（课标专用）天津市2020高考数学二轮复习专题能力训练11数列的通项与求和.docx

专题能力训练11数列的通项与求和专题能力训练第28页一、能力突破训练1.已知等差数列an的前n项和为Sn,若a1=2,a4+a10=28,则S9=()A.45B.90C.120D.75答案:B解析:因为an是等差数列,设公差为d,所以a4+a10=a1+3d+a1+9d=2a1+12d=4+12d=28,解得d=2.S9=9a1+982d=18+362=90.故选B.2.已知数列an是等差数列,满足a1+2a2=S5,下列结论错误的是()A.S9=0B.S5最小C.S3=S6D.a5=0答案:B解析:由题设可得3a1+2d=5a1+10d2a1+8d=0,即a5=0,所以D中结论正确.由等差数列的性质可得a1+a9=2a5=0,则S9=9(a1+a9)2=9a5=0,所以A中结论正确.S3-S6=3a1+3d-6a1-15d=-3(a1+4d)=-3a5=0,所以C中结论正确.B中结论是错误的.故选B.3.已知数列an的前n项和Sn=n2-2n-1,则a3+a17=()A.15B.17C.34D.398答案:C解析:Sn=n2-2n-1,a1=S1=12-2-1=-2.当n2时,an=Sn-Sn-1=n2-2n-1-(n-1)2-2(n-1)-1=n2-(n-1)2+2(n-1)-2n-1+1=n2-n2+2n-1+2n-2-2n=2n-3.an=-2,n=1,2n-3,n2.a3+a17=(23-3)+(217-3)=3+31=34.4.已知数列an满足a1=12,an+1=an+12n(nN*),则a2 019=()A.1-122018B.1-122019C.32-122018D.32-122019答案:C解析:数列an满足a1=12,an+1=an+12n(nN*),an+1-an=12n,当n2时,an=a1+a2-a1+a3-a2+an-an-1=12+121+122+12n-1=12+121-12n-11-12=32-12n-1,a2019=32-122018.故选C.5.已知数列an,构造一个新数列a1,a2-a1,a3-a2,an-an-1,此数列是首项为1,公比为13的等比数列,则数列an的通项公式为()A.an=32-3213n,nN*B.an=32+3213n,nN*C.an=1,n=1,32+3213n,n2,且nN*D.an=1,nN*答案:A解析:因为数列a1,a2-a1,a3-a2,an-an-1,是首项为1,公比为13的等比数列,所以an-an-1=13n-1,n2.所以当n2时,an=a1+(a2-a1)+(a3-a2)+(an-an-1)=1+13+132+13n-1=1-13n1-13=32-3213n.又当n=1时,an=32-3213n=1,则an=32-3213n,nN*.6.已知数列an满足a1=1,an-an+1=nanan+1(nN*),则an=.答案:2n2-n+2解析:因为an-an+1=nanan+1,所以an-an+1anan+1=1an+1-1an=n,1an=1an-1an-1+1an-1-1an-2+1a2-1a1+1a1=(n-1)+(n-2)+3+2+1+1a1=(n-1)(n-1+1)2+1=n2-n+22(n2).所以an=2n2-n+2(n2).又a1=1也满足上式,所以an=2n2-n+2.7.记Sn为数列an的前n项和.若Sn=2an+1,则S6=.答案:-63解析:Sn=2an+1,Sn-1=2an-1+1(n2).-,得an=2an-2an-1,即an=2an-1(n2).又S1=2a1+1,a1=-1.an是以-1为首项,2为公比的等比数列,则S6=-(1-26)1-2=-63.8.已知数列an中,a1=a,an+1=3an+8n+6,若an为递增数列,则实数a的取值范围为.答案:(-7,+)解析:由an+1=3an+8n+6,得an+1+4(n+1)+5=3(an+4n+5),即an+1+4(n+1)+5an+4n+5=3,所以数列an+4n+5是首项为a+9,公比为3的等比数列.所以an+4n+5=(a+9)3n-1,即an=(a+9)3n-1-4n-5.所以an+1=(a+9)3n-4n-9.因为数列an为递增数列,所以an+1an,即(a+9)3n-4n-9(a+9)3n-1-4n-5,即(a+9)3n6恒成立.因为nN*,所以(a+9)36,解得a-7.9.已知数列an中,a1=1,an-an-1+1=2n(nN*,n2).(1)求数列an的通项公式;(2)设bn=14an-1,求数列bn的通项公式及其前n项和Tn.解:(1)当n2时,因为an-an-1=2n-1,a1=1,所以an=(an-an-1)+(an-1-an-2)+(a2-a1)+a1=1+3+(2n-1)=n2.又a1=1满足上式,故an=n2(nN*).(2)bn=14an-1=14n2-1.又bn=14n2-1=1(2n+1)(2n-1)=1212n-1-12n+1,所以Tn=121-13+13-15+12n-1-12n+1=121-12n+1=n2n+1.10.已知数列an的前n项和为Sn,且a1=0,对任意nN*,都有nan+1=Sn+n(n+1).(1)求数列an的通项公式;(2)若数列bn满足an+log2n=log2bn,求数列bn的前n项和Tn.解:(1)(方法一)nan+1=Sn+n(n+1),当n2时,(n-1)an=Sn-1+n(n-1),两式相减,得nan+1-(n-1)an=Sn-Sn-1+n(n+1)-n(n-1),即nan+1-(n-1)an=an+2n,得an+1-an=2.当n=1时,1a2=S1+12,即a2-a1=2.数列an是以0为首项,2为公差的等差数列.an=2(n-1)=2n-2.(方法二)由nan+1=Sn+n(n+1),得n(Sn+1-Sn)=Sn+n(n+1),整理,得nSn+1=(n+1)Sn+n(n+1),两边同除以n(n+1),得Sn+1n+1-Snn=1.数列Snn是以S11=0为首项,1为公差的等差数列,Snn=0+n-1=n-1.Sn=n(n-1).当n2时,an=Sn-Sn-1=n(n-1)-(n-1)(n-2)=2n-2.又a1=0适合上式,数列an的通项公式为an=2n-2.(2)an+log2n=log2bn,bn=n2an=n22n-2=n4n-1.Tn=b1+b2+b3+bn-1+bn=40+241+342+(n-1)4n-2+n4n-1,4Tn=41+242+343+(n-1)4n-1+n4n,由-,得-3Tn=40+41+42+4n-1-n4n=1-4n1-4-n4n=(1-3n)4n-13.Tn=19(3n-1)4n+1.11.设数列an的前n项和为Sn.已知2Sn=3n+3.(1)求an的通项公式;(2)若数列bn满足anbn=log3an,求bn的前n项和Tn.解:(1)因为2Sn=3n+3,所以2a1=3+3,故a1=3.当n1时,2Sn-1=3n-1+3,此时2an=2Sn-2Sn-1=3n-3n-1=23n-1,即an=3n-1,所以an=3,n=1,3n-1,n1.(2)因为anbn=log3an,所以b1=13,当n1时,bn=31-nlog33n-1=(n-1)31-n.所以T1=b1=13;当n1时,Tn=b1+b2+b3+bn=13+13-1+23-2+(n-1)31-n,所以3Tn=1+130+23-1+(n-1)32-n,两式相减,得2Tn=23+(30+3-1+3-2+32-n)-(n-1)31-n=23+1-31-n1-3-1-(n-1)31-n=136-6n+323n,所以Tn=1312-6n+343n.经检验,当n=1时也适合.综上可得Tn=1312-6n+343n.二、思维提升训练12.给出数列11,12,21,13,22,31,1k,2k-1,k1,在这个数列中,第50个值等于1的项的序号是()A.4 900B.4 901C.5 000D.5 001答案:B解析:根据条件找规律,第1个1是分子、分母的和为2,第2个1是分子、分母的和为4,第3个1是分子、分母的和为6,第50个1是分子、分母的和为100,而分子、分母的和为2的有1项,分子、分母的和为3的有2项,分子、分母的和为4的有3项,分子、分母的和为99的有98项,分子、分母的和为100的项依次是:199,298,397,5050,5149,991,第50个1是其中第50项,在数列中的序号为1+2+3+98+50=98(1+98)2+50=4901.13.设Sn是数列an的前n项和,且a1=-1,an+1=SnSn+1,则Sn=.答案:-1n解析:由an+1=Sn+1-Sn=SnSn+1,得1Sn-1Sn+1=1,即1Sn+1-1Sn=-1,则1Sn为等差数列,首项为1S1=-1,公差为d=-1,1Sn=-n,Sn=-1n.14.已知等差数列an的公差为2,其前n项和Sn=pn2+2n(nN*).(1)求p的值及an;(2)若bn=2(2n-1)an,记数列bn的前n项和为Tn,求使Tn910成立的最小正整数n的值.解:(1)(方法一)an是等差数列,Sn=na1+n(n-1)2d=na1+n(n-1)22=n2+(a1-1)n.又由已知Sn=pn2+2n,p=1,a1-1=2,a1=3,an=a1+(n-1)d=2n+1,p=1,an=2n+1.(方法二)由已知a1=S1=p+2,S2=4p+4,即a1+a2=4p+4,a2=3p+2.又等差数列的公差为2,a2-a1=2,2p=2,p=1,a1=p+2=3,an=a1+(n-1)d=2n+1,p=1,an=2n+1.(方法三)当n2时,an=Sn-Sn-1=pn2+2n-p(n-1)2+2(n-1)=2pn-p+2,a2=3p+2,由已知a2-a1=2,2p=2,p=1,a1=p+2=3,an=a1+(n-1)d=2n+1,p=1,an=2n+1.(2)由(1)知bn=2(2n-1)(2n+1)=12n-1-12n+1,Tn=b1+b2+b3+bn=11-13+13-15+15-17+12n-1-12n+1=1-12n+1=2n2n+1.Tn910,2n2n+1910,20n18n+9,即n92.nN*,使Tn910成立的最小正整数n的值为5.15.已知数列an满足an+2=qan(q为实数,且q1),nN*,a1=1,a2=2,且a2+a3,a3+a4,a4+a5成等差数列.(1)求q的值和an的通项公式;(2)设bn=log2a2na2n-1,nN*,求数列bn的前n项和.解:(1)由已知,有(a3+a4)-(a2+a3)=(a4+a5)-(a3+a4),即a4-a2=a5-a3,所以a2(q-1)=a3(q-1).又因为q1,故a3=a2=2,由a3=a1q,得q=2.当n=2k-1(kN*)时,an=a2k-1=2k-1=2n-12;当n=2k(kN*)时,an=a2k=2k=2n2.所以,an的通项公式为an=2n-12,n为奇数,2n2,n为偶数.(2)由(1)得bn=log2a2na2n-1=n2n-1.设bn的前n项和为Sn,则Sn=1120+2121+3122+(n-1)12n-2+n12n-1,12Sn=1121+2122+3123+(n-1)12n-1+n12n,上述两式相减,得12Sn=1+12+122+12n-1-n2n=1-12n1-12-n2n=2-22n-n2n,整理得,Sn=4-n+22n-1.所以,数列bn的前n项和为4-n+22n-1,nN*.16.设数列A:a1,a2,aN(N2).如果对小于n(2nN)的每个正整数k都有aka1,则G(A);(3)证明:若数列A满足an-an-11(n=2,3,N),则G(A)的元素个数不小于aN-a1.(1)解G(A)的元素为2和5.(2)证明因为存在an使得ana1,所以iN*|2iN,aia1.记m=miniN*|2iN,aia1,则m2,且对任意正整数km,aka1a1.由(2)知G(A).设G(A)=n1