（課標專用）天津市2020高考數(shù)學二輪復習 專題能力訓練11 數(shù)列的通項與求和

專題能力訓練11數(shù)列的通項與求和專題能力訓練第28頁 一、能力突破訓練1.已知等差數(shù)列an的前n項和為Sn,若a1=2,a4+a10=28,則S9=()A.45B.90C.120D.75答案:B解析:因為an是等差數(shù)列,設公差為d,所以a4+a10=a1+3d+a1+9d=2a1+12d=4+12d=28,解得d=2.S9=9a1+9×82d=18+36×2=90.故選B.2.已知數(shù)列an是等差數(shù)列,滿足a1+2a2=S5,下列結論錯誤的是()A.S9=0B.S5最小C.S3=S6D.a5=0答案:B解析:由題設可得3a1+2d=5a1+10d2a1+8d=0,即a5=0,所以D中結論正確.由等差數(shù)列的性質可得a1+a9=2a5=0,則S9=9(a1+a9)2=9a5=0,所以A中結論正確.S3-S6=3a1+3d-6a1-15d=-3(a1+4d)=-3a5=0,所以C中結論正確.B中結論是錯誤的.故選B.3.已知數(shù)列an的前n項和Sn=n2-2n-1,則a3+a17=()A.15B.17C.34D.398答案:C解析:Sn=n2-2n-1,a1=S1=12-2-1=-2.當n2時,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=(2×3-3)+(2×17-3)=3+31=34.4.已知數(shù)列an滿足a1=12,an+1=an+12n(nN*),則a2 019=()A.1-122018B.1-122019C.32-122018D.32-122019答案:C解析:數(shù)列an滿足a1=12,an+1=an+12n(nN*),an+1-an=12n,當n2時,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.故選C.5.已知數(shù)列an,構造一個新數(shù)列a1,a2-a1,a3-a2,an-an-1,此數(shù)列是首項為1,公比為13的等比數(shù)列,則數(shù)列an的通項公式為()A.an=32-32×13n,nN*B.an=32+32×13n,nN*C.an=1,n=1,32+32×13n,n>2,且nN*D.an=1,nN*答案:A解析:因為數(shù)列a1,a2-a1,a3-a2,an-an-1,是首項為1,公比為13的等比數(shù)列,所以an-an-1=13n-1,n2.所以當n2時,an=a1+(a2-a1)+(a3-a2)+(an-an-1)=1+13+132+13n-1=1-13n1-13=32-32×13n.又當n=1時,an=32-32×13n=1,則an=32-32×13n,nN*.6.已知數(shù)列an滿足a1=1,an-an+1=nanan+1(nN*),則an=. 答案:2n2-n+2解析:因為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也滿足上式,所以an=2n2-n+2.7.記Sn為數(shù)列an的前n項和.若Sn=2an+1,則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為首項,2為公比的等比數(shù)列,則S6=-(1-26)1-2=-63.8.已知數(shù)列an中,a1=a,an+1=3an+8n+6,若an為遞增數(shù)列,則實數(shù)a的取值范圍為. 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,所以數(shù)列an+4n+5是首項為a+9,公比為3的等比數(shù)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.因為數(shù)列an為遞增數(shù)列,所以an+1>an,即(a+9)·3n-4n-9>(a+9)·3n-1-4n-5,即(a+9)·3n>6恒成立.因為nN*,所以(a+9)·3>6,解得a>-7.9.已知數(shù)列an中,a1=1,an-an-1+1=2n(nN*,n2).(1)求數(shù)列an的通項公式;(2)設bn=14an-1,求數(shù)列bn的通項公式及其前n項和Tn.解:(1)當n2時,因為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滿足上式,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.已知數(shù)列an的前n項和為Sn,且a1=0,對任意nN*,都有nan+1=Sn+n(n+1).(1)求數(shù)列an的通項公式;(2)若數(shù)列bn滿足an+log2n=log2bn,求數(shù)列bn的前n項和Tn.解:(1)(方法一)nan+1=Sn+n(n+1),當n2時,(n-1)an=Sn-1+n(n-1),兩式相減,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.當n=1時,1×a2=S1+1×2,即a2-a1=2.數(shù)列an是以0為首項,2為公差的等差數(shù)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),兩邊同除以n(n+1),得Sn+1n+1-Snn=1.數(shù)列Snn是以S11=0為首項,1為公差的等差數(shù)列,Snn=0+n-1=n-1.Sn=n(n-1).當n2時,an=Sn-Sn-1=n(n-1)-(n-1)·(n-2)=2n-2.又a1=0適合上式,數(shù)列an的通項公式為an=2n-2.(2)an+log2n=log2bn,bn=n·2an=n·22n-2=n·4n-1.Tn=b1+b2+b3+bn-1+bn=40+2×41+3×42+(n-1)×4n-2+n×4n-1,4Tn=41+2×42+3×43+(n-1)×4n-1+n×4n,由-,得-3Tn=40+41+42+4n-1-n×4n=1-4n1-4-n×4n=(1-3n)×4n-13.Tn=19(3n-1)×4n+1.11.設數(shù)列an的前n項和為Sn.已知2Sn=3n+3.(1)求an的通項公式;(2)若數(shù)列bn滿足anbn=log3an,求bn的前n項和Tn.解:(1)因為2Sn=3n+3,所以2a1=3+3,故a1=3.當n>1時,2Sn-1=3n-1+3,此時2an=2Sn-2Sn-1=3n-3n-1=2×3n-1,即an=3n-1,所以an=3,n=1,3n-1,n>1.(2)因為anbn=log3an,所以b1=13,當n>1時,bn=31-nlog33n-1=(n-1)·31-n.所以T1=b1=13;當n>1時,Tn=b1+b2+b3+bn=13+1×3-1+2×3-2+(n-1)×31-n,所以3Tn=1+1×30+2×3-1+(n-1)×32-n,兩式相減,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+32×3n,所以Tn=1312-6n+34×3n.經(jīng)檢驗,當n=1時也適合.綜上可得Tn=1312-6n+34×3n.二、思維提升訓練12.給出數(shù)列11,12,21,13,22,31,1k,2k-1,k1,在這個數(shù)列中,第50個值等于1的項的序號是()A.4 900B.4 901C.5 000D.5 001答案:B解析:根據(jù)條件找規(guī)律,第1個1是分子、分母的和為2,第2個1是分子、分母的和為4,第3個1是分子、分母的和為6,第50個1是分子、分母的和為100,而分子、分母的和為2的有1項,分子、分母的和為3的有2項,分子、分母的和為4的有3項,分子、分母的和為99的有98項,分子、分母的和為100的項依次是:199,298,397,5050,5149,991,第50個1是其中第50項,在數(shù)列中的序號為1+2+3+98+50=98(1+98)2+50=4901.13.設Sn是數(shù)列an的前n項和,且a1=-1,an+1=SnSn+1,則Sn=. 答案:-1n解析:由an+1=Sn+1-Sn=SnSn+1,得1Sn-1Sn+1=1,即1Sn+1-1Sn=-1,則1Sn為等差數(shù)列,首項為1S1=-1,公差為d=-1,1Sn=-n,Sn=-1n.14.已知等差數(shù)列an的公差為2,其前n項和Sn=pn2+2n(nN*).(1)求p的值及an;(2)若bn=2(2n-1)an,記數(shù)列bn的前n項和為Tn,求使Tn>910成立的最小正整數(shù)n的值.解:(1)(方法一)an是等差數(shù)列,Sn=na1+n(n-1)2d=na1+n(n-1)2×2=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.又等差數(shù)列的公差為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)當n2時,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.Tn>910,2n2n+1>910,20n>18n+9,即n>92.nN*,使Tn>910成立的最小正整數(shù)n的值為5.15.已知數(shù)列an滿足an+2=qan(q為實數(shù),且q1),nN*,a1=1,a2=2,且a2+a3,a3+a4,a4+a5成等差數(shù)列.(1)求q的值和an的通項公式;(2)設bn=log2a2na2n-1,nN*,求數(shù)列bn的前n項和.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).又因為q1,故a3=a2=2,由a3=a1·q,得q=2.當n=2k-1(kN*)時,an=a2k-1=2k-1=2n-12;當n=2k(kN*)時,an=a2k=2k=2n2.所以,an的通項公式為an=2n-12,n為奇數(shù),2n2,n為偶數(shù).(2)由(1)得bn=log2a2na2n-1=n2n-1.設bn的前n項和為Sn,則Sn=1×120+2×121+3×122+(n-1)×12n-2+n×12n-1,12Sn=1×121+2×122+3×123+(n-1)×12n-1+n×12n,上述兩式相減,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,數(shù)列bn的前n項和為4-n+22n-1,nN*.16.設數(shù)列A:a1,a2,aN(N2).如果對小于n(2nN)的每個正整數(shù)k都有ak<an,則稱n是數(shù)列A的一個“G時刻”.記G(A)是數(shù)列A的所有“G時刻”組成的集合.(1)對數(shù)列A:-2,2,-1,1,3,寫出G(A)的所有元素;(2)證明:若數(shù)列A中存在an使得an>a1,則G(A);(3)證明:若數(shù)列A滿足an-an-11(n=2,3,N),則G(A)的元素個數(shù)不小于aN-a1.(1)解G(A)的元素為2和5.(2)證明因為存在an使得an>a1,所以iN*|2iN,ai>a1.記m=miniN*|2iN,ai>a1,則m2,且對任意正整數(shù)k<m,aka1<am.因此mG(A).從而G(A).(3)證明當aNa1時,結論成立.以下設aN>a1.由(2)知G(A).設G(A)=n1,n2,np,n1<n2<<np.記n0=1.則an0<an1<an2<<anp.對i=0,1,p,記Gi=kN*|ni<kN,ak>ani.如果Gi,取mi=minGi,則對任何1k<mi,akani<ami.從而miG(A)且mi=ni+1,又因為np是G(A)中的最大元素,所以Gp=.從而對任意npkN,akanp,特別地,aNanp.對i=0,1,p-1,ani+1-1ani.因此ani+1=ani+1-1+(ani+1-ani+1-1)ani+1.所以aN-a1anp-a1=i=1p(ani-ani-1)p.因此G(A)的元素個數(shù)p不小于aN-a1.10