2020-2021學年新教材高中數(shù)學數(shù)列5.3等比數(shù)列5.3.2.2等比數(shù)列習題課課時素養(yǎng)檢測含解析

十等比數(shù)列習題課(30分鐘60分)一、選擇題(每小題5分,共30分,多選題全部選對得5分,選對但不全的得3分,有選錯的得0分)1.在等比數(shù)列{an}中,如果a1+a4=18,a2+a3=12,那么這個數(shù)列的公比為 () B.QUOTE或QUOTE 或QUOTE【解題指南】設(shè)公比為q,將a1+a4=18,a2+a3=12用首項和公比q表示,聯(lián)立即可解出q.【解析】選C.設(shè)等比數(shù)列{an}的公比為q,因為a1+a4=18,a2+a3=12,所以a1QUOTE=18,a1QUOTE=12,兩式相除化簡得(2q2-5q+2)(1+q)=0,則2q2-5q+2=0或q=-1,因為a1(1+q3)=18,得q≠-1,所以解得q=2或QUOTE.2.(2020·全國Ⅱ卷)記Sn為等比數(shù)列{an}的前n項和.若a5-a3=12,a6-a4=24,則QUOTE=()n-1 1-nn-1 1-n-1【解析】選B.設(shè)等比數(shù)列的公比為q,由a5-a3=12,a6-a4=24可得:QUOTE?QUOTE,所以an=a1qn-1=2n-1,Sn=QUOTE=QUOTE=2n-1,因此QUOTE=QUOTE=2-21-n.3.在等比數(shù)列{an}中,已知其前n項和Sn=2n+1+a,則a的值為 () B.1 【解析】選C.當n=1時,a1=S1=22+a=4+a,當n≥2時,an=Sn-Sn-1=(2n+1+a)-(2n+a)=2n,因為{an}為等比數(shù)列,所以a1也應該符合an=2n,從而可得4+a=2?a=-2.4.已知數(shù)列{an}滿足a1=1,a2=3,an+2=3an(n∈N*),則數(shù)列{an}的前2020項的和S2020等于 ()A.2(31009-1) B.2(31010-1)C.2(32020-1) D.2(32019-1)【解析】選B.由an+2=3an(n∈N*),即QUOTE=3,當n為奇數(shù)時,可得a1,a3,…,a2n-1成等比數(shù)列,首項為1,公比為3.當n為偶數(shù)時,可得a2,a4,…,a2n成等比數(shù)列,首項為3,公比為3.那么:S奇=QUOTE,S偶=QUOTE,2020=S奇+S偶=QUOTE=2×31010-2=2(31010-1).5.已知等比數(shù)列{an}的首項為8,Sn是其前n項的和,某同學經(jīng)計算得S1=8,S2=20,S3=36,S4=65,后來該同學發(fā)現(xiàn)其中一個數(shù)算錯了,則該數(shù)為()1 2 3【解析】14錯誤,則S2,S3正確,于是a1=8,a2=S2-S1=12,a3=S3-S2=16,與{an}為等比數(shù)列矛盾,故S43錯誤,則S2正確,此時,a1=8,a2=12,得q=QUOTE,a3=18,a4=27,S4=65,滿足題設(shè).6.(多選題)已知a1,a2,…,a8為各項都大于零的等比數(shù)列,公比q≠1,則下列說法不正確的是 ()1+a8>a4+a5 1+a8<a4+a51a8>a4a5 1a8【解析】選BCD.(a1+a8)-(a4+a5)=a1+a1q7-a1q3-a1q4=a1(1-q3-q4+q7)=a1(1-q3)(1-q4),若q>1,則1-q3<0,1-q4<0,a1>0,所以a1(1-q3)(1-q4)>0,若0<q<1,則1-q3>0,1-q4>0,a1>0,所以a1(1-q3)(1-q4)>0,所以a1+a8>a4+a5,故不正確的是BCD.二、填空題(每小題5分,共10分)7.已知數(shù)列QUOTE的前n項和為Sn,a1=1,2Sn=an+1-1,則Sn=__________.

【解析】當n=1時,則有2S1=a2-1,所以a2=2S1+1=2a1+1=3;當n≥2時,由2Sn=an+1-1得出2Sn-1=an-1,上述兩式相減得2an=an+1-an,所以an+1=3an,得QUOTE=3且QUOTE=3,所以,數(shù)列QUOTE是以1為首項,以3為公比的等比數(shù)列,則an=1×3n-1=3n-1,an+1=3n,那么2Sn=an+1-1=3n-1,因此,Sn=QUOTE.答案:QUOTE8.QUOTE+QUOTE+QUOTE+QUOTE+…+QUOTE-2=________.

【解析】設(shè)Sn=QUOTE+QUOTE+QUOTE+…+QUOTE,則QUOTESn=QUOTE+QUOTE+…+QUOTE+QUOTE,兩式相減得QUOTESn=QUOTE+QUOTE+…+QUOTE-QUOTE=QUOTE-QUOTE=1-QUOTE-QUOTE,所以Sn=2-QUOTE-QUOTE,所以原式=-QUOTE-QUOTE=-QUOTE.答案:-QUOTE三、解答題(每小題10分,共20分)9.已知數(shù)列{an}的前n項和Sn=k(3n-1),且a3=27.(1)求數(shù)列{an}的通項公式;(2)若bn=log3an,求數(shù)列QUOTE的前n項和Tn.【解析】(1)因為Sn=k(3n-1),a3=27.當n=3時,a3=S3-S2=k(33-32),解得k=QUOTE,當n≥2時,an=Sn-Sn-1=QUOTE(3n-1)-QUOTE(3n-1-1)=3n,又因為a1=S1=3也滿足上式,所以an=3n.(2)由(1)及已知知,bn=log33n=n,所以QUOTE=QUOTE-QUOTE,所以Tn=1-QUOTE+QUOTE-QUOTE+…+QUOTE-QUOTE=1-QUOTE=QUOTE.10.設(shè){an}是公比為正數(shù)的等比數(shù)列,a1=2,a3=a2+4.(1)求{an}的通項公式;(2)設(shè){bn}是首項為1,公差為2的等差數(shù)列,求數(shù)列{an+bn}的前n項和Sn.【解析】(1)設(shè)q為等比數(shù)列{an}的公比,則由a1=2,a3=a2+4得2q2=2q+4,即q2-q-2=0,解得q=2或q=-1(舍去),因此q=2.所以{an}的通項公式為an=2·2n-1=2n(n∈N*).(2)易知bn=2n-1,則Sn=QUOTE+n×1+QUOTE×2=2n+1+n2-2.(35分鐘70分)一、選擇題(每小題5分,共20分)1.已知函數(shù)f(x)=x+3sinQUOTE+QUOTE,則fQUOTE+fQUOTE+…+fQUOTE=()020 021040 042【解析】選A.因為f(a)+f(1-a)=a+3sinQUOTE+QUOTE+1-a+3sinQUOTE+QUOTE=2+3sinQUOTE+3sinQUOTE=2,設(shè)S=fQUOTE+fQUOTE+…+fQUOTE,①則S=fQUOTE+fQUOTE+…+fQUOTE.②①+②得2S=2020×QUOTE=4040,所以S=2020.2.已知數(shù)列{an}的通項公式是an=QUOTE,其前n項和Sn=QUOTE,則項數(shù)n等于()A.13 B.10 C.9 【解析】n=QUOTE=1-QUOTE,所以Sn=QUOTE+QUOTE+QUOTE+…+QUOTE=n-QUOTE=n-QUOTE=n-1+QUOTE,令n-1+QUOTE=QUOTE=5+QUOTE,所以n=6.3.數(shù)列1QUOTE,2QUOTE,3QUOTE,4QUOTE,…,n+QUOTE的前n項和為 ()A.QUOTE(n2+n+2)-QUOTE B.QUOTEn(n+1)+1-QUOTEC.QUOTE(n2-n+2)-QUOTE D.QUOTEn(n+1)+2QUOTE【解析】QUOTE+2QUOTE+3QUOTE+…+QUOTE=(1+2+…+n)+QUOTE=QUOTE+QUOTE=QUOTE(n2+n)+1-QUOTE=QUOTE(n2+n+2)-QUOTE.4.已知等比數(shù)列{an}滿足a3=4,且QUOTE=9,則a1+QUOTE+a3+QUOTE+a5+QUOTE+…+a19+QUOTE=()A.QUOTE B.QUOTE C.QUOTE D.QUOTE【解析】QUOTE=9,所以QUOTE=9?1+q3=9?q=2,因為a3=4,所以a1q2=4,a1=1,因此a1+QUOTE+a3+QUOTE+a5+QUOTE+…+a19+QUOTE=a1+a3+a5+…+a19+QUOTE+QUOTE+QUOTE+…+QUOTE=QUOTE+QUOTE×QUOTE=QUOTE.【補償訓練】已知數(shù)列{an}的前n項和為Sn,a1=1,Sn=2an+1,則Sn= ()n-1B.QUOTEC.QUOTE D.QUOTE【解析】選B.由Sn=2an+1,可得S1=a1=2a2,所以a2=QUOTE,當n≥2時,有Sn-1=2an,兩式作差可得QUOTE=QUOTE,故數(shù)列{an}是從第2項起構(gòu)成首項a2=QUOTE,公比q=QUOTE的等比數(shù)列.所以Sn=a1+QUOTE=1+QUOTE=QUOTE.二、填空題(每小題5分,共20分)5.若數(shù)列QUOTE滿足:2a1+22a2+23a3+…+2nan=2n(n∈N*),則數(shù)列QUOTE的前n項和Sn為__________.

【解析】由2a1+22a2+23a3+…=2nQUOTE得:2a1+22a2+23a3+…+2n-1an-1=2QUOTE,QUOTE且QUOTE,兩式作差可得:2nan=2,即an=QUOTE且QUOTE,由已知等式可得,2a1=2,解得:a1=1,適合上式,所以an=QUOTE,又QUOTE=QUOTE=QUOTE,所以數(shù)列QUOTE是以1為首項,以QUOTE為公比的等比數(shù)列,則Sn=QUOTE=2-QUOTE.答案:2-QUOTE6.設(shè)數(shù)列{an}滿足a1=1,且an+1=2QUOTE,則數(shù)列{2nan}的前n項的和Sn=________.

【解析】由題意得QUOTE-4an+1an+4(an)2=0,所以(an+1-2an)2=0,故an+1=2an,所以{an}為等比數(shù)列,an=2n-1,則2nan=n·2n,Sn=1·2+2·22+…+n·2n,2Sn=1·22+2·23+…+(n-1)2n+n·2n+1,兩式作差得-Sn=QUOTE-n·2n+1,即Sn=(n-1)2n+1+2.答案:(n-1)2n+1+27.已知cn=(2n-1)2n-1,則數(shù)列{cn}的前n項和Sn=________.

【解析】Sn=1·1+3·2+5·22+…+(2n-1)2n-1,2Sn=1·2+3·22+5·23+…+(2n-3)·2n-1+(2n-1)2n,上述兩式作差得-Sn=1+2·2+2·22+2·23+…+2·2n-1-(2n-1)2n=1+2QUOTE-(2n-1)2n,所以Sn=3-2n(3-2n).答案:3-2n(3-2n)8.已知數(shù)列{an}和{bn}中,數(shù)列{an}的前n項和為Sn.若點(n,Sn)在函數(shù)y=-x2+4x的圖像上,點(n,bn)在函數(shù)y=2x的圖像上.數(shù)列{an}的通項公式為________;數(shù)列{anbn}的前n項和Tn為________.

【解析】(1)由已知得Sn=-n2+4n,因為當n≥2時,an=Sn-Sn-1=-2n+5,又當n=1時,a1=S1=3,符合上式.所以an=-2n+5.(2)由已知得bn=2n,anbn=(-2n+5)·2n.Tn=3×21+1×22+(-1)×23+…+(-2n+5)×2n,2Tn=3×22+1×23+…+(-2n+7)×2n+(-2n+5)×2n+1.兩式相減得Tn=-6+(23+24+…+2n+1)+(-2n+5)×2n+1=QUOTE+(-2n+5)×2n+1-6=(7-2n)·2n+1-14.答案:an=-2n+5(7-2n)·2n+1-14三、解答題(每小題10分,共30分)9.已知等比數(shù)列{an}中,a2=8,a5=512.(1)求數(shù)列{an}的通項公式;(2)令bn=nan,求數(shù)列{bn}的前n項和Sn.【解析】(1)QUOTE=QUOTE=64=q3,所以q=4,所以an=a2·4n-2=8×4n-2=22n-1.(2)由bn=nan=n×22n-1知Sn=1×2+2×23+3×25+…+n×22n-1①,從而22×Sn=1×23+2×25+3×27+…+n×22n+1②,①-②得(1-22)×Sn=2+23+25+…+22n-1-n×22n+1,即Sn=QUOTE[(3n-1)22n+1+2].10.(2019·天津高考)設(shè){an}是等差數(shù)列,{bn}是等比數(shù)列.已知a1=4,b1=6,b2=2a2-2,b3=2a3+4.(1)求{an}和{bn}的通項公式.(2)設(shè)數(shù)列{cn}滿足c1=1,cn=QUOTE其中k∈N*.①求數(shù)列{QUOTE(QUOTE-1)}的通項公式.②求QUOTEaici(n∈N*).【解析】(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.依題意得QUOTE解得QUOTE故an=4+(n-1)×3=3n+1,bn=6×2n-1=3×2n.所以{an}的通項公式為an=3n+1,{bn}的通項公式為bn=3×2n.(2)①Q(mào)UOTE(QUOTE-1)=QUOTE(bn-1)=(3×2n+