數(shù)列-高考數(shù)學(xué)解答題專項(xiàng)練習(xí)

數(shù)列二、解答題型綜合訓(xùn)練1.已知數(shù)列{an}為等比數(shù)列,其前n項(xiàng)和為Sn,且an+1-an=2·3n.(1)求數(shù)列{an}的公比q和a4的值;(2)求證:-a1,Sn,an+1成等差數(shù)列.(本小題滿分10分)1.解:(1)由題可得a2-a1=6,a所以a2-a1=6又a2-a1=a1q-a1=6,所以a1=3,所以an=3n,所以a4=34=81. 5分(2)證明:因?yàn)閍n=3n,所以an+1-Sn=3n+1-3(1-3n)1-3=3Sn-(-a1)=3(1-3n)1-3+3=3n+1+32,所以Sn-(-a1)=an+1-Sn,所以2.已知Sn是數(shù)列{an}的前n項(xiàng)和,a1=1,.

(1)求an;(2)設(shè)bn=an+an+1(an·an+1①?n∈N*,an+an+1=4n;②數(shù)列Snn為等差數(shù)列,且Snn的前3項(xiàng)和為6.從以上兩個(gè)條件中任選一個(gè)補(bǔ)充在橫線處,解:(1)選擇條件①:由?n∈N*,an+an+1=4n,得an+1+an+2=4(n+1), 1分所以an+2-an=4(n+1)-4n=4,即數(shù)列{a2k-1},{a2k}(k∈N*)均是公差為4的等差數(shù)列, 2分所以a2k-1=a1+4(k-1)=4k-3=2(2k-1)-1, 3分又a1+a2=4,所以a2=3, 4分則a2k=a2+4(k-1)=4k-1=2·(2k)-1, 5分所以an=2n-1. 6分選擇條件②:因?yàn)閿?shù)列Snn為等差數(shù)列,且Snn的前3項(xiàng)和為6,所以S11+S22+S所以S22=2,所以Snn的公差d'=S22-S11=2則Snn=1+(n-1)=n,故Sn=n2.當(dāng)n≥2時(shí),an=Sn-Sn-1=n2-(n-1)2=2n-1. 5分又a1=1滿足an=2n-1,所以an=2n-1. 6分(2)因?yàn)閎n=an+an+1(a所以Tn=b1+b2+…+bn=121=121-1(3.在數(shù)列{an}中,已知a1=1,且an+1an=2n+1an-2nan+1.(1)證明:數(shù)列{2nan}為等差數(shù)列,并求出數(shù)列{an(2)若bn=nn+2an,求數(shù)列{bn}的前n項(xiàng)和S解:(1)證明:由題知an≠0,∵an+1an=2n+1an-2nan+1,∴2n+1an+1-又2a1=2,∴數(shù)列{2nan}是以 4分∴2nan=2+(n-1)=n+1,∴an=2(2)∵bn=nn+2an=n·2n(n∴Sn=b1+b2+…+bn=(223-22)+(234-223)+…+(2nn+1-2n-1n4.已知各項(xiàng)均為正數(shù)的等差數(shù)列{an}的前三項(xiàng)和為12,等比數(shù)列{bn}的前三項(xiàng)和為7b1,且a1=b1,a2=b2.(1)求數(shù)列{an}和{bn}的通項(xiàng)公式;(2)設(shè)cn=an,n=2k-1,bn,n=2k,(本小題滿分10分)4.解:(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.由題意得3a1所以an=2n,bn=2n. 5分(2)由題知數(shù)列{cn}的前20項(xiàng)和S=a1+a3+…+a19+b2+b4+…+即S=2+382×10+2×(1-25.設(shè)數(shù)列{an}滿足a1=2,2(an+1-an)=3×4n.(1)求數(shù)列{an}的通項(xiàng)公式;(2)令bn=nan,求數(shù)列{bn}的前n項(xiàng)和Sn.解:(1)因?yàn)?(an+1-an)=3×4n,所以an+1-an=12×3×4n=3×22n-1, 2當(dāng)n≥2時(shí),an-an-1=3×22(n-1)-1,…,a2-a1=3×22×1-1,累加得an-a1=3×(21+23+…+22n-3)(n≥2), 4分所以an=3×2×(1-4n-1)1-4+2又a1=2滿足上式,故an=22n-1. 6分(2)由(1)得bn=nan=n·22n-1, 7分所以Sn=1×2+2×23+…+n·22n-1①,4Sn=1×23+2×25+…+n·22n+1②, 9分由①-②得-3Sn=2+23+25+…+22n-1-n·22n+1=2×(1-4n)1-解得Sn=19[(3n-1)·22n+1+2]. 126.設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,2(Sn-n+2)=an+1,a2=10,bn=an-1.(1)求證:{bn}是等比數(shù)列;(2)設(shè)cn=bn,n為奇數(shù),1log3bn·log36.解:(1)證明:由題得2Sn=an+1+2n-4,當(dāng)n=1時(shí),2a1=a2-2=8,解得a1=4;當(dāng)n≥2時(shí),由2Sn=an+1+2n-4①,得2Sn-1=an+2n-6②,由①-②得2an=an+1-an+2,即an+1-1=3(an-1),所以bn+1=3bn,又b1=a1-1=3,所以數(shù)列{bn}是首項(xiàng)為3,公比為3的等比數(shù)列.(2)由(1)可知bn=3×3n-1=3n,所以cn=3所以T2n+1=31+12×4+33+14×6+…+32n-1+12n(2n+2)+32n+1=(3+33+…+32n-1+32n+1)+12×4+14×6+…+12n7.已知數(shù)列{an}的前n項(xiàng)和為Sn,a1=1,a2=2,且Sn+2=Sn+1+4an.(1)求數(shù)列{an}的通項(xiàng)公式;(2)求證:1a1+1+1a2+1+(本小題滿分10分)7.解:(1)由Sn+2=Sn+1+4an得an+2=4an, 1分所以當(dāng)n=2k-1(k∈N*)時(shí),a2k+1=4a2k-1,所以數(shù)列{a2k-1}是首項(xiàng)為a1=1,公比為4的等比數(shù)列,所以a2k-1=a1×4k-1=1×4k-1,即a2k-1=22k-2=2(2k-1)-1. 3分當(dāng)n=2k(k∈N*)時(shí),a2k+2=4a2k,所以數(shù)列{a2k}是首項(xiàng)為a2=2,公比為4的等比數(shù)列,所以a2k=a2×4k-1=2×4k-1=22k-1.綜上,an=2n-1. 5分(2)證明:由(1)知1an+1=12n所以1a1+1+1a2+1+…+1an+1<120+121+18.已知各項(xiàng)均為正數(shù)的數(shù)列{an}滿足a1=1,an+12-(2n+1)an+1=an2+(2(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)數(shù)列1an的前n項(xiàng)和為Tn,證明:Tn<8.解:(1)由題知an+12-an2-(2n+1)(an+1+an)=0,則(an+1+an)(an+1-an-2因?yàn)閍n>0,所以an+1-an=2n+1,即an-an-1=2n-1(n≥2),所以當(dāng)n≥2時(shí),an=a1+(a2-a1)+(a3-a2)+…+(an-an-1)=1+3+5+…+(2n-1)=n2, 4分a1=1符合上式,所以an=n2. 5分(2)證明:方法一:當(dāng)n≥2時(shí),1n2<1n(n 8分則Tn=1a1+1a2+1a3+…+1an<1+1-12+12-方法二:1n2=44n2<44n2則Tn=1a1+1a2+1a3+…+1an<2× 10分9.已知數(shù)列{an}滿足2nan+1=(n+1)an,且a1=1.(1)求數(shù)列{an}的通項(xiàng)公式;(2)若bn=ann+(-1)n·n,求數(shù)列{bn}的前2n項(xiàng)和S29.解:(1)方法一:∵2nan+1=(n+1)an,a1=1,∴an≠0,∴an+1an=12∴anan-1=12·nn-1(n≥2),…,a2a1=12×21,將上述式子累乘得an∵a1=1滿足上式,∴an=n·12n-1(n∈N*方法二:∵2nan+1=(n+1)an,∴an+1n+1=12·ann,∴an+1n+1a∴ann是首項(xiàng)為1,公比為12的等比數(shù)列∴ann=12n-1,∴an(2)bn=ann+(-1)n·n,令cn=ann=12n-1,d設(shè)數(shù)列{cn}的前2n項(xiàng)和為T2n,數(shù)列{dn}的前2n項(xiàng)和為M2n,則T2n=1×1-122n1M2n=(-1)+2+(-3)+4+…+(-1)2n-1·(2n-1)+(-1)2n·2n=n,故S2n=T2n+M2n=n+2-122n10.已知數(shù)列{an}的前n項(xiàng)和為Sn,且滿足nSn+1-(n+1)Sn=n(n+1),a3=5.(1)求a1,a2的值及數(shù)列{an}的通項(xiàng)公式;(2)設(shè)bn=anan+1,求數(shù)列{bn}的前n項(xiàng)和Tn.10.解:(1)在nSn+1-(n+1)Sn=n(n+1)中,令n=1得a1+a2-2a1=2,令n=2,得2(a1+a2+a3)-3(a1+a2)=6, 1分即a2-a1由nSn+1-(n+1)Sn=n(n+1),得Sn+1n+1則數(shù)列Snn是首項(xiàng)為1,公差為1的等差數(shù)列,則Snn=n,所以Sn=n當(dāng)n≥2時(shí),an=Sn-Sn-1=n2-(n-1)2=2n-1,又a1=1滿足上式,所以an=2n-1.綜上可得,a1=1,a2=3,數(shù)列{an}的通項(xiàng)公式為an=2n-1. 5分(2)由(1)知,當(dāng)n≥2時(shí),anan+1an+2-an-1anan+1=anan+1(an+2-an-1)=6anan+1=6bn,所以bn=16(anan+1an+2-an-1anan+1), 8所以∑k=2nbk=16(anan+1an+2-a1a2a11.已知數(shù)列{an}滿足a1=2,an+an+1=4×3n.(1)求證:數(shù)列{a2n-1+1}和{a2n-1}(2)設(shè)bn=2×3n-2anan+1,數(shù)列{bn}的前n項(xiàng)和為S證明:(1)由a2n-1+a2n=4×32n-1①,得a2n+a2n+1=4×32n②,由②-①得a2n+1-a2n-1=83×9n則a3-a1=83×9,a5-a3=83×92,a7-a5=83×93,…,a2n-1-a2n-3=83×累加得a2n-1-a1=83(9+92+…+9n-1)=83×9-9n1-9=13×9n-3,故a2即a2n-1+1=13×9n,則a2因此數(shù)列{a2n-1+1}是以3為首項(xiàng),9為公比的等比數(shù)列.將a2n-1=13×9n-1代入①得a2n-1=9n故數(shù)列{a2n-1}是以9為首項(xiàng),9(2)由(1)知a2n-1=32n-1-1,a2n=32n+1,故an=3n+(-1)n,當(dāng)n為奇數(shù)時(shí),an+1-an=2×3n+2>2×3n-2;當(dāng)n為偶數(shù)時(shí),an+1-an=2×3n-2.因此對(duì)任意n∈N*,均有an+1-an≥2×3n-2,則bn=2×3n-2ana當(dāng)n=1時(shí),S1=4a1a2=當(dāng)n≥2時(shí),Sn≤4a1a2+1a2-1a3+1a3-1a4+…12．已知公比大于1的等比數(shù)列滿足．(1)求數(shù)列的通項(xiàng)公式；(2)記為數(shù)列在區(qū)間中的項(xiàng)的個(gè)數(shù)，求數(shù)列的前100項(xiàng)和．解:（1）設(shè)等邊數(shù)列的首項(xiàng)為，公比為，則，解得：，或（舍），所以，（2）為數(shù)列在區(qū)間中的項(xiàng)的個(gè)數(shù)，由題意可知，，，且當(dāng)時(shí)，，所以.13.在等比數(shù)列{an}中,a1,a2,a3分別是下表第一、二、三行中的某一個(gè)數(shù),且a1,a2,a3中的任何兩個(gè)數(shù)不在下表的同一列.第一列第二列第三列第一行323第二行465第三行9128(1)寫出a1,a2,a3,并求數(shù)列{an}的通項(xiàng)公式;(2)若數(shù)列{bn}滿足bn=an+(-1)nlog2an,求數(shù)列{bn}的前n項(xiàng)和Sn.(本小題滿分10分)13.解:(1)根據(jù)等比數(shù)列的定義和表格中的數(shù)據(jù)可得a1=2,a2=4,a3=8, 3分則數(shù)列{an}是首項(xiàng)為2,公比為2的等比數(shù)列,所以an=2×2n-1=2n. 5分(2)因?yàn)閎n=an+(-1)nlog2an=2n+(-1)nlog22n=2n+(-1)nn, 7分當(dāng)n為偶數(shù)時(shí),Sn=(21+22+…+2n)+[-1+2-3+4-…-(n-1)+n]=2-2n+11-2+n2=2當(dāng)n為奇數(shù)時(shí),Sn=(21+22+…+2n)+[-1+2-3+4-…+(n-1)-n]=2-2n+11-2+n-12-n=2n+1+n-綜上所述,Sn=2n+114.已知{an}是等差數(shù)列,{bn}是等比數(shù)列,a2=b1=a3-1b2=a4-1b(1)求數(shù)列{an},{bn}的通項(xiàng)公式;(2)求數(shù)列{an·bn}的前n項(xiàng)和Sn,并求使不等式Sn<1成立的n的最大值.(本小題滿分10分)14.解:(1)設(shè)數(shù)列{an}的公差為d,數(shù)列{bn}的公比為q.∵a2=b1=a3-1b2=a4-1∴1+d-1q故an=a2+(n-2)d=2n-3,bn=12n-(2)由(1)知Sn=-1×120+1×121+3×122+…+(2n-5)·12n-2+(2n-3)·12n-1,∴12Sn=-1×121+1×兩式相減得12Sn=-1+2×121+2×122+…+2×12n-1-2n-32n=2×120+2×121+2×122+…+∴Sn=2-2n+12由Sn<1,得2n+1>2n-1,當(dāng)1≤n≤4時(shí),2n+1>2n-1,當(dāng)n>4時(shí),2n+1<2n-1,故所求n的最大值為4. 10分15.數(shù)列的滿足，，．(1)求數(shù)列的通項(xiàng)公式；(2)將數(shù)列中去掉數(shù)列的項(xiàng)后余下的項(xiàng)按原來的順序組成數(shù)列，求數(shù)列的前50項(xiàng)和．解:（1）因?yàn)?，所以，又因?yàn)?，所以，，所以?shù)列是以2為首項(xiàng)，2為公比的等比數(shù)列，則，即．（2）由得，，因?yàn)椋灾幸サ魯?shù)列的項(xiàng)有5項(xiàng)，所以．16.已知數(shù)列是等差數(shù)列，其前和為，，數(shù)列滿足(1)求數(shù)列，的通項(xiàng)公式;(2)若對(duì)數(shù)列，，在與之間插入個(gè)2（），組成一個(gè)新數(shù)列，求數(shù)列的前2023項(xiàng)的和.解:（1）設(shè)等差數(shù)列的首項(xiàng)為，公差為，由題意，，所以

①當(dāng)時(shí)，②，①-②可得，，當(dāng)時(shí)，適合，所以（2）因?yàn)?，所以在?shù)列中，從項(xiàng)開始到項(xiàng)為止，共有項(xiàng)數(shù)為，當(dāng)時(shí)，；當(dāng)時(shí)，，所以數(shù)列前2023項(xiàng)是項(xiàng)之后還有2023-1034=989項(xiàng)為2，所求和為.17.已知數(shù)列{an}滿足a1-1a1·a(1)求數(shù)列{an}的通項(xiàng)公式;(2)在ak和ak+1(k∈N*)中插入k個(gè)相同的數(shù)(-1)k+1·k,構(gòu)成一個(gè)新數(shù)列{bn}:a1,1,a2,-2,-2,a3,3,3,3,a4,…,求{bn}的前100項(xiàng)和S100.解:(1)當(dāng)n=1時(shí),a1-1a1=1a1,當(dāng)n≥2時(shí),由a1-1a1·a2-1a2·…·an-1a兩式相除可得an-1an=an-1an,故數(shù)列{an}是以2為首項(xiàng),1為公差的等差數(shù)列,∴an=n+1. 5分(2)方法一:將ak和插入的k個(gè)數(shù)分成一組,則前k組共有k+k(k+1)2=k2+3令k2+3k2≤100,又k∈N*當(dāng)k=12時(shí),k2+3k2=90∴{bn}的前100項(xiàng)中包含前12組數(shù)和第13組的前10個(gè)數(shù), 8分∴S100=(a1+1)+(a2-22)+(a3+32)+…+(a11+112)+(a12-122)+(a13+13×9)=(a1+a2+…+a13)+(1-22+32-42+…+112-122)+117=13×(2+14)2-(3+7+11+15+19+23)+117=104-6×(3+23)2+117=104-方法二:設(shè)插入的所有數(shù)構(gòu)成數(shù)列{cn}.∵1+2+3+…+12=78,78+13=91<100,1+2+3+…+12+13=91,91+14=105>100,∴b1,b2,…,b100中包含{an}的前13項(xiàng)及{cn}的前87項(xiàng), 8分∴S100=(a1+a2+…+a13)+(c1+c2+…+c87)=(2+3+…+14)+(1-22+32-42+…+112-122+13×9)=13×(2+14)2-(3+7+11+15+19+23)+13×9=13×8-78+13×9=18.已知數(shù)列{an}的前n項(xiàng)和為Sn,a1=-94,且4Sn+1=3Sn-9(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)數(shù)列{bn}滿足3bn+(n-4)an=0(n∈N*),記{bn}的前n項(xiàng)和為Tn,若Tn≤λbn對(duì)任意n∈N*恒成立,求實(shí)數(shù)λ的取值范圍.18.解:(1)當(dāng)n=1時(shí),4(a1+a2)=3a1-9,則4a2=94-9=-274,∴a2=-2716當(dāng)n≥2時(shí),由4Sn+1=3Sn-9①,得4Sn=3Sn-1-9②,①-②得4an+1=3an, 3分又a2=-2716≠0,∴an≠0,∴an+1an又a2a1=34,∴{an}是首項(xiàng)為-94故an=-94·34n-1=-(2)由3bn+(n-4)an=0,得bn=-n-43an=(n-4)3所以Tn=-3×34-2×342-1×343+0×344+…+(n-5)·34Tn=-3×342-2×343-1×344+0×345+…+(兩式相減得14Tn=-3×34+342+343+344+…+34n-(n-4)·34n+1=-94+9161-34n-所以Tn=-4n·34n+1由Tn≤λbn對(duì)任意n∈N*恒成立,得-4n·34n+1≤λ(n-4)·34n對(duì)任意n∈N即λ(n-4)+3n≥0對(duì)任意n∈N*恒成立.當(dāng)n=4時(shí),不等式恒成立;當(dāng)n<4時(shí),λ≤-3nn-4=-3-12n-當(dāng)n>4時(shí),λ≥-3nn-4=-3-12n-4,得所以-3≤λ≤1. 12分19.在國家一系列利好政策的支持下，我國新能源汽車產(chǎn)業(yè)發(fā)展迅速.某汽車企業(yè)計(jì)劃大力發(fā)展新能源汽車