（福建專版）2019高考數(shù)學(xué)一輪復(fù)習(xí) 課時(shí)規(guī)范練29 等比數(shù)列及其前n項(xiàng)和 文.docx

課時(shí)規(guī)范練29等比數(shù)列及其前n項(xiàng)和基礎(chǔ)鞏固組1.已知等比數(shù)列an滿足a1=14,a3a5=4(a4-1),則a2=() A.2B.1C.12D.182.在正項(xiàng)等比數(shù)列an中,a2,a48是方程2x2-7x+6=0的兩個(gè)根,則a1a2a25a48a49的值為()A.212B.93C.93D.353.(2017安徽黃山市二模)已知數(shù)列an的前n項(xiàng)和為Sn,且a1=2,an+1=Sn+1(nN*),則S5=()A.31B.42C.37D.474.設(shè)首項(xiàng)為1,公比為23的等比數(shù)列an的前n項(xiàng)和為Sn,則()A.Sn=2an-1B.Sn=3an-2C.Sn=4-3anD.Sn=3-2an5.(2017全國(guó))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.86.設(shè)等比數(shù)列an的前n項(xiàng)和為Sn.若S2=3,S4=15,則S6=()A.31B.32C.63D.647.設(shè)數(shù)列an是首項(xiàng)為a1,公差為-1的等差數(shù)列,Sn為其前n項(xiàng)和.若S1,S2,S4成等比數(shù)列,則a1的值為.8.(2017北京)若等差數(shù)列an和等比數(shù)列bn滿足a1=b1=-1,a4=b4=8,則a2b2=.9.(2017江蘇,9)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=634,則a8=.10.(2017河南新鄉(xiāng)二模,文17)在數(shù)列an中,a1=12,an的前n項(xiàng)和Sn滿足Sn+1-Sn=12n+1(nN*).(1)求數(shù)列an的通項(xiàng)公式an以及前n項(xiàng)和Sn;(2)若S1+S2,S1+S3,m(S2+S3)成等差數(shù)列,求實(shí)數(shù)m的值.導(dǎo)學(xué)號(hào)24190754綜合提升組11.(2017四川廣元二診)已知數(shù)列an的前n項(xiàng)和為Sn,且對(duì)任意正整數(shù)n都有an=34Sn+2成立.若bn=log2an,則b1 008=()A.2 017B.2 016C.2 015D.2 01412.設(shè)等比數(shù)列an滿足a1+a3=10,a2+a4=5,則a1a2a3an的最大值為.13.已知an是公差為3的等差數(shù)列,數(shù)列bn滿足b1=1,b2=13,anbn+1+bn+1=nbn.(1)求an的通項(xiàng)公式;(2)求bn的前n項(xiàng)和.創(chuàng)新應(yīng)用組14.已知數(shù)列an的前n項(xiàng)和為Sn,在數(shù)列bn中,b1=a1,bn=an-an-1(n2),且an+Sn=n.(1)設(shè)cn=an-1,求證:cn是等比數(shù)列;(2)求數(shù)列bn的通項(xiàng)公式.答案：1.Ca3a5=4(a4-1),a42=4(a4-1),解得a4=2.又a4=a1q3,且a1=14,q=2,a2=a1q=12.2.Ba2,a48是方程2x2-7x+6=0的兩個(gè)根,a2a48=3.又a1a49=a2a48=a252=3,a25>0,a1a2a25a48a49=a255=93.3.Dan+1=Sn+1(nN*),Sn+1-Sn=Sn+1(nN*),Sn+1+1=2(Sn+1)(nN*),數(shù)列Sn+1是首項(xiàng)為3,公比為2的等比數(shù)列.則S5+1=324,解得S5=47.4.DSn=a1(1-qn)1-q=a1-anq1-q=1-23an1-23=3-2an,故選D.5.A設(shè)等差數(shù)列的公差為d,則d0,a32=a2a6,即(1+2d)2=(1+d)(1+5d),解得d=-2,所以S6=61+652(-2)=-24,故選A.6.CS2=3,S4=15,由等比數(shù)列前n項(xiàng)和的性質(zhì),得S2,S4-S2,S6-S4成等比數(shù)列,(S4-S2)2=S2(S6-S4),即(15-3)2=3(S6-15),解得S6=63,故選C.7.-12由已知得S1=a1,S2=a1+a2=2a1-1,S4=4a1+432(-1)=4a1-6,而S1,S2,S4成等比數(shù)列,(2a1-1)2=a1(4a1-6),整理,得2a1+1=0,解得a1=-12.8.1設(shè)等差數(shù)列an的公差為d,等比數(shù)列bn的公比為q,由題意知-1+3d=-q3=8,即-1+3d=8,-q3=8,解得d=3,q=-2.故a2b2=-1+3-1(-2)=1.9.32設(shè)該等比數(shù)列的公比為q,則S6-S3=634-74=14,即a4+a5+a6=14.S3=74,a1+a2+a3=74.由得(a1+a2+a3)q3=14,q3=1474=8,即q=2.a1+2a1+4a1=74,a1=14,a8=a1q7=1427=32.10.解 (1)an+1=Sn+1-Sn=12n+1,當(dāng)n2時(shí),an=12n.又a1=12,當(dāng)n=1時(shí)上式也成立.an=12n,Sn=121-12n1-12=1-12n.(2)由(1)可得:S1=12,S2=34,S3=78.S1+S2,S1+S3,m(S2+S3)成等差數(shù)列,12+34+m34+78=212+78,解得m=1213.11.A在an=34Sn+2中,令n=1得a1=8,an=34Sn+2成立,an+1=34Sn+1+2成立,兩式相減得an+1-an=34an+1,an+1=4an,又a10,數(shù)列an為等比數(shù)列,an=84n-1=22n+1,bn=log2an=2n+1,b1 008=2 017,故選A.12.64由已知a1+a3=10,a2+a4=(a1+a3)q=5,得q=12,所以a1=8,所以a1a2a3an=8n121+2+(n-1)=2-12n2+7n2,所以當(dāng)n=3或n=4時(shí),a1a2a3an取最大值為2-1232+732=26=64.13.解 (1)由已知,得a1b2+b2=b1,因?yàn)閎1=1,b2=13,所以a1=2.所以數(shù)列an是首項(xiàng)為2,公差為3的等差數(shù)列,通項(xiàng)公式為an=3n-1.(2)由(1)和anbn+1+bn+1=nbn,得bn+1=bn3,因此bn是首項(xiàng)為1,公比為13的等比數(shù)列.記bn的前n項(xiàng)和為Sn,則Sn=1-13n1-13=32-123n-1.14.(1)證明 an+Sn=n,an+1+Sn+1=n+1.-得an+1-an+an+1=1,2an+1=an+1,2(an+1-1)=an-1,an+1-1an-1=12,an-1是等比數(shù)列.又a1+a1=1,a1=12,首項(xiàng)c1=a1-1,c1=-12,公比q=12.又cn=an-1,cn是以-12為首項(xiàng),以12為公比的等比數(shù)列.(2)解 由(1)可知cn=-1212n-1=-12n,an=cn+1=1-12n.當(dāng)n2時(shí),bn=an-an-1=1-12n-1-12n-1=12n-1-12n=12n.又b1=a1=12代入上式也符合,bn=12n.