（全國(guó)通用版）2019版高考數(shù)學(xué)一輪復(fù)習(xí) 第五章 數(shù)列 課時(shí)分層作業(yè) 三十一 5.3 等比數(shù)列及其前n項(xiàng)和 文.doc

課時(shí)分層作業(yè) 三十一 等比數(shù)列及其前n項(xiàng)和一、選擇題(每小題5分,共25分)1.(2018重慶模擬)已知各項(xiàng)均為正數(shù)的等比數(shù)列an的前n項(xiàng)和為Sn,且S3=14,a3=8,則a6=()A.16B.32C.64D.128【解析】選C.由題意得,等比數(shù)列的公比為q,由S3=14,a3=8,則解得a1=2,q=2,所以a6=a1q5=225=64,故選C.2.(2017全國(guó)卷)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【解析】選A.設(shè)等差數(shù)列的公差為d,d0,=a2a6,即(1+2d)2=(1+d)(1+5d),d2=-2d(d0),所以d=-2,所以S6=61+(-2)=-24.3.(2017全國(guó)卷)我國(guó)古代數(shù)學(xué)名著算法統(tǒng)宗中有如下問題:“遠(yuǎn)望巍巍塔七層,紅光點(diǎn)點(diǎn)倍加增,共燈三百八十一,請(qǐng)問尖頭幾盞燈?”意思是:一座7層塔共掛了381盞燈,且相鄰兩層中的下一層燈數(shù)是上一層燈數(shù)的2倍,則塔的頂層共有燈 ()A.1盞B.3盞C.5盞D.9盞【解析】選B.設(shè)塔的頂層共有燈x盞,則各層的燈數(shù)構(gòu)成一個(gè)公比為2的等比數(shù)列,由=381可得x=3.4.(2018臨沂模擬)已知等比數(shù)列an的前n項(xiàng)和為Sn=a2n-1+,則a的值為()A.-B.C.-D.【解析】選A.當(dāng)n2時(shí),an=Sn-Sn-1=a2n-1-a2n-2=a2n-2,當(dāng)n=1時(shí),a1=S1=a+,又因?yàn)閍n是等比數(shù)列,所以a+=,所以a=-.5.在公比為的等比數(shù)列an中,若sin(a1a4)=,則cos(a2a5)的值是()A.-B.C.D.【解析】選B.由等比數(shù)列的通項(xiàng)公式可知a2a5=(a1a4)q2=2(a1a4),cos(a2a5)=1-2sin2(a1a4)=1-2=.二、填空題(每小題5分,共15分)6.(2017北京高考)若等差數(shù)列an和等比數(shù)列bn滿足a1=b1=-1,a4=b4=8,則=_.【解析】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-1+3d=-q3=8d=3,q=-2=1.答案:17.已知數(shù)列an是等比數(shù)列,a2=2,a5=,則a1a2a3+a2a3a4+anan+1an+2=_.【解析】設(shè)數(shù)列an的公比為q,則q3=,解得q=,a1=4.易知數(shù)列anan+1an+2是首項(xiàng)為a1a2a3=421=8,公比為q3=的等比數(shù)列,所以a1a2a3+a2a3a4+anan+1an+2=(1-2-3n).答案:(1-2-3n)8.(2015湖南高考)設(shè)Sn為等比數(shù)列的前n項(xiàng)和,若a1=1且3S1,2S2,S3成等差數(shù)列,則an=_.【解題指南】由3S1,2S2,S3成等差數(shù)列,可求得公比q=3,然后求an.【解析】因?yàn)?S1,2S2,S3成等差數(shù)列,所以22(a1+a2)=3a1+a1+a2+a3a3=3a2q=3,所以an=a1qn-1=3n-1.答案:3n-1三、解答題(每小題10分,共20分)9.(2018煙臺(tái)模擬)已知等差數(shù)列an中,a1=1,且a1,a2,a4+2成等比數(shù)列.(1)求數(shù)列an的通項(xiàng)公式及其前n項(xiàng)和Sn.(2)設(shè)bn=,求數(shù)列bn的前2n項(xiàng)和T2n.【解析】(1)設(shè)等差數(shù)列an的公差為d,因?yàn)閍1=1,且a1,a2,a4+2成等比數(shù)列.所以=a1(a4+2),即(1+d)2=1(1+3d+2),解得d=2或-1.其中d=-1時(shí),a2=0,舍去.所以d=2,可得an=1+2(n-1)=2n-1.Sn=n2.(2)bn=.所以當(dāng)n為偶數(shù)時(shí),=16.當(dāng)n為奇數(shù)時(shí),=.所以數(shù)列bn的奇數(shù)項(xiàng)是以為首項(xiàng),為公比的等比數(shù)列;偶數(shù)項(xiàng)是以8為首項(xiàng),16為公比的等比數(shù)列.所以數(shù)列bn的前2n項(xiàng)和T2n=(b1+b3+b2n-1)+(b2+b4+b2n)=+=(16n-16-n).10.(2015廣東高考改編)設(shè)數(shù)列an的前n項(xiàng)和為Sn,nN*.已知a1=1,a2=,a3=,且當(dāng)n2時(shí),4Sn+2+5Sn=8Sn+1+Sn-1. (1)求a4的值.(2)證明:為等比數(shù)列.【解析】(1)當(dāng)n=2時(shí),4S4+5S2=8S3+S1,即4+5=8+1,解得a4=.(2)由4Sn+2+5Sn=8Sn+1+Sn-1(n2),4Sn+2-4Sn+1+Sn-Sn-1=4Sn+1-4Sn(n2),即4an+2+an=4an+1(n2).因?yàn)?a3+a1=4+1=6=4a2,所以4an+2+an=4an+1,所以=,所以數(shù)列是以a2-a1=1為首項(xiàng),為公比的等比數(shù)列.1.(5分)(2018福州模擬)已知數(shù)列an滿足log3an+1=log3an+1(nN*),且a2+a4+a6=9,則lo(a5+a7+a9)的值是()A.-5B.-C.5D.【解析】選A.因?yàn)閘og3an+1=log3an+1,所以an+1=3an.所以數(shù)列an是公比q=3的等比數(shù)列,所以a2+a4+a6=a2(1+q2+q4)=9.所以a5+a7+a9=a5(1+q2+q4)=a2q3(1+q2+q4)=35.所以lo35=-5.【變式備選】等比數(shù)列an滿足an>0,nN*,且a3a2n-3=22n(n2),則當(dāng)n1時(shí),log2a1+log2a2+log2a2n-1=_.【解析】由等比數(shù)列的性質(zhì),得a3a2n-3=22n,從而得an=2n.所以log2a1+log2a2+log2a2n-1=log2(a1a2n-1)(a2a2n-2)(an-1an+1)an=log22n(2n-1)=n(2n-1)=2n2-n.答案:2n2-n2.(5分)已知數(shù)列an為等比數(shù)列,若a4+a6=10,則a7(a1+2a3)+a3a9的值為()A.10B.20C.100D.200【解析】選C.a7(a1+2a3)+a3a9=a7a1+2a7a3+a3a9=+2a4a6+=(a4+a6)2=102=100.3.(5分)(2016全國(guó)卷)設(shè)等比數(shù)列an滿足a1+a3=10,a2+a4=5,則a1a2an的最大值為_.【解析】由于an是等比數(shù)列,設(shè)an=a1qn-1,其中a1是首項(xiàng),q是公比.所以解得:故an=,所以a1a2an=.當(dāng)n=3或4時(shí),取到最小值-6,此時(shí)取到最大值26=64.所以a1a2an的最大值為64.答案:644.(12分)(2016全國(guó)卷)已知數(shù)列an的前n項(xiàng)和Sn=1+an,其中0. (1)證明an是等比數(shù)列,并求其通項(xiàng)公式.(2)若S5=,求.【解析】(1)由題意得a1=S1=1+a1,故a1=,由Sn=1+an,Sn+1=1+an+1得an+1=an+1-an,所以=,因此數(shù)列an是以a1=為首項(xiàng),以為公比的等比數(shù)列,an=.(2)由(1)得Sn=1-,又因?yàn)镾5=,所以=1-,即=,解得=-1.5.(13分)(2018鄭州模擬)已知數(shù)列an滿足a1=5,a2=5,an+1=an+6an-1(n2). (1)求證:an+1+2an是等比數(shù)列.(2)求數(shù)列an的通項(xiàng)公式.【解析】(1)因?yàn)閍n+1=an+6an-1(n2),所以an+1+2an=3an+6an-1=3(an+2an-1)(n2).因?yàn)閍1=5,a2=5,所以a2+2a1=15,所以an+2an-10(n2),所以=3(n2),所以數(shù)列an+1+2an是以15為首項(xiàng),3為公比的等比數(shù)列.(2)由(1)得an+1+2an=153n-1=53n,則an+1=-2an+53n,所以an+1-3n+1=-2(an-3n).又因?yàn)閍1-3=2,所以an-3n0,所以an-3n是以2為首項(xiàng),-2為公比的等比數(shù)列.所以an-3n=2(-2)n-1,即an=2(-2)n-1+3n.