2022高考數(shù)學(xué)大二輪復(fù)習(xí) 專題四 數(shù)列 專題能力訓(xùn)練12 數(shù)列的通項(xiàng)與求和 理

2022高考數(shù)學(xué)大二輪復(fù)習(xí) 專題四 數(shù)列 專題能力訓(xùn)練12 數(shù)列的通項(xiàng)與求和 理1.已知等差數(shù)列an的前n項(xiàng)和為Sn,若a1=2,a4+a10=28,則S9=()A.45B.90C.120D.752.已知數(shù)列an是等差數(shù)列,滿足a1+2a2=S5,下列結(jié)論錯(cuò)誤的是()A.S9=0B.S5最小C.S3=S6D.a5=03.已知數(shù)列an的前n項(xiàng)和Sn=n2-2n-1,則a3+a17=()A.15B.17C.34D.3984.已知函數(shù)f(x)滿足f(x+1)= +f(x)(xR),且f(1)=,則數(shù)列f(n)(nN*)前20項(xiàng)的和為()A.305B.315C.325D.3355.已知數(shù)列an,構(gòu)造一個(gè)新數(shù)列a1,a2-a1,a3-a2,an-an-1,此數(shù)列是首項(xiàng)為1,公比為的等比數(shù)列,則數(shù)列an的通項(xiàng)公式為()A.an=,nN*B.an=,nN*C.an=D.an=1,nN*6.已知數(shù)列an滿足a1=1,an-an+1=nanan+1(nN*),則an=. 7.(2018全國(guó),理14)記Sn為數(shù)列an的前n項(xiàng)和.若Sn=2an+1,則S6=. 8.已知Sn是等差數(shù)列an的前n項(xiàng)和,若a1=-2 017,=6,則S2 017=. 9.已知在數(shù)列an中,a1=1,an+1=an+2n+1,且nN*.(1)求數(shù)列an的通項(xiàng)公式;(2)令bn=,數(shù)列bn的前n項(xiàng)和為Tn.如果對(duì)于任意的nN*,都有Tn>m,求實(shí)數(shù)m的取值范圍.10.已知數(shù)列an的前n項(xiàng)和為Sn,且a1=0,對(duì)任意nN*,都有nan+1=Sn+n(n+1).(1)求數(shù)列an的通項(xiàng)公式;(2)若數(shù)列bn滿足an+log2n=log2bn,求數(shù)列bn的前n項(xiàng)和Tn.11.設(shè)數(shù)列an的前n項(xiàng)和為Sn .已知2Sn=3n+3.(1)求an的通項(xiàng)公式;(2)若數(shù)列bn滿足anbn=log3an,求bn的前n項(xiàng)和Tn.二、思維提升訓(xùn)練12.給出數(shù)列, ,在這個(gè)數(shù)列中,第50個(gè)值等于1的項(xiàng)的序號(hào)是()A.4 900B.4 901C.5 000D.5 00113.設(shè)Sn是數(shù)列an的前n項(xiàng)和,且a1=-1,an+1=SnSn+1,則Sn=. 14.已知等差數(shù)列an的公差為2,其前n項(xiàng)和Sn=pn2+2n(nN*).(1)求p的值及an;(2)若bn=,記數(shù)列bn的前n項(xiàng)和為Tn,求使Tn>成立的最小正整數(shù)n的值.15.已知數(shù)列an滿足an+2=qan(q為實(shí)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(xiàng)公式;(2)設(shè)bn=,nN*,求數(shù)列bn的前n項(xiàng)和.16.設(shè)數(shù)列A:a1,a2,aN(N2).如果對(duì)小于n(2nN)的每個(gè)正整數(shù)k都有ak<an,則稱n是數(shù)列A的一個(gè)“G時(shí)刻”.記G(A)是數(shù)列A的所有“G時(shí)刻”組成的集合.(1)對(duì)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(gè)數(shù)不小于aN-a1.專題能力訓(xùn)練12數(shù)列的通項(xiàng)與求和一、能力突破訓(xùn)練1.B解析 因?yàn)閍n是等差數(shù)列,設(shè)公差為d,所以a4+a10=a1+3d+a1+9d=2a1+12d=4+12d=28,解得d=2.所以S9=9a1+d=18+36×2=90.故選B.2.B解析 由題設(shè)可得3a1+2d=5a1+10d2a1+8d=0,即a5=0,所以D中結(jié)論正確.由等差數(shù)列的性質(zhì)可得a1+a9=2a5=0,則S9=9a5=0,所以A中結(jié)論正確.S3-S6=3a1+3d-6a1-15d=-3(a1+4d)=-3a5=0,所以C中結(jié)論正確.B中結(jié)論是錯(cuò)誤的.故選B.3.C解析 Sn=n2-2n-1,a1=S1=12-2-1=-2.當(dāng)n2時(shí),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=a3+a17=(2×3-3)+(2×17-3)=3+31=34.4.D解析 f(1)=,f(2)=,f(3)=,f(n)=+f(n-1),f(n)是以為首項(xiàng),為公差的等差數(shù)列.S20=20=335.5.A解析 因?yàn)閿?shù)列a1,a2-a1,a3-a2,an-an-1,是首項(xiàng)為1,公比為的等比數(shù)列,所以an-an-1=,n2.所以當(dāng)n2時(shí),an=a1+(a2-a1)+(a3-a2)+(an-an-1)=1+=又當(dāng)n=1時(shí),an=1,則an=,nN*.6解析 因?yàn)閍n-an+1=nanan+1,所以=n,+=(n-1)+(n-2)+3+2+1+=+1=(n2).所以an=(n2).又a1=1也滿足上式,所以an=7.-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(xiàng),2為公比的等比數(shù)列,則S6=-63.8.-2 017解析 Sn是等差數(shù)列an的前n項(xiàng)和,是等差數(shù)列,設(shè)其公差為d.=6,6d=6,d=1.a1=-2 017,=-2 017.=-2 017+(n-1)×1=-2 018+n.S2 017=(-2 018+2 017)×2 017=-2 017.故答案為-2 017.9.解 (1)an+1=an+2n+1,an+1-an=2n+1,an-an-1=2n-1,an=a1+(a2-a1)+(a3-a2)+(an-an-1)=1+3+5+(2n-1)=n2.(2)由(1)知,bn=,Tn=+=1-,數(shù)列Tn是遞增數(shù)列,最小值為1-,只需要>m,m的取值范圍是10.解 (1)(方法一)nan+1=Sn+n(n+1),當(dāng)n2時(shí),(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(dāng)n=1時(shí),1×a2=S1+1×2,即a2-a1=2.數(shù)列an是以0為首項(xiàng),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=1.數(shù)列是以=0為首項(xiàng),1為公差的等差數(shù)列,=0+n-1=n-1.Sn=n(n-1).當(dāng)n2時(shí),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(xiàng)公式為an=2n-2.(2)an+log2n=log2bn,bn=n=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=-n×4n=Tn=(3n-1)×4n+1.11.解 (1)因?yàn)?Sn=3n+3,所以2a1=3+3,故a1=3.當(dāng)n>1時(shí),2Sn-1=3n-1+3,此時(shí)2an=2Sn-2Sn-1=3n-3n-1=2×3n-1,即an=3n-1,所以an=(2)因?yàn)閍nbn=log3an,所以b1=,當(dāng)n>1時(shí),bn=31-nlog33n-1=(n-1)·31-n.所以T1=b1=;當(dāng)n>1時(shí),Tn=b1+b2+b3+bn=+(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=+(30+3-1+3-2+32-n)-(n-1)×31-n=-(n-1)×31-n=,所以Tn=經(jīng)檢驗(yàn),當(dāng)n=1時(shí)也適合.綜上可得Tn=二、思維提升訓(xùn)練12.B解析 根據(jù)條件找規(guī)律,第1個(gè)1是分子、分母的和為2,第2個(gè)1是分子、分母的和為4,第3個(gè)1是分子、分母的和為6,第50個(gè)1是分子、分母的和為100,而分子、分母的和為2的有1項(xiàng),分子、分母的和為3的有2項(xiàng),分子、分母的和為4的有3項(xiàng),分子、分母的和為99的有98項(xiàng),分子、分母的和為100的項(xiàng)依次是:,第50個(gè)1是其中第50項(xiàng),在數(shù)列中的序號(hào)為1+2+3+98+50=+50=4 901.13.-解析 由an+1=Sn+1-Sn=SnSn+1,得=1,即=-1,則為等差數(shù)列,首項(xiàng)為=-1,公差為d=-1,=-n,Sn=-14.解 (1)(方法一)an是等差數(shù)列,Sn=na1+d=na1+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(dāng)n2時(shí),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=,Tn=b1+b2+b3+bn=+=1-Tn>,20n>18n+9,即n>nN*,使Tn>成立的最小正整數(shù)n的值為5.15.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?yàn)閝1,故a3=a2=2,由a3=a1·q,得q=2.當(dāng)n=2k-1(kN*)時(shí),an=a2k-1=2k-1=;當(dāng)n=2k(kN*)時(shí),an=a2k=2k=所以,an的通項(xiàng)公式為an=(2)由(1)得bn=設(shè)bn的前n項(xiàng)和為Sn,則Sn=1+2+3+(n-1)+n,Sn=1+2+3+(n-1)+n,上述兩式相減,得Sn=1+=2-,整理得,Sn=4-所以,數(shù)列bn的前n項(xiàng)和為4-,nN*.16.(1)解 G(A)的元素為2和5.(2)證明 因?yàn)榇嬖赼n使得an>a1,所以iN*|2iN,ai>a1.記m=miniN*|2iN,ai>a1,則m2,且對(duì)任意正整數(shù)k<m,aka1<am.因此mG(A).從而G(A).(3)證明 當(dāng)aNa1時(shí),結(jié)論成立.以下設(shè)aN>a1.由(2)知G(A).設(shè)G(A)=n1,n2,np,n1<n2<<np.記n0=1.則<<對(duì)i=0,1,p,記Gi=kN*|ni<kN,ak>.如果Gi,取mi=minGi,則對(duì)任何1k<mi,ak從而miG(A)且mi=ni+1,又因?yàn)閚p是G(A)中的最大元素,所以Gp=.從而對(duì)任意npkN,ak,特別地,aN對(duì)i=0,1,p-1,因此+()+1.所以aN-a1-a1=)p.因此G(A)的元素個(gè)數(shù)p不小于aN-a1.