2020-2021学年新教材高中数学数列5.3等比数列5.3.2.2等比数列习题课课时素养检测含解析

十等比数列习题课(30分钟60分)一、选择题(每小题5分,共30分,多选题全部选对得5分,选对但不全的得3分,有选错的得0分)1.在等比数列{an}中,如果a1+a4=18,a2+a3=12,那么这个数列的公比为 () B.QUOTE或QUOTE 或QUOTE【解题指南】设公比为q,将a1+a4=18,a2+a3=12用首项和公比q表示,联立即可解出q.【解析】选C.设等比数列{an}的公比为q,因为a1+a4=18,a2+a3=12,所以a1QUOTE=18,a1QUOTE=12,两式相除化简得(2q2-5q+2)(1+q)=0,则2q2-5q+2=0或q=-1,因为a1(1+q3)=18,得q≠-1,所以解得q=2或QUOTE.2.(2020·全国Ⅱ卷)记Sn为等比数列{an}的前n项和.若a5-a3=12,a6-a4=24,则QUOTE=()n-1 1-nn-1 1-n-1【解析】选B.设等比数列的公比为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.在等比数列{an}中,已知其前n项和Sn=2n+1+a,则a的值为 () B.1 【解析】选C.当n=1时,a1=S1=22+a=4+a,当n≥2时,an=Sn-Sn-1=(2n+1+a)-(2n+a)=2n,因为{an}为等比数列,所以a1也应该符合an=2n,从而可得4+a=2⇒a=-2.4.已知数列{an}满足a1=1,a2=3,an+2=3an(n∈N*),则数列{an}的前2020项的和S2020等于 ()A.2(31009-1) B.2(31010-1)C.2(32020-1) D.2(32019-1)【解析】选B.由an+2=3an(n∈N*),即QUOTE=3,当n为奇数时,可得a1,a3,…,a2n-1成等比数列,首项为1,公比为3.当n为偶数时,可得a2,a4,…,a2n成等比数列,首项为3,公比为3.那么:S奇=QUOTE,S偶=QUOTE,2020=S奇+S偶=QUOTE=2×31010-2=2(31010-1).5.已知等比数列{an}的首项为8,Sn是其前n项的和,某同学经计算得S1=8,S2=20,S3=36,S4=65,后来该同学发现其中一个数算错了,则该数为()1 2 3【解析】14错误,则S2,S3正确,于是a1=8,a2=S2-S1=12,a3=S3-S2=16,与{an}为等比数列矛盾,故S43错误,则S2正确,此时,a1=8,a2=12,得q=QUOTE,a3=18,a4=27,S4=65,满足题设.6.(多选题)已知a1,a2,…,a8为各项都大于零的等比数列,公比q≠1,则下列说法不正确的是 ()1+a8>a4+a5 1+a8<a4+a51a8>a4a5 1a8【解析】选BCD.(a1+a8)-(a4+a5)=a1+a1q7-a1q3-a1q4=a1(1-q3-q4+q7)=a1(1-q3)(1-q4),若q>1,则1-q3<0,1-q4<0,a1>0,所以a1(1-q3)(1-q4)>0,若0<q<1,则1-q3>0,1-q4>0,a1>0,所以a1(1-q3)(1-q4)>0,所以a1+a8>a4+a5,故不正确的是BCD.二、填空题(每小题5分,共10分)7.已知数列QUOTE的前n项和为Sn,a1=1,2Sn=an+1-1,则Sn=__________.

【解析】当n=1时,则有2S1=a2-1,所以a2=2S1+1=2a1+1=3;当n≥2时,由2Sn=an+1-1得出2Sn-1=an-1,上述两式相减得2an=an+1-an,所以an+1=3an,得QUOTE=3且QUOTE=3,所以,数列QUOTE是以1为首项,以3为公比的等比数列,则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】设Sn=QUOTE+QUOTE+QUOTE+…+QUOTE,则QUOTESn=QUOTE+QUOTE+…+QUOTE+QUOTE,两式相减得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、解答题(每小题10分,共20分)9.已知数列{an}的前n项和Sn=k(3n-1),且a3=27.(1)求数列{an}的通项公式;(2)若bn=log3an,求数列QUOTE的前n项和Tn.【解析】(1)因为Sn=k(3n-1),a3=27.当n=3时,a3=S3-S2=k(33-32),解得k=QUOTE,当n≥2时,an=Sn-Sn-1=QUOTE(3n-1)-QUOTE(3n-1-1)=3n,又因为a1=S1=3也满足上式,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.设{an}是公比为正数的等比数列,a1=2,a3=a2+4.(1)求{an}的通项公式;(2)设{bn}是首项为1,公差为2的等差数列,求数列{an+bn}的前n项和Sn.【解析】(1)设q为等比数列{an}的公比,则由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}的通项公式为an=2·2n-1=2n(n∈N*).(2)易知bn=2n-1,则Sn=QUOTE+n×1+QUOTE×2=2n+1+n2-2.(35分钟70分)一、选择题(每小题5分,共20分)1.已知函数f(x)=x+3sinQUOTE+QUOTE,则fQUOTE+fQUOTE+…+fQUOTE=()020 021040 042【解析】选A.因为f(a)+f(1-a)=a+3sinQUOTE+QUOTE+1-a+3sinQUOTE+QUOTE=2+3sinQUOTE+3sinQUOTE=2,设S=fQUOTE+fQUOTE+…+fQUOTE,①则S=fQUOTE+fQUOTE+…+fQUOTE.②①+②得2S=2020×QUOTE=4040,所以S=2020.2.已知数列{an}的通项公式是an=QUOTE,其前n项和Sn=QUOTE,则项数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.数列1QUOTE,2QUOTE,3QUOTE,4QUOTE,…,n+QUOTE的前n项和为 ()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.已知等比数列{an}满足a3=4,且QUOTE=9,则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,因为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.【补偿训练】已知数列{an}的前n项和为Sn,a1=1,Sn=2an+1,则Sn= ()n-1B.QUOTEC.QUOTE D.QUOTE【解析】选B.由Sn=2an+1,可得S1=a1=2a2,所以a2=QUOTE,当n≥2时,有Sn-1=2an,两式作差可得QUOTE=QUOTE,故数列{an}是从第2项起构成首项a2=QUOTE,公比q=QUOTE的等比数列.所以Sn=a1+QUOTE=1+QUOTE=QUOTE.二、填空题(每小题5分,共20分)5.若数列QUOTE满足:2a1+22a2+23a3+…+2nan=2n(n∈N*),则数列QUOTE的前n项和Sn为__________.

【解析】由2a1+22a2+23a3+…=2nQUOTE得:2a1+22a2+23a3+…+2n-1an-1=2QUOTE,QUOTE且QUOTE,两式作差可得:2nan=2,即an=QUOTE且QUOTE,由已知等式可得,2a1=2,解得:a1=1,适合上式,所以an=QUOTE,又QUOTE=QUOTE=QUOTE,所以数列QUOTE是以1为首项,以QUOTE为公比的等比数列,则Sn=QUOTE=2-QUOTE.答案:2-QUOTE6.设数列{an}满足a1=1,且an+1=2QUOTE,则数列{2nan}的前n项的和Sn=________.

【解析】由题意得QUOTE-4an+1an+4(an)2=0,所以(an+1-2an)2=0,故an+1=2an,所以{an}为等比数列,an=2n-1,则2nan=n·2n,Sn=1·2+2·22+…+n·2n,2Sn=1·22+2·23+…+(n-1)2n+n·2n+1,两式作差得-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,则数列{cn}的前n项和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,上述两式作差得-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.已知数列{an}和{bn}中,数列{an}的前n项和为Sn.若点(n,Sn)在函数y=-x2+4x的图像上,点(n,bn)在函数y=2x的图像上.数列{an}的通项公式为________;数列{anbn}的前n项和Tn为________.

【解析】(1)由已知得Sn=-n2+4n,因为当n≥2时,an=Sn-Sn-1=-2n+5,又当n=1时,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+1.两式相减得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、解答题(每小题10分,共30分)9.已知等比数列{an}中,a2=8,a5=512.(1)求数列{an}的通项公式;(2)令bn=nan,求数列{bn}的前n项和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,从而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)设{an}是等差数列,{bn}是等比数列.已知a1=4,b1=6,b2=2a2-2,b3=2a3+4.(1)求{an}和{bn}的通项公式.(2)设数列{cn}满足c1=1,cn=QUOTE其中k∈N*.①求数列{QUOTE(QUOTE-1)}的通项公式.②求QUOTEaici(n∈N*).【解析】(1)设等差数列{an}的公差为d,等比数列{bn}的公比为q.依题意得QUOTE解得QUOTE故an=4+(n-1)×3=3n+1,bn=6×2n-1=3×2n.所以{an}的通项公式为an=3n+1,{bn}的通项公式为bn=3×2n.(2)①QUOTE(QUOTE-1)=QUOTE(bn-1)=(3×2n+