022·山東·理T18文T19)已知數(shù)列{an}的前n項(xiàng)和Sn=3n2

+8n,{bn}是等差數(shù)列,且an=bn+bn+1.(1)求數(shù)列{bn}的通項(xiàng)公式;(2)令cn=

(an+1)

n+1(bn+2)

n

,求數(shù)列{cn}的前n項(xiàng)和Tn.

27.(2022·天津·理T18)已知{an}是各項(xiàng)均為正數(shù)的等差數(shù)列,公差為d.對(duì)隨意的n∈N*

,bn是an和an+1的等比中項(xiàng).

(1)設(shè)cn=bn+12?bn2

,n∈N*

,求證:數(shù)列{cn}是等差數(shù)列;

(2)設(shè)a1=d,Tn=∑k=1

2n

(-1)

k

bk2

,n∈N*,求證:∑

k=1n

1Tk

0,an2

+2an=4Sn+3.

(1)求{an}的通項(xiàng)公式;

(2)設(shè)bn=1

anan+1

,求數(shù)列{bn}的前n項(xiàng)和.

36.(2022·安徽·文T18)已知數(shù)列{an}是遞增的等比數(shù)列,且a1+a4=9,a2a3=8.(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)Sn為數(shù)列{an}的前n項(xiàng)和,bn=

an+1

SnSn+1

,求數(shù)列{bn}的前n項(xiàng)和Tn.

37.(2022·天津·理T18)已知數(shù)列{an}滿足an+2=qan(q為實(shí)數(shù),且q≠1),n∈N*

,a1=1,a2=2,且a2+a3,a3+a4,a4+a5成等差數(shù)列.

(1)求q的值和{an}的通項(xiàng)公式;

(2)設(shè)bn=log2a

2na2n-1

,n∈N*

,求數(shù)列{bn}的前n項(xiàng)和.

38.(2022·山東·文T19)已知數(shù)列{an}是首項(xiàng)為正數(shù)的等差數(shù)列,數(shù)列{1an·an+1}的前n項(xiàng)和為n

2n+1.

(1)求數(shù)列{an}的通項(xiàng)公式;

(2)設(shè)bn=(an+1)·2an,求數(shù)列{bn}的前n項(xiàng)和Tn.

39.(2022·浙江·文T17)已知數(shù)列{an}和{bn}滿足a1=2,b1=1,an+1=2an(n∈N*

),b1+12b2+13b3+…+1

nbn=bn+1-1(n∈N*

).

(1)求an與bn;

(2)記數(shù)列{anbn}的前n項(xiàng)和為Tn,求Tn.

40.(2022·天津·文T18)已知{an}是各項(xiàng)均為正數(shù)的等比數(shù)列,{bn}是等差數(shù)列,且a1=b1=1,b2+b3=2a3,a5-3b2=7.

(1)求{an}和{bn}的通項(xiàng)公式;

(2)設(shè)cn=anbn,n∈N*,求數(shù)列{cn}的前n項(xiàng)和.

41.(2022·湖北·文T19)設(shè)等差數(shù)列{an}的公差為d,前n項(xiàng)和為Sn,等比數(shù)列{bn}的公比為q,已知b1=a1,b2=2,q=d,S10=100.

(1)求數(shù)列{an},{bn}的通項(xiàng)公式;

(2)當(dāng)d>1時(shí),記cn=an

bn

,求數(shù)列{cn}的前n項(xiàng)和Tn.

42.(2022·全國2·理T17)已知數(shù)列{an}滿足a1=1,an+1=3an+1.

(1)證實(shí):{an+1

2

}是等比數(shù)列,并求{an}的通項(xiàng)公式;

(2)證實(shí):1

a1+1

a2

+…+1

an

10B.當(dāng)b=1

4時(shí),a10>10C.當(dāng)b=-2時(shí),a10>10D.當(dāng)b=-4時(shí),a10>10

【答案】A

【解析】當(dāng)b=12時(shí),a2=a12+12≥12,a3=a22+12≥34,a4=a32+12≥1716≥1,當(dāng)n≥4時(shí),an+1=an2+1

2≥an2≥1,則

log1716

an+1>2log1716

an?log1716

an+1>2n-1

,則

an+1≥（17

16）

2n-1

(n≥4),則a10≥（17

16）26

=（1+116）64

=1+64

16+

64×632×1

162

+…>1+4+7>10,故選A.3.(2022·全國1·理T4)記Sn為等差數(shù)列{an}的前n項(xiàng)和,若3S3=S2+S4,a1=2,則a5=()A.-12B.-10C.10D.12

【答案】B

【解析】由于3S3=S2+S4,所以3S3=(S3-a3)+(S3+a4),即S3=a4-a3.設(shè)公差為d,則3a1+3d=d,又由a1=2,得d=-3,所以a5=a1+4d=-10.

4.(2022·浙江·T10)已知a1,a2,a3,a4成等比數(shù)列,且a1+a2+a3+a4=ln(a1+a2+a3).若a1>1,則()A.a1a3,a2a4D.a1>a3,a2>a4【答案】B

【解析】設(shè)等比數(shù)列的公比為q,則a1+a2+a3+a4=

a1(1-q4)1-q,a1+a2+a3=a1(1-q3)

1-q

.∵a1+a2+a3+a4=ln(a1+a2+a3),

∴a1+a2+a3=ea1+a2+a3+a4,即a1(1+q+q2

)=ea1(1+q+q

2+q3)

.

又a1>1,∴q1,即q+q2

>0,解得q0舍去).由a1>1,可知a1(1+q+q2

)>1,∴a1(1+q+q2

+q3

)>0,即1+q+q2

+q3

>0,

即(1+q)+q2

(1+q)>0,即(1+q)(1+q2

)>0,這與qa3,a25.(2022·北京·理T4文T5)“十二平均律”是通用的音律體系,明代朱載堉最早用數(shù)學(xué)辦法計(jì)算出半音比例,為這個(gè)理論的進(jìn)展做出了重要貢獻(xiàn).十二平均律將一個(gè)純八度音程分成十二份,依次得到十三個(gè)單音,從其次個(gè)單音起,每一個(gè)單音的頻率與它的前一個(gè)單音的頻率的比都等于√212

.若第一個(gè)單音的頻率為f,則第八個(gè)單音的頻率為()A.√23

fB.√223

f

C.√2512

f

D.√2712

f

【答案】D

【解析】設(shè)第n個(gè)單音的頻率為an,由題意,an

an-1

=√212

(n≥2),所以{an}為等比數(shù)列,由于a1=f,所以

a8=a1×(√212)7

=√2712

f,故選D.

6.(2022·全國1·理T12)幾位高校生響應(yīng)國家的創(chuàng)業(yè)號(hào)召,開發(fā)了一款應(yīng)用軟件.為激發(fā)大家學(xué)習(xí)數(shù)學(xué)的愛好,他們推出了“解數(shù)學(xué)題獵取軟件激活碼”的活動(dòng).這款軟件的激活碼為下面數(shù)知識(shí)題的答案:已知數(shù)列

1,1,2,1,2,4,1,2,4,8,1,2,4,8,16,…,其中第一項(xiàng)是20,接下來的兩項(xiàng)是20,21,再接下來的三項(xiàng)是20,21,22

,依此類推.求滿足如下條件的最小整數(shù)N:N>100且該數(shù)列的前N項(xiàng)和為2的整數(shù)冪.那么該款軟件的激活碼是()

A.440

B.330

C.220

D.110【答案】A

【解析】設(shè)數(shù)列的首項(xiàng)為第1組,接下來兩項(xiàng)為第2組,再接下來三項(xiàng)為第3組,以此類推,設(shè)第n組的項(xiàng)數(shù)為n,則前n組的項(xiàng)數(shù)和為n(1+n)

2

.第n組的和為1-2n

1-2=2n

-1,前n組總共的和為2(1-2n)

1-2-n=2n+1

-2-n.

由題意,N>100,令

n(1+n)

2

>100,得n≥14且n∈N*

,即N浮現(xiàn)在第13組之后.若要使最小整數(shù)N滿足:N>100且

前N項(xiàng)和為2的整數(shù)冪,則SN-Sn(1+n)2

應(yīng)與-2-n互為相反數(shù),即2k

-1=2+n(k∈N*

,n≥14),所以k=log2(n+3),解

得n=29,k=5.所以N=

29×(1+29)

2

+5=440,故選A.7.(2022·全國3·理T9)等差數(shù)列{an}的首項(xiàng)為1,公差不為0.若a2,a3,a6成等比數(shù)列,則{an}前6項(xiàng)的和為()A.-24B.-3C.3D.8

【答案】A

【解析】設(shè)等差數(shù)列的公差為d,則d≠0,a32=a2·a6,即(1+2d)2

=(1+d)(1+5d),解得d=-2,

所以S6=6×1+6×5

2×(-2)=-24,故選A.

8.(2022·全國1·理T3)已知等差數(shù)列{an}前9項(xiàng)的和為27,a10=8,則a100=()A.100B.99C.98D.97

【答案】C【解析】由于S9=

(a1+a9)×9

2

=27,a1+a9=2a5,所以a5=3.又由于a10=8,所以d=a10-a

5

10-5

=1.故a100=a10+(100-10)×1=98.

9.(2022·浙江·理T13)已知{an}是等差數(shù)列,公差d不為零,前n項(xiàng)和是Sn,若a3,a4,a8成等比數(shù)列,則()A.a1d>0,dS4>0

B.a1d0,dS40

【答案】B

【解析】設(shè){an}的首項(xiàng)為a1,公差為d,則a3=a1+2d,a4=a1+3d,a8=a1+7d.∵a3,a4,a8成等比數(shù)列,

∴(a1+3d)2

=(a1+2d)(a1+7d),即3a1d+5d2

=0.∵d≠0,∴a1d=-5

3

d2

c1,b1+c1=2a1,an+1=an,bn+1=cn+a

n2,cn+1=bn+a

n2,則()

A.{Sn}為遞減數(shù)列

B.{Sn}為遞增數(shù)列

C.{S2n-1}為遞增數(shù)列,{S2n}為遞減數(shù)列

D.{S2n-1}為遞減數(shù)列,{S2n}為遞增數(shù)列【答案】B

【解析】由于b1>c1,不妨設(shè)b1=4a13,c1=2a13,p=12(a1+b1+c1)=3

2

a1,則S1=√

3a12·a12·a16·5a

16

=

√1512a1

2

;a2=a1,b2=2

3a1+a1

2

=

5

6a1,c2=4

3a1+a12

=7

6

a1,

S2=√

3a1·a1·2a1·a

1=

√6a1

2

;明顯S2>S1.

同理,a3=a1,b3=7

6a1+a1

2

=

13

12

a1,c3=

5

6a1+a1

2

=

1112a1,S3=√3a12·a12·512a1·712

a1=

√10524a1

2

,明顯S3>S2.

19.(2022·全國1·理T7)設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn,若Sm-1=-2,Sm=0,Sm+1=3,則m=()A.3B.4C.5D.6【答案】C

【解析】∵Sm-1=-2,Sm=0,Sm+1=3,∴am=Sm-Sm-1=2,am+1=Sm+1-Sm=3.∴d=am+1-am=3-2=1.∵Sm=

m(a1+am)

2

=

m(a1+2)

2

=0,∴a1=-2,am=-2+(m-1)×1=2.∴m=5.

20.(2022·全國·理T5)已知{an}為等比數(shù)列,a4+a7=2,a5a6=-8,則a1+a10=()A.7B.5C.-5D.-7

【答案】D

【解析】∵{an}為等比數(shù)列,∴a5a6=a4a7=-8.聯(lián)立{a4+a7=2,a4a7=-8可解得{a4=4,a7=-2或{a4=-2,a7=4,

當(dāng){a4=4,a7=-2

時(shí),q3

=-12,故a1+a10=a

4

3+a7q3

=-7;

當(dāng){a4=-2,a7=4

時(shí),q3=-2,同理,有a1+a10=-7.

21.(2022·全國·文T12)數(shù)列{an}滿足an+1+(-1)n

an=2n-1,則{an}的前60項(xiàng)和為()A.3690B.3660C.1845D.1830

【答案】D

【解析】∵an+1+(-1)n

an=2n-1,∴

a2=1+a1,a3=2-a1,a4=7-a1,a5=a1,a6=9+a1,a7=2-a1,a8=15-a1,a9=a1,a10=17+a1,a11=2-a1,a12=23-a1,…,a57=a1,a58=113+a1,a59=2-a1,a60=119-a1,

∴a1+a2+…+a60=(a1+a2+a3+a4)+(a5+a6+a7+a8)+…+(a57+a58+a59+a60)=10+26+42+…+234=15×(10+234)

=1830.二、填空題

1.(2022·全國3·文T14)記Sn為等差數(shù)列{an}的前n項(xiàng)和.若a3=5,a7=13,則S10=.【答案】100

【解析】設(shè)等差數(shù)列{an}的公差為d,則{a3=a1+2d=5,a7=a1+6d=13,解得{a1=1,d=2.故S10=10a1+10×9

2d=10×1+10×9

2×2=100.

2.(2022·全國3·理T14)記Sn為等差數(shù)列{an}的前n項(xiàng)和.若a1≠0,a2=3a1,則S10

S5

=.

【答案】4

【解析】設(shè)等差數(shù)列{an}的公差為d.∵a1≠0,a2=3a1,∴a1+d=3a1,即d=2a1.

∴S10

S5

=

10a1+10×9

2d

5a1+5×42d

=

100a1

25a1

=4.3.(2022·江蘇·T8)已知數(shù)列{an}(n∈N*

)是等差數(shù)列,Sn是其前n項(xiàng)和.若a2a5+a8=0,S9=27,則S8的值是.【答案】16

【解析】∵{an}為等差數(shù)列,設(shè)公差為d,a2a5+a8=0,S9=27,∴

{(a1+d)(a1+4d)+a1+7d=0,①9a1+9×82

d=27,②

收拾②得a1+4d=3,即a1=3-4d,③

把③代入①解得d=2,∴a1=-5.∴S8=8a1+28d=16.

4.(2022·北京·理T10)設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn.若a2=-3,S5=-10,則a5=,Sn的最小值為.【答案】0-10

【解析】等差數(shù)列{an}中,由S5=5a3=-10,得a3=-2,又a2=-3,公差d=a3-a2=1,a5=a3+2d=0,由等差數(shù)列{an}的性質(zhì)得當(dāng)n≤5時(shí),an≤0,當(dāng)n≥6時(shí),an大于0,所以Sn的最小值為S4或S5,即為-10.

5.(2022·全國1·文T14)記Sn為等比數(shù)列{an}的前n項(xiàng)和.若a1=1,S3=3

,則S4=.【答案】5

【解析】設(shè)等比數(shù)列{an}的公比為q.S3=a1+a1q+a1q2

=1+q+q2

=3

4,即q2

+q+14=0.解得q=-12

.

故S4=a1(1-q4)1-q

=

1-(-12)4

1+12=5

8.

6.(2022·全國1·理T14)記Sn為等比數(shù)列{an}的前n項(xiàng)和.若a1=13

,a42

=a6,則S5=________.

【答案】121

3

【解析】設(shè)等比數(shù)列{an}的公比為q,則a4=a1q3

=1

3

q3,a6=a1q5

=13

q5

.

∵a42=a6,∴19q6

=13

q5

.∵q≠0,∴q=3.

∴S5=a1(1-q5)1-q

=

13(1-35)

1-3

=121

3.

7.(2022·全國1·理T14)記Sn為數(shù)列{an}的前n項(xiàng)和.若Sn=2an+1,則S6=.【答案】-63

【解析】∵Sn=2an+1,①∴Sn-1=2an-1+1(n≥2).②

①-②,得an=2an-2an-1,即an=2an-1(n≥2).

又S1=2a1+1,∴a1=-1.∴{an}是以-1為首項(xiàng),2為公比的等比數(shù)列,則

S6=-1(1-26)

1-2=-63.

8.(2022·北京·理T9)設(shè){an}是等差數(shù)列,且a1=3,a2+a5=36,則{an}的通項(xiàng)公式為.【答案】an=6n-3

【解析】∵{an}為等差數(shù)列,設(shè)公差為d,∴a2+a5=2a1+5d=36.

∵a1=3,∴d=6.∴an=3+(n-1)×6=6n-3.

9.(2022·上?！10)設(shè)等比數(shù)列{an}的通項(xiàng)公式為an=qn-1

(n∈N*

),前n項(xiàng)和為Sn,若limn→∞Snan+1

=1

2,則

q=.【答案】3

【解析】由an=qn-1

,得an+1=qn

.當(dāng)q=1時(shí),不滿足題意;當(dāng)q≠1時(shí),Sn=a1(1-qn)1-q

=

1-qn

1-q

.若01,

則limn→∞Sn

an+1

=lim

n→∞1-qn

(1-q)qn

=limn→∞

1(1-q)·(1qn-1)=-11-q=1

2,解得q=3.10.(2022·江蘇·T14)已知集合A={x|x=2n-1,n∈N*

},B={x|x=2n

,n∈N*

}.將A∪B的全部元素從小到大依次羅列構(gòu)成一個(gè)數(shù)列{an}.記Sn為數(shù)列{an}的前n項(xiàng)和,則使得Sn>12an+1成立的n的最小值為.【答案】27

【解析】①若an+1=2k

(k∈N*

),則Sn=21

+22

+…+2k-1

+1+3+(2)

-1=2k

-2+(2k-1)2

?(2k-1)2

+2k

-2>12·2k

.令2k

=t?1

4t2

+t-2>12t?t(t-44)>8.∴t≥64?k≥6.此時(shí),n=k-1+2k-1

=37.②若an+1=2k+1(k∈N*

),

則Sn=21

+22

+(2)

+1+3+…+2k-1(2t

12(2k+1)?2t+1

>-k2

+24k+14.∴-k2

+24k+1412.

取k=21,此時(shí)772

0,q>0)的兩個(gè)不同的零點(diǎn),且a,b,-2這三個(gè)數(shù)可適當(dāng)排序后成等差數(shù)列,也可適當(dāng)排序后成等比數(shù)列,則p+q的值等于.【答案】9

【解析】由題意,得{a+b=p>0,ab=q>0,∴{a>0,

b>0.

不妨設(shè)a0,∴a3=2.

∴a1a2a3a4a5=(a1a5)·(a2·a4)·a3=25

.

∴l(xiāng)og2a1+log2a2+log2a3+log2a4+log2a5=log2(a1a2a3a4a5)=log225

=5.

26.(2022·安徽·理T12)數(shù)列{an}是等差數(shù)列,若a1+1,a3+3,a5+5構(gòu)成公比為q的等比數(shù)列,則q=.【答案】1

【解析】設(shè)數(shù)列{an}的公差為d,則a1=a3-2d,a5=a3+2d,

由題意得,(a1+1)(a5+5)=(a3+3)2

,即(a3-2d+1)(a3+2d+5)=(a3+3)2

,收拾,得(d+1)2

=0,∴d=-1,則a1+1=a3+3,故q=1.

27.(2022·全國2·文T16)數(shù)列{an}滿足an+1=1

1-an

,a8=2,則a1=____________.

【答案】12

【解析】將a8=2代入an+1=

1

1-an

,可求得a7=12

;

將a7=1

2代入an+1=1

1-an

,可求得a6=-1;

將a6=-1代入an+1=1

1-an

,可求得a5=2.

由此可知數(shù)列{an}是一個(gè)周期數(shù)列,且周期為3,所以a1=a7=1

2

.

28.(2022·北京·理T12)若等差數(shù)列{an}滿足a7+a8+a9>0,a7+a100,即a8>0;而a7+a10=a8+a9203時(shí),f'(n)>0,00.由于ck≤bk≤ck+1,

所以qk-1

≤k≤qk

,其中k=1,2,3,…,m.當(dāng)k=1時(shí),有q≥1;當(dāng)k=2,3,…,m時(shí),有

lnkk≤lnq≤lnk

k-1

.設(shè)f(x)=lnx

x(x>1),則f'(x)=1-lnx

x2.令f'(x)=0,得x=e.列表如下:

由于ln22=ln8

60,取bn=an,可得bn+1-bn=an+1-an=d>0,

則b2-b1,b3-b2,…,b201-b200中有200個(gè)正數(shù),符合題意;②若d=0,取bn=a1-1n

,

則|bn-an|=|a1-1

n-a1|=1

n0,則b2-b1,b3-b2,…,b201-b200中有200個(gè)正數(shù),符合題意;

③若-20,則b2-b1,b3-b2,…,b201-b200中恰有100個(gè)正數(shù),符合題意;④若d≤-2,假設(shè)存在數(shù)列{bn}滿足:{bn}與{an}臨近,則為an-1≤bn≤an+1,an+1-1≤bn+1≤an+1+1,可得bn+1-bn≤an+1+1-(an-1)=2+d≤0,

b2-b1,b3-b2,…,b201-b200中沒有正數(shù),與已知沖突.故d≤-2不符合題意.

綜上可得,d的取值范圍是(-2,+∞).

9.(2022·江蘇·T20)設(shè){an}是首項(xiàng)為a1,公差為d的等差數(shù)列,{bn}是首項(xiàng)為b1,公比為q的等比數(shù)列.(1)設(shè)a1=0,b1=1,q=2,若|an-bn|≤b1對(duì)n=1,2,3,4均成立,求d的取值范圍;

(2)若a1=b1>0,m∈N*

,q∈(1,√2m

],證實(shí):存在d∈R,使得|an-bn|≤b1對(duì)n=2,3,…,m+1均成立,并求d的取值范圍(用b1,m,q表示).

【解析】(1)由條件知,an=(n-1)d,bn=2n-1

.由于|an-bn|≤b1對(duì)n=1,2,3,4均成立,即|(n-1)d-2n-1

|≤1對(duì)n=1,2,3,4均成立,即1≤1,1≤d≤3,3≤2d≤5,7≤3d≤9,得73≤d≤5

2.因此,d的取值范圍為[73,5

2].(2)由條件知,an=b1+(n-1)d,bn=b1qn-1

.

若存在d,使得|an-bn|≤b1(n=2,3,…,m+1)成立,即|b1+(n-1)d-b1qn-1

|≤b1(n=2,3,…,m+1),即當(dāng)

n=2,3,…,m+1時(shí),d滿足qn-1-2n-1b1≤d≤qn-1

n-1b1.由于

q∈(1,√2m

],則10,對(duì)

n=2,3,…,m+1均成立.

因此,取d=0時(shí),|an-bn|≤b1對(duì)n=2,3,…,m+1均成立.下面研究數(shù)列{

qn-1-2n-1}的最大值和數(shù)列{qn-1

n-1}的最小值(n=2,3,…,m+1).①當(dāng)2≤n≤m時(shí),qn-2n?qn-1-2

n-1

=

nqn-qn-nqn-1+2

n(n-1)

=

n(qn-qn-1)-qn+2

n(n-1)

,

當(dāng)

10.

因此,當(dāng)2≤n≤m+1時(shí),數(shù)列{qn-1-2

n-1}單調(diào)遞增,故數(shù)列{qn-1-2n-1}的最大值為qm-2

m.

②設(shè)f(x)=2x

(1-x),當(dāng)x>0時(shí),f'(x)=(ln2-1-xln2)2x

0,可得q=2,故bn=2n-1

.所

以,Tn=1-2n1-2

=2n

-1.

設(shè)等差數(shù)列{an}的公差為d.由b4=a3+a5,可得a1+3d=4.由b5=a4+2a6,可得3a1+13d=16,從而a1=1,d=1,故an=n.所以,Sn=

n(n+1)

2

.(2)由(1),有T1+T2+…+Tn=(21

+22

+(2)

)-n=2×(1-2n)1-2

-n=2n+1

-n-2.

由Sn+(T1+T2+…+Tn)=an+4bn可得,

n(n+1)2

+2n+1-n-2=n+2n+1

,收拾得n2

-3n-4=0,解得n=-1(舍),或n=4.所以,n的值為4.

11.(2022·天津·理T18)設(shè){an}是等比數(shù)列,公比大于0,其前n項(xiàng)和為Sn(n∈N*

),{bn}是等差數(shù)列.已知

a1=1,a3=a2+2,a4=

b3+b5,a5=b4+2b6.(1)求{an}和{bn}的通項(xiàng)公式;

(2)設(shè)數(shù)列{Sn}的前n項(xiàng)和為Tn(n∈N*

),①求Tn;

②證實(shí)∑k=1

n

(Tk+bk+2)bk

(k+1

)(k+2)=

2n+2n+2

-2(n∈N*

).【解析】(1)解設(shè)等比數(shù)列{an}的公比為q.由a1=1,a3=a2+2,可得q2

-q-2=0.由于q>0,可得q=2,故an=2n-1

.設(shè)等差數(shù)列{bn}的公差為d.由a4=b3+b5,可得b1+3d=4.由a5=b4+2b6,可得3b1+13d=16,從而b1=1,d=1,故bn=n.所以,數(shù)列{an}的通項(xiàng)公式為an=2n-1

,數(shù)列{bn}的通項(xiàng)公式為bn=n.(2)①解由(1),有

Sn=1-2n1-2

=2n

-1,故Tn=∑k=1

n

(2k

-1)=∑k=1

n

2k

-n=

2×(1-2n)1-2

-n=2n+1

-n-2.②證實(shí)由于(Tk+bk+2)b

k

(k+1)(k+2)=(2k+1-k-2+k+2)k(k+1)(k+2)=

k·2k+1

(k+1)(k+2)

=

2k+2k+2

?2k+1

k+1,所以,∑k=1n

(Tk+bk+2)bk

(k+1)(k+2)

=

(233-22

2

)+(244-233)+…+(2n+2n+2-2n+1n+1)

=

2n+2

n+2

-2.12.(2022·全國2·理T17文T17)記Sn為等差數(shù)列{an}的前n項(xiàng)和,已知a1=-7,S3=-15.(1)求{an}的通項(xiàng)公式;(2)求Sn,并求Sn的最小值.

【解析】(1)設(shè){an}的公差為d,由題意得3a1+3d=-15.由a1=-7得d=2.

所以{an}的通項(xiàng)公式為an=2n-9.(2)由(1)得Sn=n2

-8n=(n-4)2

-16.

所以當(dāng)n=4時(shí),Sn取得最小值,最小值為-16.

13.(2022·全國1·文T17)已知數(shù)列{an}滿足a1=1,nan+1=2(n+1)an.設(shè)bn=a

n

n.

(1)求b1,b2,b3;

(2)推斷數(shù)列{bn}是否為等比數(shù)列,并說明理由;(3)求{an}的通項(xiàng)公式.【解析】(1)由條件可得an+1=

2(n+1)

n

an.將n=1代入得,a2=4a1,而a1=1,所以a2=4.將n=2代入得,a3=3a2,所以a3=12.

從而b1=1,b2=2,b3=4.

(2){bn}是首項(xiàng)為1,公比為2的等比數(shù)列.由條件可得

an+1n+1=

2an

n

,即bn+1=2bn,又b1=1,所以{bn}是首項(xiàng)為1,公比為2的等比數(shù)列.

(3)由(2)可得ann

=2n-1

,所以an=n·2n-1

.

14.(2022·全國3·理T17文T17)等比數(shù)列{an}中,a1=1,a5=4a3.(1)求{an}的通項(xiàng)公式;

(2)記Sn為{an}的前n項(xiàng)和,若Sm=63,求m.【解析】(1)設(shè){an}的公比為q,由題設(shè)得an=qn-1

.由已知得q4

=4q2

,解得q=0(舍去),q=-2或q=2.故an=(-2)n-1

或an=2n-1

.(2)若an=(-2)n-1,則

Sn=

1-(-2)n

3

.

由Sm=63得(-2)m=-188,此方程沒有正整數(shù)解.若an=2n-1

,則Sn=2n

-1.由Sm=63得2m

=64,解得m=6.綜上,m=6.

15.(2022·全國1·文T17)設(shè)Sn為等比數(shù)列{an}的前n項(xiàng)和,已知S2=2,S3=-6.(1)求{an}的通項(xiàng)公式;

(2)求Sn,并推斷Sn+1,Sn,Sn+2是否成等差數(shù)列.【解析】(1)設(shè){an}的公比為q.由題設(shè)可得{a1(1+q)=2,

a1(1+q+q2)=-6.

解得q=-2,a1=-2.

故{an}的通項(xiàng)公式為an=(-2)n

.(2)由(1)可得Sn=a1(1-qn)1-q=-23+(-1)n2n+1

3

.

因?yàn)?/p>

Sn+2+Sn+1=-4+(-1)n2

n+3-2n+2=2[-2+(-1)n2n+1

]=2Sn,故Sn+1,Sn,Sn+2成等差數(shù)列.

16.(2022·全國2·文T17)已知等差數(shù)列{an}的前n項(xiàng)和為Sn,等比數(shù)列{bn}的前n項(xiàng)和為Tn,a1=-1,b1=1,a2+b2=2.

(1)若a3+b3=5,求{bn}的通項(xiàng)公式;(2)若T3=21,求S3.

【解析】設(shè){an}的公差為d,{bn}的公比為q,

則an=-1+(n-1)d,bn=qn-1

.由a2+b2=2得d+q=3.①(1)由a3+b3=5,得2d+q2

=6.②

聯(lián)立①和②解得{d=3,q=0(舍去),{d=1,

q=2.

因此{(lán)bn}的通項(xiàng)公式為bn=2n-1

.(2)由b1=1,T3=21得q2

+q-20=0,解得q=-5或q=4.

當(dāng)q=-5時(shí),由①得d=8,則S3=21.當(dāng)q=4時(shí),由①得d=-1,則S3=-6.

17.(2022·全國3·文T17)設(shè)數(shù)列{an}滿足a1+3a2+…+(2n-1)an=2n.(1)求{an}的通項(xiàng)公式;(2)求數(shù)列{

an

2n+1

}的前n項(xiàng)和.

【解析】(1)由于a1+3a2+…+(2n-1)an=2n,故當(dāng)n≥2時(shí),a1+3a2+…+(2n-3)an-1=2(n-1).兩式相減得(2n-1)an=2.所以an=2

2n-1(n≥2).又由題設(shè)可得a1=2,從而{an}的通項(xiàng)公式為an=2

2n-1

.(2)記{

an

2n+1}的前n項(xiàng)和為Sn.

由(1)知

an2n+1=

2

(2n+1)(2n-1)=

12n-1?1

2n+1

.則Sn=11

?13

+13

?15

+…+

12n-1?

1

2n+1

=

2n

2n+1

.18.(2022·天津·理T18)已知{an}為等差數(shù)列,前n項(xiàng)和為Sn(n∈N*

),{bn}是首項(xiàng)為2的等比數(shù)列,且公比大于0,b2+b3=12,b3=a4-2a1,S11=11b4.(1)求{an}和{bn}的通項(xiàng)公式;

(2)求數(shù)列{a2nb2n-1}的前n項(xiàng)和(n∈N*

).

【解析】(1)設(shè)等差數(shù)列{an}的公差為d,等比數(shù)列{bn}的公比為q.由已知b2+b3=12,得b1(q+q2

)=12,而b1=2,所以q2

+q-6=0.又由于q>0,解得q=2.所以,bn=2n

.由b3=a4-2a1,可得3d-a1=8.①由S11=11b4,可得a1+5d=16,②

聯(lián)立①②,解得a1=1,d=3,由此可得an=3n-2.

所以,數(shù)列{an}的通項(xiàng)公式為an=3n-2,數(shù)列{bn}的通項(xiàng)公式為bn=2n

.

(2)設(shè)數(shù)列{a2nb2n-1}的前n項(xiàng)和為Tn,由a2n=6n-2,b2n-1=2×4n-1

,有a2nb2n-1=(3n-1)×4n

,故Tn=2×4+5×42

+8×43

+…+(3n-1)×4n

,

4Tn=2×42

+5×43

+8×44

+…+(3n-4)×4n

+(3n-1)×4n+1

,上述兩式相減,得-3Tn=2×4+3×42

+3×43

+…+3×4n

-(3n-1)×4n+1

=12×(1-4n)1-4-4-(3n-1)×4n+1

=-(3n-2)×4n+1

-8.得Tn=

3n-23×4n+1+8

3

.所以,數(shù)列{a2nb2n-1}的前n項(xiàng)和為

3n-23×4n+1+8

3

.19.(2022·山東·理T19)已知{xn}是各項(xiàng)均為正數(shù)的等比數(shù)列,且x1+x2=3,x3-x2=2.(1)求數(shù)列{xn}的通項(xiàng)公式;

(2)如圖,在平面直角坐標(biāo)系xOy中,依次銜接點(diǎn)P1(x1,1),P2(x2,2)…Pn+1(xn+1,n+1)得到折線P1P2…Pn+1,求由該折線與直線y=0,x=x1,x=xn+1所圍成的區(qū)域的面積Tn.

【解析】(1)設(shè)數(shù)列{xn}的公比為q,由已知q>0.由題意得{x1+x1q=3,x1q2

-x1q=2.所以3q2

-5q-2=0.由于q>0,所以q=2,x1=1,

因此數(shù)列{xn}的通項(xiàng)公式為xn=2n-1

.

(2)過P1,P2,…,Pn+1向x軸作垂線,垂足分離為Q1,Q2,…,Qn+1.由(1)得xn+1-xn=2n

-2n-1

=2n-1

,記梯形PnPn+1Qn+1Qn的面積為bn,由題意bn=

(n+n+1)2

×2n-1=(2n+1)×2n-2

,所以Tn=b1+b2+…+bn

=3×2-1

+5×20

+7×21

+…+(2n-1)×2n-3

+(2n+1)×2n-2

.

①

又2Tn=3×20

+5×21

+7×22

+…+(2n-1)×2n-2

+(2n+1)×2n-1

,②

①-②得

-Tn=3×2-1+(2+22+…+2n-1)-(2n+1)×2n-1

=32+2(1-2n-1)1-2

-(2n+1)×2n-1

.所以Tn=

(2n-1)×2n+1

2

.20.(2022·山東·文T19)已知{an}是各項(xiàng)均為正數(shù)的等比數(shù)列,且a1+a2=6,a1a