2025版新教材高中數(shù)學(xué)第4章數(shù)列本章達(dá)標(biāo)檢測(cè)含解析蘇教版選擇性必修第一冊(cè)

PAGEPAGE18本章達(dá)標(biāo)檢測(cè)(滿分:150分;時(shí)間:120分鐘)一、單項(xiàng)選擇題(本大題共8小題,每小題5分,共40分.在每小題給出的四個(gè)選項(xiàng)中,只有一項(xiàng)是符合題目要求的)1.已知數(shù)列{an}是等比數(shù)列,Sn為其前n項(xiàng)和,若a1+a2+a3=4,a4+a5+a6=8,則S12= ()A.40B.60C.32D.502.已知Sn是等差數(shù)列{an}的前n項(xiàng)和,若S15=45,則5a5-3a3的值為 ()A.6B.15C.34D.173.已知數(shù)列{an}的前n項(xiàng)和為Sn,且Sn=nn+1,則1a5A.56B.65C.14.已知數(shù)列{an}的前n項(xiàng)和為Sn,若Sn+2=2an(n∈N*),則S4a2=A.2B.132C.1525.已知兩個(gè)等差數(shù)列{an}與{bn}的前n項(xiàng)和分別為An和Bn,且AnBn=7n+45n+3,則使得A.2B.3C.5D.46.在△ABC中,a,b,c分別是角A,B,C的對(duì)邊,A=π3,b,a,c成等差數(shù)列,且AB·AC=9,則a= (A.23B.32C.22D.337.定義n∑i=1nui為n個(gè)正數(shù)u1,u2,u3,…,un的“歡樂數(shù)”.若已知正項(xiàng)數(shù)列{an}的前n項(xiàng)的“歡樂數(shù)”為1A.20182019B.20192020C.201920188.對(duì)于數(shù)列{an},若存在常數(shù)M,使對(duì)隨意n∈N*,都有|an|≤M成立,則稱數(shù)列{an}有界.若有數(shù)列{an}滿意a1=1,則下列條件中,能使{an}有界的是 ()A.an+an+1=1+nB.an+1-an=1-1C.anan+1=1+2nD.an+1二、多項(xiàng)選擇題(本大題共4小題,每小題5分,共20分.在每小題給出的選項(xiàng)中,有多個(gè)選項(xiàng)符合題目要求,全部選對(duì)的得5分,部分選對(duì)的得3分,有選錯(cuò)的得0分)9.若數(shù)列{an}是等比數(shù)列,則 ()A.{an+an+1}是等比數(shù)列B.{anan+1}是等比數(shù)列C.{an2+D.{Sn}不是等比數(shù)列(Sn為數(shù)列{an}的前n項(xiàng)和)10.已知數(shù)列{an}不是常數(shù)列,其前n項(xiàng)和為Sn,則下列選項(xiàng)正確的是 ()A.若數(shù)列{an}為等差數(shù)列,Sn>0恒成立,則{an}為遞增數(shù)列B.若數(shù)列{an}為等差數(shù)列,a1>0,S3=S10,則Sn的最大值在n=6或7時(shí)取得C.若數(shù)列{an}為等比數(shù)列,則S2021·a2021>0恒成立D.若數(shù)列{an}為等比數(shù)列,則{2a11.已知數(shù)列{an}的前n項(xiàng)和為S,a1=1,Sn+1=Sn+2an+1,數(shù)列2nanan+1的前n項(xiàng)和為Tn,nA.數(shù)列{an+1}是等差數(shù)列B.數(shù)列{an+1}是等比數(shù)列C.數(shù)列{an}的通項(xiàng)公式為an=2n-1D.Tn<112.在數(shù)列{an}中,若an2-an-12=p(n≥2,n∈N*,p為常數(shù)),則稱{an}為等方差數(shù)列.A.若{an}是等差數(shù)列,則{anB.{(-1)n}是等方差數(shù)列C.若{an}是等方差數(shù)列,則{akn}(k∈N*,k為常數(shù))也是等方差數(shù)列D.若{an}既是等方差數(shù)列,又是等差數(shù)列,則該數(shù)列為常數(shù)列三、填空題(本大題共4小題,每小題5分,共20分,將答案填在題中的橫線上)13.將數(shù)列{3n+1}中的項(xiàng)數(shù)為奇數(shù)的項(xiàng)依據(jù)從小到大排列得到數(shù)列{an},則{an}的前n項(xiàng)和為.

14.用數(shù)學(xué)歸納法證明“當(dāng)n∈N*,1+2+22+23+…+25n-1能被31整除”時(shí),從k到k+1,等式左邊需添加的項(xiàng)是.

15.無窮數(shù)列{an}滿意:只要ap=aq(p,q∈N*),必有ap+1=aq+1,則稱{an}為“和諧遞進(jìn)數(shù)列”.若{an}為“和諧遞進(jìn)數(shù)列”,且a1=1,a2=3,a4=1,a8a9=23,則a7=,S2021=.(第一個(gè)空2分,其次個(gè)空3分)

16.如圖是一個(gè)數(shù)表,第1行依次寫著從小到大的正整數(shù),然后把每行相鄰的兩個(gè)數(shù)的和寫在這兩數(shù)正中間的下方,得到下一行,數(shù)表從上到下與從左到右均有無限項(xiàng),則這個(gè)數(shù)表中的第11行的第7個(gè)數(shù)為(用詳細(xì)數(shù)字作答).

四、解答題(本大題共6小題,共70分,解答應(yīng)寫出文字說明、證明過程或演算步驟)17.(本小題滿分10分)在①Sn=2bn-1;②-4bn=bn-1(n≥2);③bn=bn-1+2(n≥2)這三個(gè)條件中任選一個(gè),補(bǔ)充在下面的橫線上.若問題中的k存在,求出k的值;若k不存在,說明理由.已知數(shù)列{an}為等比數(shù)列,a1=23,a3=a1a2,數(shù)列{bn}的首項(xiàng)b1=1,其前n項(xiàng)和為Sn,,是否存在k∈N*,使得對(duì)隨意n∈N*,anbn≤akbk恒成立?注:假如選擇多個(gè)條件分別解答,按第一個(gè)解答計(jì)分.18.(本小題滿分12分)已知等差數(shù)列{an}的前n項(xiàng)和為Sn,S4=16,a3=3a2.(1)求{an}的通項(xiàng)公式;(2)設(shè)bn=1anan+1,求{bn}的前2n

19.(本小題滿分12分)設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,且對(duì)隨意的正整數(shù)n都滿意(Sn-1)2(1)求S1,S2,S3的值,猜想Sn的表達(dá)式;(2)用數(shù)學(xué)歸納法證明(1)中猜想的Sn的表達(dá)式的正確性.

20.(本小題滿分12分)甲、乙兩超市同時(shí)開業(yè),第一年的全年銷售額均為a萬元.由于經(jīng)營(yíng)方式不同,甲超市前n年的總銷售額為(n2-n+2)a(1)求甲、乙兩超市第n年銷售額的表達(dá)式;(2)若其中某一超市的年銷售額不足另一超市的年銷售額的50%,則該超市將被另一超市收購,推斷哪個(gè)超市有可能被收購.假如有這種狀況,將會(huì)出現(xiàn)在第幾年?

21.(本小題滿分12分)已知數(shù)列{an}的前n項(xiàng)積Tn滿意條件:①1Tn是首項(xiàng)為2的等差數(shù)列;②T2-T5=(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)數(shù)列{bn}滿意bn=nn+2-an,其前n項(xiàng)和為Sn,求證:對(duì)隨意正整數(shù)n,都有0<Sn<

22.(本小題滿分12分)若無窮數(shù)列a1,a2,a3,…滿意:對(duì)隨意兩個(gè)正整數(shù)i,j(j-i≥3),ai-1+aj+1=ai+aj與ai+1+aj-1=ai+aj至少有一個(gè)成立,則稱這個(gè)數(shù)列為“和諧數(shù)列”.(1)求證:若數(shù)列{an}為等差數(shù)列,則{an}為“和諧數(shù)列”;(2)求證:若數(shù)列{an}為“和諧數(shù)列”,則數(shù)列{an}從第3項(xiàng)起為等差數(shù)列;(3)若{an}是各項(xiàng)均為整數(shù)的“和諧數(shù)列”,滿意a1=0,且存在p∈N*,使得ap=p,a1+a2+a3+…+ap=-p,求p的全部可能的值.

答案全解全析一、單項(xiàng)選擇題1.B由等比數(shù)列的性質(zhì)可知,數(shù)列S3,S6-S3,S9-S6,S12-S9是等比數(shù)列,即數(shù)列4,8,S9-S6,S12-S9是等比數(shù)列,因此S12=4+8+16+32=60,故選B.2.A設(shè)等差數(shù)列{an}的公差為d.∵S15=45,∴15(a1+∴a8=3,∴5a5-3a3=5×(a1+4d)-3×(a1+2d)=2(a1+7d)=2a8=6.故選A.3.Dan=Sn-Sn-1=1n(n+1)(n≥2),又a1=S1=12,符合上式,∴an=1∴1a5=5×6=30,故選4.CSn+2=2an(n∈N*),①當(dāng)n=1時(shí),可得a1=2,當(dāng)n≥2,n∈N*時(shí),Sn-1+2=2an-1,②①-②得an=2an-2an-1,即an=2an-1(n≥2),∴數(shù)列{an}是以2為首項(xiàng),2為公比的等比數(shù)列,∴an=2·2n-1=2n(n∈N*),∴S4a2=2×(1-5.C∵數(shù)列{an}和{bn}均為等差數(shù)列,且其前n項(xiàng)和An和Bn滿意AnBn∴A2n-1=7×(2n-=7+242n+2閱歷證知,當(dāng)n=1,2,3,5,11時(shí),anbn為整數(shù)6.B由題知A=π3,b,a,c成等差數(shù)列,則2a=b+c,由AB·AC=9得cbcosA=9,所以bc結(jié)合余弦定理得a2=b2+c2-2bccosA=(b+c)2-3bc=4a2-54,所以a=32(負(fù)值舍去),故選B.7.B設(shè)Sn為數(shù)列{an}的前n項(xiàng)和.由“歡樂數(shù)”的定義可知nSn=13n+1,即Sn當(dāng)n=1時(shí),a1=S1=4;當(dāng)n≥2且n∈N*時(shí),an=Sn-Sn-1=6n-2,閱歷證可知a1=4滿意an=6n-2,∴an=6n-2(n∈N*).∴36(an+2)(an+1+2∴數(shù)列36(an+2)(an+1+2)的前2024項(xiàng)和為1-12+12-18.D對(duì)于A選項(xiàng),假設(shè){an}有界,即存在常數(shù)M,對(duì)隨意n∈N*,都有an+1≤M,an≤M,則1+n=an+1+an≤M+M=2M.由于左邊1+n遞增到無窮大,而右邊為常數(shù),因此A選項(xiàng)錯(cuò)誤;同理,對(duì)于C選項(xiàng),an+1an=1+2n≤M2,錯(cuò)誤;對(duì)于B選項(xiàng),n=1時(shí),a1=a2,n≥2且n∈N*時(shí),an+1-an=1-1n≥12,累加可得,an-a2≥12(n-2),又a2=a1=1,所以an≥n2,n=1時(shí),符合上式,所以an≥n2(n∈N對(duì)于D選項(xiàng),a2=2,an+1an=1+1n2=n2+1n累乘可得anan-1×an-1an-2×…×a3即ana2=2n·(n-1)<2,從而an<4,D正確二、多項(xiàng)選擇題9.BCD設(shè){an}的公比為q.A.設(shè)an=(-1)n,則an+an+1=0,明顯{an+an+1}不是等比數(shù)列.B.an+2an+1anan+1=q2,C.an+22+an+12an2+anD.當(dāng)q=1時(shí),Sn=na1,{Sn}明顯不是等比數(shù)列;當(dāng)q≠1時(shí),若{Sn}是等比數(shù)列,則Sn2=Sn-1·Sn+1(n即a1(1-qn)1-所以q=1,與q≠1沖突.綜上,{Sn}不是等比數(shù)列.故選BCD.10.ABC對(duì)于A:若數(shù)列{an}為等差數(shù)列,Sn>0恒成立,則公差d>0,故{an}為遞增數(shù)列,故A正確;對(duì)于B:若數(shù)列{an}為等差數(shù)列,a1>0,設(shè)其公差為d,由S3=S10,得3a1+3×22d=10a1+10×92d,即a1=-6d,故an=(n-7)d,所以當(dāng)n≤7時(shí),an≥0,a7=0,故Sn的最大值在n=6或7時(shí)取得,故B對(duì)于C:若數(shù)列{an}為等比數(shù)列,則S2024·a2024=a1(1-q2021)1-q·a1·q2024=a12對(duì)于D:若數(shù)列{an}為等比數(shù)列,則2an=2a1·qn-1,所以2an+12an=211.BCD因?yàn)镾n+1=Sn+2an+1,所以an+1=Sn+1-Sn=2an+1,可化為an+1+1=2(an+1),由S1=a1=1,可得數(shù)列{an+1}是首項(xiàng)為2,公比為2的等比數(shù)列,則an+1=2n,即an=2n-1(n∈N*),又2nanan+1=可得Tn=1-122-1+122-1-123-1+…+故選BCD.12.BCD在選項(xiàng)A中,取an=n,則{an}是等差數(shù)列,且an2=n2,則an+12-an2=(n+1)2-n2=2n+1,不是常數(shù),所以{a在選項(xiàng)B中,an2-an-12=[(-1)n]2-[(-1)n-1]2=1-1=0,是常數(shù)在選項(xiàng)C中,由{an}是等方差數(shù)列,得an2-an-12=p,從而an2=a12+(n-1)p,所以ak(n+1)2-akn2=[a12+(kn+k-1)p]-[a12+(kn-1)p]=kp在選項(xiàng)D中,由{an}是等差數(shù)列,可設(shè)其公差為d,則an-an-1=d,又{an}是等方差數(shù)列,所以an2-an-12=p,所以an2-an-12=(an+an-1)(an-an-1)=(an+an-1)d=p①,從而(②-①得,(d+d)d=0,解得d=0,所以{an}是常數(shù)列,故D正確.故選BCD.三、填空題13.答案3n2+n解析設(shè){an}的前n項(xiàng)和為Sn.令bn=3n+1,則an=b2n-1=3(2n-1)+1=6n-2,因?yàn)閍n+1-an=6(n+1)-2-6n+2=6,所以{an}是以6為公差,a1=4為首項(xiàng)的等差數(shù)列,所以Sn=4n+n(n-1)2×6=3n2+n.14.答案25k+25k+1+25k+2+25k+3+25k+4解析依據(jù)數(shù)學(xué)歸納法,當(dāng)n=k時(shí),原式為1+2+22+23+…+25k-1;當(dāng)n=k+1時(shí),原式為1+2+22+23+…+25k-1+25k+25k+1+25k+2+25k+3+25k+4.故從k到k+1,等式左邊需添加的項(xiàng)是25k+25k+1+25k+2+25k+3+25k+4.15.答案1;2024+6743解析數(shù)列{an}滿意:只要ap=aq(p,q∈N*),必有ap+1=aq+1,由a1=1,a2=3,a4=1,得a1=a4=1,所以a2=a5=3,a3=a6,a4=a7=1,a5=a8=3,a6=a9,則a7=1.又a8a9=23,所以a9=2,即a1=1,a2=3,a3=2,a4=1,a5=3,a6=2,……所以數(shù)列{an}是以3為周期的周期數(shù)列,又2024=673×3+2,所以S2024=673×(3+3)+a1+a2=2024+6733+1+3=2024+6743.16.答案12288解析設(shè)am,n表示第m行的第n個(gè)數(shù),由題中數(shù)表可知,每一行均成等差數(shù)列,且第m行的公差為2m-1,則am,n=am,1+(n-1)·2m-1,am,1=am-1,1+am-1,2=2am-1,1+2m-2,則am,12m-am-1,12m則am,12m=12+(m-1)4所以am,n=(m+1)2m-2+(n-1)2m-1=(m+2n-1)2m-2,故a11,7=(11+14-1)×29=24×29=12288.四、解答題17.解析設(shè)等比數(shù)列{an}的公比為q,因?yàn)閍1=23,所以a3=23a所以q=a3a2故an=23n. (2選①,Sn=2bn-1,則Sn-1=2bn-1-1(n≥2),兩式相減并整理得bnbn-又b1=1,所以{bn}是首項(xiàng)為1,公比為2的等比數(shù)列,所以bn=2n-1, (5分)所以anbn=23n·2n-1=124由指數(shù)函數(shù)的性質(zhì)知,數(shù)列{anbn}單調(diào)遞增,沒有最大值,所以不存在k∈N*,使得對(duì)隨意n∈N*,anbn≤akbk恒成立. (10分)選②,由-4bn=bn-1(n≥2),b1=1知數(shù)列{bn}是首項(xiàng)為1,公比為-14的等比數(shù)列所以bn=-14n-所以anbn=23n·-14n-1=(-4)×-16n≤4×16n≤4×1所以存在k=1,使得對(duì)隨意n∈N*,anbn≤akbk恒成立. (10分)選③,由bn=bn-1+2(n≥2)可知{bn}是以2為公差的等差數(shù)列,又b1=1,所以bn=2n-1. (5分)設(shè)cn=anbn=(2n-1)23則cn+1-cn=(2n+1)23n+1-(2n-1)·23n=所以當(dāng)n≤2時(shí),cn+1>cn,當(dāng)n≥3時(shí),cn+1<cn,則c1<c2<c3>c4>c5>…,所以存在k=3,使得對(duì)隨意n∈N*,anbn≤akbk恒成立. (10分)18.解析(1)設(shè)等差數(shù)列{an}的公差為d, (1分)由題意得S4=4a解得a1=-所以an=-2+4(n-1)=4n-6. (6分)(2)由(1)得,bn=1anan=1812n所以T2n=b1+b2+b3+…+b2n-1+b2n=18(-1-1)+1-13+13-15+…+14n-5=18-1所以{bn}的前2n項(xiàng)和T2n=n2(1-19.解析(1)當(dāng)n=1時(shí),(S1-∴S1=12當(dāng)n≥2時(shí),(Sn-1)2=(Sn-S∴Sn=12∴S2=23,S3=34, (3猜想Sn=nn+1,n∈N*. (6(2)下面用數(shù)學(xué)歸納法證明:①當(dāng)n=1時(shí),S1=12,nn+1=12,猜想正確②假設(shè)n=k(k≥1,k∈N*)時(shí),猜想正確,即Sk=kk那么當(dāng)n=k+1時(shí),可得Sk+1=12-Sk=即n=k+1時(shí),猜想也成立. (10分)由①②可知,對(duì)隨意的正整數(shù)n,Sn=nn+1都成立. (1220.信息提取(1)甲、乙兩超市第一年的全年銷售額均為a萬元.(2)甲超市前n年的總銷售額為(n2-n+2)a2萬元.數(shù)學(xué)建模本題是以超市的銷售額為背景的實(shí)際問題,由于年份n為正整數(shù),因此可以轉(zhuǎn)化為數(shù)列問題來求解.已知甲超市前n年的總銷售額為(n2-n+2)a2萬元(實(shí)際問題)相當(dāng)于已知其前n項(xiàng)和Sn=(n2-n+2)a2,從而利用Sn求甲超市第n年的銷售額an(數(shù)學(xué)問題).已知乙超市第n年的銷售額比前一年的銷售額多23n-1a萬元(實(shí)際問題)相當(dāng)于已知其bn-bn-1=23n-1a,從而利用累加法求乙超市第n年的銷售額解析(1)設(shè)甲、乙兩個(gè)超市第n年全年的銷售額分別為an萬元、bn萬元,甲超市前n年的總銷售額為Sn萬元,則Sn=(n2-n+2當(dāng)n=1時(shí),S1=a1=a;當(dāng)n≥2時(shí),an=Sn-Sn-1=a2·[n2-n+2-(n-1)2+(n-1)-2]=(n-1)a. (3分經(jīng)檢驗(yàn),a1=a不滿意上式,故an=a又b1=a,n≥2時(shí),bn-bn-1=23n所以bn=b1+(b2-b1)+(b3-b2)+…+(bn-bn-1)=a+23a+232a+…+23n-1a明顯b1=a也滿意上式,所以bn=3-223n-1a(n∈(2)當(dāng)n=2時(shí),a2=a,b2=53a,則a2>12b2;當(dāng)n=3時(shí),a3=2a,b3=199a,則a3>12b3;當(dāng)n≥4時(shí),an≥3a,bn<3a,故乙超市有可能被甲超市收購當(dāng)n≥4時(shí),令12an>bn,則12(n-1)a>3即n-1>6-423所以n>7-423即n+423n-1>7,即第7年起先乙超市的年銷售額不足甲超市的年銷售額的50%,乙超市將被甲超市收購. (12分)21.解析(1)設(shè)等差數(shù)列1Tn由已知得1T1=2,所以1Tn=2+(所以1T2=2+d,1T5=2+4d因?yàn)門2-T5=16,所以12+d-1解得d=1,所以1Tn=n+1,即Tn=1n+1又Tn=a1·a2·…·an=1n所以當(dāng)n=1時(shí),a1=T1=12當(dāng)n≥2時(shí),Tn-1=a1·a2·…·an-1=1n所以an=TnTn當(dāng)n=1時(shí)也符合上式,所以an=nn+1(n∈N*). (6(2)證明:證法一:由(1)知an=nn所以bn=nn+2-因?yàn)閚+1n+2所以nn+2=nn+1×n+1n所以nn+2>所以bn=nn+2-所以Sn>0. (8分)因?yàn)閚n+2=nn所以bn=nn+2-nn+1<nn+1+n+1n所以Sn<12×12-13+13-14+…+1綜上,對(duì)隨意正整數(shù)n,都有0<Sn<14.(12分證法二:由(1)知an=nn所