廣西2020版高考數(shù)學(xué)一輪復(fù)習(xí) 考點(diǎn)規(guī)范練30 等比數(shù)列及其前n項(xiàng)和 文.docx

考點(diǎn)規(guī)范練30等比數(shù)列及其前n項(xiàng)和一、基礎(chǔ)鞏固1.已知等比數(shù)列an滿足a1=14,a3a5=4(a4-1),則a2=()A.2B.1C.12D.18答案C解析a3a5=4(a4-1),a42=4(a4-1),解得a4=2.又a4=a1q3,且a1=14,q=2.a2=a1q=12.2.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.35答案B解析a2,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.故選B.3.(2018北京,文5)“十二平均律”是通用的音律體系,明代朱載堉最早用數(shù)學(xué)方法計(jì)算出半音比例,為這個(gè)理論的發(fā)展做出了重要貢獻(xiàn).十二平均律將一個(gè)純八度音程分成十二份,依次得到十三個(gè)單音,從第二個(gè)單音起,每一個(gè)單音的頻率與它的前一個(gè)單音的頻率的比都等于122.若第一個(gè)單音的頻率為f,則第八個(gè)單音的頻率為()A.32fB.322fC.1225fD.1227f答案D解析由題意知,這十三個(gè)單音的頻率構(gòu)成首項(xiàng)為f,公比為122的等比數(shù)列,則第八個(gè)單音的頻率為(122)7f=1227f.4.已知an為等比數(shù)列,a4+a7=2,a5a6=-8,則a1+a10=()A.7B.5C.-5D.-7答案D解析an為等比數(shù)列,a5a6=a4a7=-8.聯(lián)立a4+a7=2,a4a7=-8,可解得a4=4,a7=-2或a4=-2,a7=4,當(dāng)a4=4,a7=-2時(shí),q3=-12,故a1+a10=a4q3+a7q3=-7;當(dāng)a4=-2,a7=4時(shí),q3=-2,故a1+a10=a4q3+a7q3=-7.綜上可知,a1+a10=-7.5.等差數(shù)列an的公差為2,若a2,a4,a8成等比數(shù)列,則an的前n項(xiàng)和Sn=()A.n(n+1)B.n(n-1)C.n(n+1)2D.n(n-1)2答案A解析a2,a4,a8成等比數(shù)列,a42=a2a8,即(a1+6)2=(a1+2)(a1+14),解得a1=2.Sn=na1+n(n-1)2d=2n+n2-n=n2+n=n(n+1).故選A.6.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.答案-12解析由已知得S1=a1,S2=a1+a2=2a1-1,S4=4a1+432(-1)=4a1-6.S1,S2,S4成等比數(shù)列,(2a1-1)2=a1(4a1-6),整理,得2a1+1=0,解得a1=-12.7.設(shè)數(shù)列an的前n項(xiàng)和為Sn,若S2=4,an+1=2Sn+1,nN*,則a1=,S5=.答案1121解析由題意,可得a1+a2=4,a2=2a1+1,所以a1=1,a2=3.再由an+1=2Sn+1,an=2Sn-1+1(n2),得an+1-an=2an,即an+1=3an(n2).又因?yàn)閍2=3a1,所以數(shù)列an是以1為首項(xiàng),3為公比的等比數(shù)列.所以S5=1-351-3=121.8.(2018遼寧部分重點(diǎn)高中聯(lián)考)已知數(shù)列an是等差數(shù)列,a2,a4,a8成等比數(shù)列,則該等比數(shù)列的公比為.答案1或2解析設(shè)an的公差為d.a2,a4,a8成等比數(shù)列,a42=a2a8,(a1+3d)2=(a1+d)(a1+7d),即d2=a1d,d=0或d=a1.當(dāng)d=0時(shí),a2=a4,公比為1;當(dāng)d=a1時(shí),a2=2d,a4=4d,公比為2.故等比數(shù)列的公比為1或2.9.(2018全國(guó),文17)已知數(shù)列an滿足a1=1,nan+1=2(n+1)an.設(shè)bn=ann.(1)求b1,b2,b3;(2)判斷數(shù)列bn是否為等比數(shù)列,并說(shuō)明理由;(3)求an的通項(xiàng)公式.解(1)由條件可得an+1=2(n+1)nan.將n=1代入得,a2=4a1,而a1=1,所以a2=4.將n=2代入得,a3=3a2,所以a3=12.從而b1=1,b2=2,b3=4.(2)bn是首項(xiàng)為1,公比為2的等比數(shù)列.由條件可得an+1n+1=2ann,即bn+1=2bn,又b1=1,所以bn是首項(xiàng)為1,公比為2的等比數(shù)列.(3)由(2)可得ann=2n-1,所以an=n2n-1.10.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.解(1)由已知,a1b2+b2=b1,b1=1,b2=13,得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-1.11.已知等差數(shù)列an的前n項(xiàng)和為Sn,且S4=4(a3+1),3a3=5a4,數(shù)列bn是等比數(shù)列,且b1b2=b3,2b1=a5.(1)求數(shù)列an,bn的通項(xiàng)公式;(2)求數(shù)列|an|的前n項(xiàng)和Tn.解(1)設(shè)等差數(shù)列an的公差為d.S4=4(a3+1),3a3=5a4,4a1+6d=4(a1+2d+1),3a1+6d=5a1+15d,解得a1=9,d=-2.an=11-2n.設(shè)數(shù)列bn的公比為q.b1b2=b3,2b1=a5,b12q=b1q2,2b1=1,解得b1=12,q=12.bn=12n.(2)由(1)知,Sn=10n-n2.由an=11-2n0可知n5.5,即a1>0,a2>0,a5>0,a6<0,a7<0,an<0.故當(dāng)n5時(shí),Tn=Sn=10n-n2;當(dāng)n6時(shí),Tn=2S5-Sn=n2-10n+50.于是Tn=10n-n2,n5,n2-10n+50,n6.二、能力提升12.若a,b是函數(shù)f(x)=x2-px+q(p>0,q>0)的兩個(gè)不同的零點(diǎn),且a,b,-2這三個(gè)數(shù)可適當(dāng)排序后成等差數(shù)列,也可適當(dāng)排序后成等比數(shù)列,則p+q的值等于()A.6B.7C.8D.9答案D解析a,b是函數(shù)f(x)=x2-px+q(p>0,q>0)的兩個(gè)不同的零點(diǎn),a+b=p,ab=q.p>0,q>0,a>0,b>0.又a,b,-2這三個(gè)數(shù)可適當(dāng)排序后成等差數(shù)列,也可適當(dāng)排序后成等比數(shù)列,2b=a-2,ab=4或2a=b-2,ab=4.解得a=4,b=1;解得a=1,b=4.p=a+b=5,q=14=4.p+q=9.故選D.13.(2018北京石景山一模)如圖,正方形上連接著等腰直角三角形,等腰直角三角形的腰上再連接正方形,如此繼續(xù)下去得到一個(gè)樹形圖形,稱為“勾股樹”.若某“勾股樹”含有1 023個(gè)正方形,且其最大的正方形的邊長(zhǎng)為22,則其最小的正方形的邊長(zhǎng)為.答案132解析由題意,得各正方形的邊長(zhǎng)構(gòu)成以22為首項(xiàng),22為公比的等比數(shù)列.已知共得到1023個(gè)正方形,則1+2+2n-1=1023,解得n=10,故最小的正方形的邊長(zhǎng)為22229=132.14.設(shè)等比數(shù)列an滿足a1+a3=10,a2+a4=5,則a1a2an的最大值為.答案64解析設(shè)an的公比為q.由已知a1+a3=10,a2+a4=a1q+a3q=5,兩式相除得a1+a3q(a1+a3)=105,解得q=12,a1=8,所以a1a2an=8n121+2+(n-1)=2-12n2+7n2,拋物線f(n)=-12n2+72n的對(duì)稱軸為n=-722-12=3.5,又nN*,所以當(dāng)n=3或n=4時(shí),a1a2an取最大值為2-1232+732=26=64.15.(2018廣東東莞二模)已知等比數(shù)列an與等差數(shù)列bn,a1=b1=1,a1a2,a1,a2,b3成等差數(shù)列,b1,a2,b4成等比數(shù)列.(1)求an,bn的通項(xiàng)公式;(2)設(shè)Sn,Tn分別是數(shù)列an,bn的前n項(xiàng)和,若Sn+Tn>100,求n的最小值.解(1)設(shè)數(shù)列an的公比為q,數(shù)列bn的公差為d,則2q=2+2d,q2=1+3d,解得d=0,q=1(舍)或d=1,q=2,故an=2n-1,bn=n.(2)由(1)易知Sn=1-2n1-2=2n-1,Tn=n(n+1)2.由Sn+Tn>100,得2n+n(n+1)2>101.2n+n(n+1)2是單調(diào)遞增數(shù)列,且26+672=85<101,27+782=156>101,n的最小值為7.三、高考預(yù)測(cè)16.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.(1)證明an+1=an+6an-1(n2),an+1+2an=3an+6an-1=3(an+2an-1)(n2).又a1=5,a2=5,a2+2a1=15,an+2an-10(n2),an+1+2anan+2an-1=3(n2),數(shù)列an+1+2an是以15為首項(xiàng),3為公比的等比數(shù)列.(2)解由(1)得an+1+2an=153n-1=53n,則an+1=-2an+53n,an+1-3n+1=-2(an-3n).又a1-3=2,an-3n0,an-3n是以2為首項(xiàng),-2為公比的等比數(shù)列.an-3n=2(-2)n-1,即an=2(-2)n-1+3n=3n-(-2)n.