高中數(shù)學(xué)專題突破練習(xí)《等比數(shù)列前n項(xiàng)和及其應(yīng)用》含詳細(xì)答案解析

4.3.2等比數(shù)列的前n項(xiàng)和公式第1課時(shí)等比數(shù)列前n項(xiàng)和及其應(yīng)用基礎(chǔ)過(guò)關(guān)練題組一求等比數(shù)列的前n項(xiàng)和1.(2019北京西城高二期末)數(shù)列{an}的前n項(xiàng)和為Sn,且a1=3,an+1=2an(n∈N*),則S5=()A.32 B.48 C.62 D.932.在等比數(shù)列{an}中,a1=1,且a4+a5+A.31 B.32 C.63 D.643.設(shè)數(shù)列{(-1)n}的前n項(xiàng)和為Sn,則Sn=()A.n[(-1)C.(-1)n4.等比數(shù)列1,x,x2,x3,…的前n項(xiàng)和Sn=()A.1-xnC.1-xn5.(2020天津津南高三上期末)在數(shù)列{an}中,a1=1,2an+1=an(n∈N*),記{an}的前n項(xiàng)和為Sn,則()A.Sn=2an-1 B.Sn=1-2anC.Sn=an-2 D.Sn=2-an6.(2020廣西柳州高級(jí)中學(xué)高二上期末)在等比數(shù)列{an}中,公比q=12,a2a4=2a5,則數(shù)列{an}的前5項(xiàng)和S5=7.在等比數(shù)列{an}中,a2-a1=2,且2a2為3a1和a3的等差中項(xiàng),求數(shù)列{an}的首項(xiàng)、公比及其前n項(xiàng)和.8.已知等差數(shù)列{an}和等比數(shù)列{bn}滿足a1=b1=1,a2+a4=10,b2b4=a5.(1)求{an}的通項(xiàng)公式;(2)求和:b1+b3+b5+…+b2n-1.題組二等比數(shù)列前n項(xiàng)和的應(yīng)用9.設(shè)正項(xiàng)等比數(shù)列{an}的前n項(xiàng)和為Sn,且S20=(210+1)S10,則數(shù)列{an}的公比為()A.4 B.2 C.1 D.110.設(shè)等比數(shù)列{an}的前n項(xiàng)和為Sn,若S3+S6=S9,則公比q=()A.1或-1 B.1C.-1 D.111.(2020廣東中山高二上期末統(tǒng)考)各項(xiàng)均為正數(shù)的等比數(shù)列{an}的前n項(xiàng)和為Sn,若a2=2,S6-S4=6a4,則a5=()A.4 B.10 C.16 D.3212.已知等比數(shù)列{an}的前n項(xiàng)和為Sn,a4-a1=78,S3=39,設(shè)bn=log3an,則數(shù)列{bn}的前10項(xiàng)和為()A.log371 B.69C.50 D.5513.(2020天津耀華中學(xué)高二上期中)等比數(shù)列{an}中,Sn為其前n項(xiàng)和,若Sn=3×2n+a,則a=.

14.(2020北京昌平高三上期末)各項(xiàng)均為正數(shù)的等比數(shù)列{an}中,a1=1,a2+a3=6,則S6S315.(2020山東臨沂羅莊高二上期末)記Sn為等比數(shù)列{an}的前n項(xiàng)和.若a1=1,S3=34,則S4=能力提升練題組一求等比數(shù)列的前n項(xiàng)和1.(2020湖南師大附中高二期末,)已知{an}是等比數(shù)列,a2=2,a5=14,則a1a2+a2a3+…+anan+1=()A.16(1-4-n) B.16(1-2-n)C.323(1-4-n) D.323(1-22.(2020天津?yàn)I海高二上期末,)數(shù)列1,1+2,1+2+22,…,1+2+22+…+2n-1,…的前n項(xiàng)和為T(mén)n,則Tn=()A.2n+1-n B.2n+1-n-2 C.2n-n D.2n3.(2020浙江杭州七校高二上聯(lián)考,)在等比數(shù)列{an}中,已知a4=3a3,則a2a1+a4a2+aA.3-n-C.3n-34.(2020遼寧鞍山一中高二期中,)設(shè)f(n)=2+23+25+27+…+22n+7(n∈N*),則f(n)=(易錯(cuò))A.23(4n-1) B.23(4C.23(4n+3-1) D.23(45.(2019重慶九校高二上聯(lián)考,)已知數(shù)列{an+81}是公比為3的等比數(shù)列,若a1=-78,則數(shù)列{|an|}的前100項(xiàng)和S100=()A.3101-16C.3101-156.(2020河北邢臺(tái)高三上期末,)在等比數(shù)列{an}中,a1+a2=5,且a2+a3=20.(1)求{an}的通項(xiàng)公式;(2)求數(shù)列{3an+an}的前n項(xiàng)和Sn7.(2020山東濟(jì)寧高二上期末質(zhì)量檢測(cè),)已知公差不為0的等差數(shù)列{an}的前n項(xiàng)和為Sn,且S4=20,a1,a2,a4成等比數(shù)列.(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)bn=2an(n∈N*),求數(shù)列{bn}的前n項(xiàng)和T題組二等比數(shù)列前n項(xiàng)和的應(yīng)用8.(2020遼寧省實(shí)驗(yàn)中學(xué)高二上期中,)已知數(shù)列{an},{bn}滿足a1=b1=1,an+1-an=bn+1bn=2(n∈N*),則數(shù)列{banA.43(4n-1-1) B.43(4n-1) C.13(4n-1-1) D.19.(2020江西新余一中高二上第二次段考,)數(shù)列{an}的通項(xiàng)公式為an=2n+1,其前n項(xiàng)和為T(mén)n,若不等式nlog2(Tn+4)-λ(n+1)+7≥3n對(duì)任意n∈N*恒成立,則實(shí)數(shù)λ的取值范圍為()A.λ≤3 B.λ≤4 C.2≤λ≤3 D.3≤λ≤410.(多選)(2020山東淄博高二上期末,)在遞增的等比數(shù)列{an}中,已知公比為q,Sn是其前n項(xiàng)和,若a1a4=32,a2+a3=12,則下列說(shuō)法正確的是()A.q=1B.數(shù)列{Sn+2}是等比數(shù)列C.S8=510D.數(shù)列{lgan}是公差為2的等差數(shù)列11.(2020北京西城高二上期末,)已知等比數(shù)列{an}的公比為2,且a3,a4+4,a5成等差數(shù)列.(1)求{an}的通項(xiàng)公式;(2)設(shè){an}的前n項(xiàng)和為Sn,且Sn=62,求n的值.

答案全解全析基礎(chǔ)過(guò)關(guān)練1.D由a1=3,an+1=2an(n∈N*)可知,數(shù)列{an}是以3為首項(xiàng),2為公比的等比數(shù)列,所以S5=3(1-2.A設(shè)等比數(shù)列{an}的公比為q,∵a4+a5+a8a又a1=1,∴S5=a1(1故選A.3.D易知數(shù)列{(-1)n}是首項(xiàng)為-1,公比為-1的等比數(shù)列,所以Sn=-1×[1-(-14.C當(dāng)x=1時(shí),Sn=n;當(dāng)x≠1且x≠0時(shí),Sn=1-xn1-x5.D∵2an+1=an(n∈N*),∴an+1=12an又a1=1,∴數(shù)列{an}是以1為首項(xiàng),12為公比的等比數(shù)列,∴an=1∴Sn=1-12n1-16.答案31解析設(shè)等比數(shù)列{an}的首項(xiàng)為a1,由a2a4=2a5,得a12q4=2a1q又q=12,a1≠0,∴a1∴S5=a1(=41-1327.解析設(shè)等比數(shù)列{an}的公比為q(q≠0).由已知可得a即a1q解②得q=3或q=1.由于a1(q-1)=2,因此q=1不合題意,舍去.所以公比q=3,首項(xiàng)a1=1.所以數(shù)列{an}的前n項(xiàng)和Sn=a1(1-qn)1-8.解析(1)設(shè)等差數(shù)列{an}的公差為d,因?yàn)閍2+a4=2a3=10,所以a3=5=1+2d,所以d=2,所以an=2n-1(n∈N*).(2)設(shè){bn}的公比為q.由b2b4=a5得q·q3=2×5-1,所以q2=3,所以{b2n-1}是以b1=1為首項(xiàng),q2=3為公比的等比數(shù)列,所以b1+b3+b5+…+b2n-1=1×(1-3n)19.B設(shè)等比數(shù)列{an}的首項(xiàng)為a1,公比為q,由題得q>0且q≠1,所以a1(1-q20)1-q=(210+1)×a1(1-q10)1-q,所以1-q10.A設(shè)等比數(shù)列{an}的首項(xiàng)為a1,由題意可知,當(dāng)q=1時(shí),S3+S6=S9,顯然成立;當(dāng)q≠1時(shí),由S3+S6=S9得a1(1-q3)1-q+a1(1-q11.C設(shè)等比數(shù)列{an}的首項(xiàng)為a1,公比為q,由題知q>0,由S6-S4=6a4得,a6+a5=6a4,又a4≠0,∴q2+q-6=0,解得q=2或q=-3(舍去).∴a5=a2q3=2×23=16,故選C.12.D設(shè)等比數(shù)列{an}的公比為q,由a4-∴a1(q3∴q-1=2,即q=3.∴a1=78q3-1=3,∴a∴bn=log3an=log33n=n,∴數(shù)列{bn}是以b1=1為首項(xiàng),1為公差的等差數(shù)列,其前10項(xiàng)和為10×(13.答案-3解析解法一:∵Sn=3×2n+a,∴當(dāng)n=1時(shí),a1=S1=6+a;當(dāng)n≥2時(shí),an=Sn-Sn-1=(3×2n+a)-(3×2n-1+a)=3×2n-1,∴a2=6,a3=12.又{an}是等比數(shù)列,∴a22=a1a∴62=(6+a)×12,解得a=-3.此時(shí)a1=3,符合an=3×2n-1,且{an}是等比數(shù)列.∴a=-3.解法二:設(shè)等比數(shù)列{an}的首項(xiàng)為a1,公比為q,易知q≠1,由Sn=a1(1-qn)1-q又Sn=3×2n+a,∴a=-3.14.答案9解析設(shè)等比數(shù)列{an}的公比為q,且q>0,∵a1=1,a2+a3=a1q+a1q2=6,∴q2+q-6=0,解得q=2或q=-3(舍去).∴S6S3=a1(15.答案5解析設(shè)等比數(shù)列的公比為q,由已知得S3=a1+a1q+a1q2=1+q+q2=34即q2+q+14=0,解得q=-1所以S4=a1(1-q能力提升練1.C設(shè)等比數(shù)列{an}的公比為q.解法一:∵a2=2,a5=14∴a1q∴a1a2+a2a3+…+anan+1=a12q+a12q3+…+a12q2n-1=a12(q+q3+解法二:同解法一得,a1=4,q=12∴a1a2=4×2=8,∴數(shù)列{anan+1}是首項(xiàng)為8,公比為14的等比數(shù)列∴a1a2+a2a3+…+anan+1=81-14n2.B設(shè)該數(shù)列為{an},由已知得數(shù)列的通項(xiàng)公式為an=1-2n1-2=2n-1,則Tn=a1+a2+…+an=(2-1)+(22-1)+…+(2n-1)=2+22+…+23.D設(shè)等比數(shù)列{an}的公比為q.∵a4=3a3,∴q=3,∴a2a1+a4a2+a6a3+…+a2nan=q+q24.D易知1,3,5,7,…是首項(xiàng)為1,公差為2的等差數(shù)列,設(shè)該數(shù)列為{am},則am=2m-1,設(shè)an=2n+7,即2m-1=2n+7,∴m=n+4,∴f(n)是以2為首項(xiàng),22=4為公比的等比數(shù)列的前n+4項(xiàng)的和,∴f(n)=2(1-4n+4)易錯(cuò)警示數(shù)列求和要弄清數(shù)列的特征,特別要注意數(shù)列的項(xiàng)數(shù),如本題中求的不是前n項(xiàng)和,而是前n+4項(xiàng)和,解題時(shí)要防止項(xiàng)數(shù)弄錯(cuò)導(dǎo)致解題錯(cuò)誤.5.C∵a1=-78,∴a1+81=3.又∵數(shù)列{an+81}是公比為3的等比數(shù)列,∴an+81=3×3n-1=3n,可得an=3n-81.易得當(dāng)n≤4時(shí),an≤0,當(dāng)n≥5時(shí),an>0,∴數(shù)列{|an|}的前100項(xiàng)和S100=|a1|+|a2|+|a3|+|a4|+a5+a6+…+a100=81-3+81-9+81-27+0+(35-81)+(36-81)+…+(3100-81)=204+35×(6.解析(1)設(shè)等比數(shù)列{an}的公比為q,因?yàn)閝=a2所以a1+a2=a1+4a1=5a1=5,即a1=1.故an=4n-1.(2)由(1)及題知,3an+an=3·4n-1+2n-1所以Sn=3×1-4n1-4+1-2n7.解析(1)設(shè)等差數(shù)列{an}的公差為d(d≠0),∵S4=20,a1,a2,a4成等比數(shù)列,∴4a1∴an=a1+(n-1)d=2+(n-1)×2=2n(n∈N*).(2)由(1)得,bn=2an=22n=4∴b1=4,bn+1b∴數(shù)列{bn}是首項(xiàng)為4,公比為4的等比數(shù)列.∴Tn=4×(1-4n8.D依題意得{an}是以a1=1為首項(xiàng),2為公差的等差數(shù)列,{bn}是以b1=1為首項(xiàng),2為公比的等比數(shù)列,∴an=1+2(n-1)=2n-1,bn=1×2n-1=2n-1,∴ban=b2n-1=22n-2=4∴{ban}是以1為首項(xiàng),4為公比的等比數(shù)列,設(shè)其前n項(xiàng)和為S則Sn=1×(1-4n)19.A依題意得,Tn=4(1-∴不等式nlog2(Tn+4)-λ(n+1)+7≥3n可化為nlog22n+2-λ(n+1)+7≥3n,即n2-n+7≥λ(n+1).又n∈N*,∴λ≤n2-n+7n+1對(duì)任意只需滿足λ≤n2-設(shè)n+1=t,則t∈N*,t≥2,∴λ≤n2-n∵t+9t-3≥2t當(dāng)且僅當(dāng)t=3,即n=2時(shí)等號(hào)成立,∴n2∴λ≤3,故選A.10.BC∵a1a解得a2=4∵{an}為遞增數(shù)列,∴a∴q=a3a2=2,a1∴an=2n,Sn=2×(1-∴S8=29-2=510,Sn+2=