2024年高考指導(dǎo)數(shù)學(xué)(人教A版理科第一輪復(fù)習(xí))單元質(zhì)檢卷6　數(shù)列

單元質(zhì)檢卷六數(shù)列(時(shí)間:100分鐘滿(mǎn)分:130分)一、選擇題:本題共12小題,每小題5分,共60分.在每小題給出的四個(gè)選項(xiàng)中,只有一項(xiàng)符合題目要求.1.已知{an}為等差數(shù)列,首項(xiàng)a1=2,公差d=3,若an+an+2=28,則n=()A.1 B.2 C.3 D.42.已知等比數(shù)列{an}的各項(xiàng)均為正數(shù),若log2a3+log2a9=4,則log2a6=()A.±1 B.±2C.2 D.43.在數(shù)列{an}中,a1=2,an+1=11-an,則a2021A.-2 B.-1C.2 D.14.已知數(shù)列{an}是等差數(shù)列,其前n項(xiàng)和為Sn,有下列四個(gè)命題:甲:a18=0;乙:S35=0;丙:a17-a19=0;丁:S19-S16=0.如果只有一個(gè)是假命題,則該命題是()A.甲 B.乙C.丙 D.丁5.設(shè){an}是等比數(shù)列,前n項(xiàng)和為Sn,若S2S2+S4A.15 B.1C.13 D.6.已知數(shù)列{an}的前n項(xiàng)和為Sn,且a1=1,Sn=an+1-3,若Sk≥125,k∈N*,則k的最小值為()A.5 B.6C.7 D.87.已知數(shù)列{an}的首項(xiàng)為14,數(shù)列{bn}為等比數(shù)列,且bn=an+1an,若b1b20=2,則aA.64 B.128 C.256 D.5128.在等差數(shù)列{an}和等比數(shù)列{bn}中,有a1=b1=t>0,且a2n+1=b2n+1,則下列關(guān)系式中正確的是()A.an+1<bn+1 B.an+1≥bn+1C.an+1=bn+1 D.an+1>bn+19.已知數(shù)列{an}的前n項(xiàng)和為Sn,且滿(mǎn)足Sn=4n2+n.若數(shù)列{bn}滿(mǎn)足bn=an+34,則1b1b2+A.5052020 C.2019210.意大利著名數(shù)學(xué)家斐波那契在研究兔子的繁殖問(wèn)題時(shí),發(fā)現(xiàn)有這樣的一列數(shù):1,1,2,3,5,8,…,該數(shù)列的特點(diǎn)是前兩個(gè)數(shù)均為1,從第三個(gè)數(shù)起,每一個(gè)數(shù)都等于它前面兩個(gè)數(shù)的和.人們把這樣的一列數(shù)所組成的數(shù)列{an}稱(chēng)為“斐波那契數(shù)列”,數(shù)列{an}的前n項(xiàng)和為Sn,則下列結(jié)論錯(cuò)誤的是()A.S8=54B.a1+a3+a5+a7+…+a2019=a2020C.a2+a4+a6+a8+…+a2020=a2021D.S2020+S2019-S2018-S2017=a202211.有這樣一道題目:“戴氏善屠,日益功倍.初日屠五兩,今三十日屠訖,問(wèn)共屠幾何?”其意思為:“有一個(gè)姓戴的人善于屠肉,每一天屠完的肉是前一天的2倍,第一天屠了5兩肉,共屠了30天,問(wèn)一共屠了多少兩肉?”在這個(gè)問(wèn)題中,該屠夫前5天所屠肉的總兩數(shù)為()A.35 B.75C.155 D.31512.已知數(shù)列{an}與{bn}滿(mǎn)足bn+1an+bnan+1=(-3)n+1,bn=2,n為偶數(shù),1,n為奇數(shù),A.a3-a1=8B.a4-a2=18C.{a2n+2-a2n}是等差數(shù)列D.{a2n+1-a2n-1}是等比數(shù)列二、填空題:本題共4小題,每小題5分,共20分.13.各項(xiàng)均為正數(shù)的等比數(shù)列{an},其公比q≠1,且a3a7=4,請(qǐng)寫(xiě)出一個(gè)符合條件的通項(xiàng)公式an=.

14.已知數(shù)列{an}的前n項(xiàng)和為Sn,且Sn+2an=n,則an=.

15.《九章算術(shù)》中的“兩鼠穿墻題”是我國(guó)數(shù)學(xué)的古典名題:“今有垣厚若干尺,兩鼠對(duì)穿,大鼠日一尺,小鼠也日一尺,大鼠日自倍,小鼠日自半.問(wèn)何日相逢,各穿幾何?”大意是:有兩只老鼠從墻的兩邊打洞穿墻,大老鼠第一天進(jìn)一尺,以后每天加倍,小老鼠第一天也進(jìn)一尺,以后每天減半.如果墻足夠厚,Sn為前n天兩只老鼠打洞長(zhǎng)度之和,則S3=.

16.已知數(shù)列{an}滿(mǎn)足a1=2,an+1=2(n+2)n+1an三、解答題:共50分.解答應(yīng)寫(xiě)出文字說(shuō)明、證明過(guò)程或演算步驟.17.(12分)記Sn為數(shù)列{an}的前n項(xiàng)和,bn為數(shù)列{Sn}的前n項(xiàng)積.已知2Sn+(1)證明:數(shù)列{bn}是等差數(shù)列;(2)求{an}的通項(xiàng)公式.18.(12分)已知數(shù)列{an}滿(mǎn)足a1=1,an+1=a(1)記bn=a2n,寫(xiě)出b1,b2,并求數(shù)列{bn}的通項(xiàng)公式;(2)求{an}的前20項(xiàng)和.19.(12分)數(shù)列{an}滿(mǎn)足an+1=an,a1=12,n∈N(1)證明:0<an(2)若數(shù)列{bn}滿(mǎn)足bn=an+1an?anan+1,設(shè)數(shù)列{20.(14分)已知公比q>1的等比數(shù)列{an}和等差數(shù)列{bn}滿(mǎn)足:a1=2,b1=1,a2=b4,且a2是b2和b8的等比中項(xiàng).(1)求數(shù)列{an}與{bn}的通項(xiàng)公式;(2)記數(shù)列{anbn}的前n項(xiàng)和為T(mén)n,若當(dāng)n∈N*時(shí),等式(-1)nλ-Tn<0恒成立,求實(shí)數(shù)λ的取值范圍.答案：?jiǎn)卧|(zhì)檢卷六數(shù)列1.D因?yàn)閧an}為等差數(shù)列,a1=2,公差d=3,所以an=2+3(n-1)=3n-1,an+an+2=3n-1+3n+5=28,解得n=4.故選D.2.C由題意得a3>0,a6>0,a9>0,a3a9=a6所以log2a3+log2a9=log2(a3a9)=log2a62=2log2a6=4,則log2a6=3.B由a1=2,an+1=11-an知,a2=-1,a3=12,a4=2,a∴{an}是周期為3的周期數(shù)列,而2021=3×673+2,∴a2021=a2=-1.4.C設(shè)等差數(shù)列{an}的公差為d,若S35=0,則S35=35(a1+a35)若a17-a19=0,所以-2d=0,即d=0;若S19-S16=a17+a18+a19=0,所以a18=0.又因?yàn)橹挥幸粋€(gè)是假命題,所以丙是假命題.5.B設(shè)等比數(shù)列{an}的公比為q,由S2S2+S4=15,可得S4=4S2,整理得a3所以(a1+a2)q2=3(a1+a2),解得q2=3,所以a6.BS1=a1=1,Sn=an+1-3=Sn+1-Sn-3,則Sn+1+3=2(Sn+3),S1+3=4,所以{Sn+3}是等比數(shù)列,首項(xiàng)為4,公比為2,所以Sn+3=4×2n-1=2n+1,Sn=2n+1-3,由Sk=2k+1-3≥125,得k≥6.所以k的最小值為6.7.C由bn=an+1an,得an+1=anbn,所以a2=14b1,a3=a2b2=14b1b2,a4=14b1b2b3,…,a21=14b1b2b3…b20=14(b18.B設(shè)等比數(shù)列的公比為q,則b2n+1=b1q2n=tq2n>0,故a2n+1>0,因?yàn)閧an}為等差數(shù)列,故a1+a2n+1=2an+1=t+a2n+1,因?yàn)閧bn}為等比數(shù)列,故b1b2n+1=bn+12,故b1b結(jié)合題設(shè)條件有ta2n+1=|bn+1|,由基本不等式可得t+a2n+1≥2ta2n+1=2|bn+1|,當(dāng)且僅當(dāng)故2an+1≥2|bn+1|,而|bn+1|≥bn+1,故an+1≥bn+1,故選B.9.DSn=4n2+n,當(dāng)n≥2時(shí),Sn-1=4(n-1)2+n-1=4n2-7n+3,則an=Sn-Sn-1=8n-3(n≥2),當(dāng)n=1時(shí),a1=S1=5,適合上式,所以an=8n-3,所以bn=an+34=故1bn1b1b2+1b2b3+…+1b2020b202110.C斐波那契數(shù)列滿(mǎn)足遞推關(guān)系an+2=an+1+an(n∈N*),對(duì)于選項(xiàng)A,斐波那契數(shù)列的前8項(xiàng)為1,1,2,3,5,8,13,21,∴S8=1+1+2+3+5+8+13+21=54,A正確;對(duì)于選項(xiàng)B,a1=a2,a3=a4-a2,a5=a6-a4,…,a2019=a2020-a2018,各式相加得a1+a3+a5+a7+…+a2019=a2020,B正確;對(duì)于選項(xiàng)C,a2=a3-a1,a4=a5-a3,a6=a7-a5,…,a2020=a2021-a2019,各式相加得a2+a4+a6+a8+…+a2020=a2021-a1=a2021-1,C錯(cuò)誤;對(duì)于選項(xiàng)D,S2020+S2019-S2018-S2017=(S2020-S2018)+(S2019-S2017)=(a2020+a2019)+(a2019+a2018)=a2021+a2020=a2022,D正確.故選C.11.C由題意可得該屠夫每天屠的肉成等比數(shù)列,記首項(xiàng)為a1,公比為q,前n項(xiàng)和為Sn,所以a1=5,q=2,因此前5天所屠肉的總兩數(shù)為a1(1-q512.D因?yàn)閿?shù)列{an}與{bn}滿(mǎn)足bn+1an+bnan+1=(-3)n+1,令n=1,得b2a1+b1a2=(-3)1+1=-2,又因?yàn)閍1=2,b1=1,b2=2,所以a2=-6,令n=2,得b3a2+b2a3=(-3)2+1=10,又因?yàn)閍2=-6,b3=1,b2=2,所以a3=8,所以a3-a1=6,故A錯(cuò)誤;令n=3,得b4a3+b3a4=(-3)3+1=-26,又因?yàn)閍3=8,b3=1,b4=2,所以a4=-42,所以a4-a2=-42+6=-36,故B錯(cuò)誤;由已知得b2n+1a2n+b2na2n+1=(-3)2n+1,b2n=2,b2n+1=1,所以a2n+2a2n+1=32n+1;b2na2n-1+b2n-1a2n=(-3)2n-1+1,b2n-1=1,b2n=2,所以2a2n-1+a2n=-32n-1+1,兩式相減得a2n+1-a2n-1=32n+32n-12=6所以{a2n+1-a2n-1}是以6為首項(xiàng),9為公比的等比數(shù)列,故D正確;由a2n+1-a2n-1=6×9n-1得a2n-1=a1+(a3-a1)+(a5-a3)+…+(a2n-1-a2n-3)=2+6×(1+9+92+…+9n-2)=2+6×1-9n由2a2n-1+a2n=254+34×9n-1+a2n=-32n-1+1,得a2n=-12×所以a2n+2-a2n=-12×9n+1-32--12×9n-32=-4所以a2n+4-a2n+2-(a2n+2-a2n)=-4×9n+1+4×9n不是常數(shù),所以{a2n+2-a2n}不是等差數(shù)列,故C錯(cuò)誤.13.2n-4(答案不唯一)因?yàn)閍3a7=a52=4,an>0,所以a5=2,又q≠1,不妨令q=2,所以an=a1qn-1=a5qn-5=2×2n-5=2n-14.1-23n當(dāng)n=1時(shí),a1+2a1=1,則a1=13,當(dāng)n≥2時(shí),Sn+2an=n,Sn-1+2an-1=n-1,兩式相減得3an-2an-1=1,即an=23an-1+1即an-1=23(an-1-1),所以數(shù)列{an-1}是首項(xiàng)為a1-1=-23,公比為23的等比數(shù)列,則an-1=-23所以an=1-23n.15.354由題意知,大老鼠每天打洞的距離是以1為首項(xiàng),2為公比的等比數(shù)列,所以大老鼠前n天打洞長(zhǎng)度之和為1-2n同理小老鼠前n天打洞長(zhǎng)度之和為1-(12)所以Sn=2n-1+2-12n-1=2n所以S3=23-123-116.10111010由題意an+1an=2(n+2)n+1,∴當(dāng)n≥2時(shí),an=a1·a2a1·a3a2·…·anan-1=2×2×32×2×43·令S=a1+a2+…+a2020,則S=2×20+3×21+4×22+…+2021×22019,可得2S=2×21+3×22+…+2020×22019+2021×22020,∴-S=2+21+22+…+22019-2021×22020=2+2×(1-22019)1-2-2021×22∴a1+a2+…+a2020=S=2020×22020,因此a17.(1)證明當(dāng)n=1時(shí),b1=S1,易得b1=3當(dāng)n≥2時(shí),bnbn-1=Sn,代入2Sn+1bn=2消去Sn,得故{bn}是以32為首項(xiàng),12(2)解易得a1=S1=b1=3由(1)可得bn=n+22,由2Sn+1b當(dāng)n≥2時(shí),an=Sn-Sn-1=n+2n+1?n+1n=-故an=318.解(1)b1=a2=a1+1=2,b2=a4=a3+1=a2+2+1=5.由bn+1=a2n+2=a2n+1+1=a2n+2+1=a2n+3,得bn+1-bn=a2n+3-a2n=3.所以bn是首項(xiàng)為2,公差為3的等差數(shù)列所以bn=2+(n-1)×3=3n-1.(2)由(1)知,數(shù)列an的奇數(shù)列與偶數(shù)列都是以3為公差的等差數(shù)列,設(shè)數(shù)列an的前n項(xiàng)和為Sn,則S20=(a1+a3+a5+…+a19)+(a2+a4+…+a20)=10+10×92×3+20+10×92×19.證明(1)由題意可知,an+12?an2∵an+1=an,a1=12,n∈N∴an>0,兩邊同時(shí)取對(duì)數(shù)得lgan+1=12lgan∴數(shù)列{lgan}是以首項(xiàng)為lg12,公比為12∴l(xiāng)gan=lg12·12n-1,∴an=12∵an=12

∴an+12?an2=a綜上所述,0<a(2)∵an∴1a由(1)知0<an+12?an2≤14,∴bn=(∴b1+b2+b3+…+bn≤141a12?1a22+1a22?1a32+1a32?∴Sn≤144-1an+12,又0<an+1<1,∴0∴Sn<320.解(1)設(shè)等差數(shù)列{bn}的公差為d,由題意得,b2b8=a2所以(1+3d)2=(1+d)(1+7d),整理可得d2-d=0,解得d=0或d=1.若d=0,則a2=b4=1,可得q=a2a1若d=1,則a2=b4=1+3d=4,可得q=a2a1=所以an=2×2n-