課時(shí)規(guī)范練28數(shù)列求和基礎(chǔ)鞏固組1.(2021廣東汕頭二中高三模擬)設(shè)數(shù)列{an}為等比數(shù)列,數(shù)列{bn}為等差數(shù)列,且Sn為數(shù)列{bn}的前n項(xiàng)和,若a2=1,a10=16,且a6=b6,則S11=()A.20 B.30 C.44 D.882.(2021江西南昌高三月考)Sn=12+222+3A.2n+1-nC.2n-n+13.(2021廣東肇慶高三月考)在數(shù)列{an}中,a1=2,(1-an)·an+1=1,Sn是數(shù)列{an}的前n項(xiàng)和,則S2021=()A.4042 B.2021C.20232 D.4.(2021河北衡水高三期末)在數(shù)列{an}中,an+an+1=2n,Sn為其前n項(xiàng)和,若a1=a4,則S101=()A.4882 B.5100C.5102 D.52125.(2021江蘇南京高三月考)已知數(shù)列{an}的通項(xiàng)公式為an=n(n+1)!,則其前A.1-1(n+1)! BC.2-1n! D.26.(2021天津和平高三期中)函數(shù)y=x2在點(diǎn)(n,n2)(n∈N*)處的切線記為ln,直線ln,ln+1及x軸圍成的三角形的面積記為Sn,則1S1+1S2+17.(2021湖南長(zhǎng)郡中學(xué)高三模擬)在數(shù)列{an}中,a1=1,a2=2,且an+2=2an+1+3an,設(shè)bn=an+1+an.(1)求證:數(shù)列{bn}是等比數(shù)列,并求數(shù)列{bn}的通項(xiàng)公式;(2)若數(shù)列{bn}的前n項(xiàng)和為Sn,數(shù)列94bnSnSn+1的前n項(xiàng)和為Tn8.(2021廣東廣州高三模擬)已知數(shù)列{an}的前n項(xiàng)和Sn滿足Sn=3n2-n2,數(shù)列{log3bn}是公差為-1的等差數(shù)列,(1)求數(shù)列{an},{bn}的通項(xiàng)公式;(2)設(shè)cn=a2n+1+b2n+1,求數(shù)列{cn}的前n項(xiàng)和Tn.綜合提升組9.(2021陜西西安高三期中)已知數(shù)列{an}的前n項(xiàng)和為Sn=2-3n,則此數(shù)列奇數(shù)項(xiàng)的前m項(xiàng)和為()A.94?9mC.94?9m-10.(2021貴溪實(shí)驗(yàn)中學(xué)高三月考)對(duì)于實(shí)數(shù)x,[x]表示不超過(guò)x的最大整數(shù).已知數(shù)列{an}的通項(xiàng)公式為an=1n+1+n,前n項(xiàng)和為Sn,則[S1]+[S2]+…+[S40]A.105 B.120 C.125 D.13011.在數(shù)列{an}中,a1=1,nan+1=(n+1)an+n(n+1),若bn=ancos2nπ3,且數(shù)列{bn}的前n項(xiàng)和為Sn,則S11=(A.64 B.80 C.-64 D.-8012.在數(shù)列{an}中,a1=2,ap+q=apaq(p,q∈N*),記bm為數(shù)列{an}中在區(qū)間(0,m](m∈N*)內(nèi)的項(xiàng)的個(gè)數(shù),則數(shù)列{bm}的前150項(xiàng)和S150=.

13.(2021山東聊城一中高三一模)已知等差數(shù)列{an}的首項(xiàng)為2,公差為d,前n項(xiàng)和為Sn,各項(xiàng)均為正數(shù)的等比數(shù)列{bn}的首項(xiàng)為1,公比為q,且滿足前n項(xiàng)和為a3=2b2,S5=b2+b4.(1)求數(shù)列{an},數(shù)列{bn}的通項(xiàng)公式;(2)設(shè)cn=(-1)nlog3Sn+log3bn,求數(shù)列{cn}的前26項(xiàng)和.14.(2021江蘇揚(yáng)州中學(xué)高三模擬)在等差數(shù)列{an}中,a1+3,a3,a4成等差數(shù)列,且a1,a3,a8成等比數(shù)列.(1)求數(shù)列{an}的通項(xiàng)公式;(2)在任意相鄰兩項(xiàng)ak與ak+1(k=1,2,…)之間插入2k個(gè)2,使它們和原數(shù)列的項(xiàng)構(gòu)成一個(gè)新的數(shù)列{bn},求數(shù)列{bn}的前200項(xiàng)和T200.創(chuàng)新應(yīng)用組15.(2021浙江高三開學(xué)考試)在數(shù)列{an}中,a1=3,an+1=3an-4n,若bn=4n2+8n+5anan+1,且數(shù)列{bn}的前nA.n1+22n+3 B.C.n1+16n+9 D.n1+2616.(2021河北滄州高三期中)在數(shù)列{an}中,a1=-1,an+1+(-1)n·an=11-2n,記數(shù)列{an}的前n項(xiàng)和為Sn.(1)求S101的值;(2)求Sn的最大值.17.(2021天津靜海一中高三月考)已知等比數(shù)列{an}的前n項(xiàng)和為Sn,公比q>0,S2=2a2-2,S3=a4-2,數(shù)列{an}滿足a2=4b1,nbn+1-(n+1)bn=n2+n(n∈N*).(1)求數(shù)列{an}的通項(xiàng)公式;(2)證明數(shù)列bnn為等差數(shù)列;(3)設(shè)數(shù)列{cn}的通項(xiàng)公式為cn=-anbn2,n為奇數(shù),an

課時(shí)規(guī)范練28數(shù)列求和1.C解析:∵數(shù)列{an}為等比數(shù)列,∴a62=a2a10=16.又a6=a2q4>0,∴b6=a6=4.又?jǐn)?shù)列{bn}為等差數(shù)列,∴S11=a1+a112×11=11a2.A解析:由Sn=12+222+323+…+n2n,可得12Sn=122+223+…+n-12n3.D解析:因?yàn)閍1=2,(1-an)an+1=1,所以由(1-a1)a2=1得a2=-1,進(jìn)而得a3=12,a4=2=a1,可得an+3=an,所以a1+a2+…+a2021=673(a1+a2+a3)+a1+a2=673×32+1=20212.4.C解析:因?yàn)閍n+an+1=2n,①所以an+1+an+2=2n+2,②由②-①得an+2-an=2,所以數(shù)列{an}奇數(shù)項(xiàng)與偶數(shù)項(xiàng)均成公差為2的等差數(shù)列.當(dāng)n為奇數(shù)時(shí),an=a1+n-12×2=n+a1-1;當(dāng)an=a2+n-22×2=n+a2-2=n+(2-a1)-2=n-a1.又因?yàn)閍1=a4,所以a1=4-a1,得a所以an=n+1,n為奇數(shù),n-2,n為偶數(shù),所以S101=(a1+…+a101=512(2+102)+502(0+98)=5102.故選5.A解析:因?yàn)閍n=n(n+1)!=n+1-1(n+1)!=1n!?16.4nn+1解析:因?yàn)閥'=2x,所以在點(diǎn)(n,n2)(n∈N*)處的切線的斜率為k=2n,所以切線方程為y-n2=2n(x-n),即ln的方程為y=2nx-n2,令y=0,得x=n2.ln+1:y=2(n+1)x-(n+1)2,令y=0,得x=n+12.由y=2(n+1)x-(n+1)2,y=2nx-n2得x=2n+12,y=n2+n,直線ln,ln+1的交點(diǎn)坐標(biāo)為2n+12,n2+n,所以直線ln,ln+1及x軸圍成的三角形的面積為Sn=12n+12?n2(n2+n)=14n(n+1),7.證明(1)由題意得an+2+an+1=3an+1+3an,bn≠0,∴bn+1=3bn,即bn+1bn=3,且b1=a2+a1=2∴數(shù)列{bn}是以3為首項(xiàng),3為公比的等比數(shù)列,∴bn=3n.(2)由(1)得,Sn=3(∴94b∴Tn=1213-1=1212?8.解(1)當(dāng)n=1時(shí),a1=S1=3-1當(dāng)n≥2,an=Sn-Sn-1=3n2-n所以n=1滿足n≥2時(shí)的情況,所以an=3n-2(n∈N*).因?yàn)閘og3bn=log3b1+(n-1)×(-1)=1-n,所以bn=31-n.(2)因?yàn)閏n=a2n+1+b2n+1=3(2n+1)-2+31-(2n+1)=6n+1+19n,所以Tn=(7+6n+1)n2+19[1-(19)

n]9.B解析:當(dāng)n≥2時(shí),an=Sn-Sn-1=(2-3n)-(2-3n-1)=-2·3n-1.因?yàn)楫?dāng)n=1時(shí),a1=-1不滿足,所以數(shù)列{an}從第2項(xiàng)開始成等比數(shù)列.又a3=-18,則數(shù)列{an}的奇數(shù)項(xiàng)構(gòu)成的數(shù)列的前m項(xiàng)和Tm=-18(1-9m-110.B解析:因?yàn)閍n=1n+1+n=n+1?n,所以Sn[S1]=[1+1-1]=0,[S2]=[2+1-1]=0,[S3]=[3+1-1]=1,[S4]=[4+1-1]=1,…,[S7]=[7+1-1]=1,[S8]=[8+1-1]=2,[S9]=[9+1-1]=2,…,[S14]=[14+1-1]=2,[S15]=[15+1-1]=3,[S16]=[16+1-1]=3,…,[S23]=[23+1-1]=3,[S24]=[24+1-1]=4,[S25]=[25+1-1]=4,…,[S34]=[34+1-1]=4,[S35]=[35+1-1]=5,[S36]=[36+1-1]=5,…,[S40]=[40+1-1]=5.故[S1]+[S2]+…+[S40]=0×2+1×5+2×7+3×9+4×11+5×6=120.故選B.11.C解析:已知nan+1=(n+1)an+n(n+1),則an+1n+1=ann+1,可得數(shù)列ann是首項(xiàng)為1,公差為1的等差數(shù)列,即有ann=n,即an=n2,則b則S11=-12(12+22+42+52+72+82+102+112)+(32+62+92=-12(12+22-32-32+42+52-62-62+72+82-92-92+102+112=-12×(5+23+41+59)=-64.故選C12.803解析:令p=1,q=n,則a1+n=a1an=2an,所以數(shù)列{an}是首項(xiàng)為2,公比為2的等比數(shù)列,所以an=2n.當(dāng)m=1時(shí),b1=0.當(dāng)2n≤m<2n+1時(shí),bm=n,所以S150=b1+(b2+b3)+(b4+b5+b6+b7)+…+(b64+b65+…+b127)+(b128+b129+…+b150)=0+1×2+2×22+3×23+4×24+5×25+6×26+7×(150-127)=803.13.解(1)由題意得a∴q3-9q=0.∵{bn}是各項(xiàng)均為正數(shù)的等比數(shù)列,∴q=3,則d=2,∴an=2+2(n-1)=2n,bn=1·3n-1=3n-1.(2)由(1)得,Sn=12n(2+2n)=n(n+則cn=(-1)nlog3[n(n+1)]+log33n-1=(-1)nlog3n+(-1)nlog(n+1)+n-1,∴數(shù)列{cn}的前26項(xiàng)和為T26=(-log31-log32+0)+(log32+log33+1)+(-log33-log34+2)+…+(-log325-log326+24)+(log326+log327+25)=-log31+log327+26×(0+25)2=3+14.解(1)設(shè)等差數(shù)列{an}的公差為d.由題意得a1+3+a4=2a3,即2a1+3+3d=2a1+4d,解得d=3.又a1a8=a32,即a1(a1+7×3)=(a1+2×3)2,解得a1所以an=3n+1.(2)在數(shù)列{bn}中,ak+1前面(包括ak+1)共有2+22+23+…+2k+(k+1)=2k+1+k-1項(xiàng).令2k+1+k-1≤200(k=1,2,…),則k≤6,所以a1,a2,a3,a4,a5,a6,a7出現(xiàn)在數(shù)列{bn}的前200項(xiàng)中.當(dāng)k=6時(shí),2k+1+k-1=133,所以a7前面(包括a7)共有133項(xiàng),所以a7后面(不包括a7)還有67個(gè)2.所以T200=(4+7+…+22)+2(2+22+23+…+26+67)=91+386=477.15.D解析:由an+1=3an-4n,可得an+1-[2(n+1)+1]=3[an-(2n+1)].∵a1=3,∴a1-(2×1+1)=0,則可得數(shù)列{an-(2n+1)}為常數(shù)列0,即an-(2n+1)=0,∴an=2n+1,∴bn=4n2+8n+5(2n+1∴Sn=n+13?15+15?17+…+12n+1?16.解(1)當(dāng)n=2k(k∈N*)時(shí),由an+1+(-1)n·an=11-2n可得,a2k+1+a2k=11-4k,①所以a2+a3=11-4×1,a4+a5=11-4×2,…,a100+a101=11-4×50,因此S101=a1+11×50-4×(1+50)×50(2)當(dāng)n=2k-1(k∈N*)時(shí),a2k-a2k-1=13-4k.②①式減去②式得a2k+1+a2k-1=-2.又a1=-1,所以a2k-1=-1(k∈N*),a2k=12-4k,可得a2=8,a4=4,a6=0.當(dāng)k>3(k∈N*)時(shí),a2k<0,a2k-1=-1<0,則當(dāng)n>6(n∈N*)時(shí),an<0.又S1=-1,S2=7,S3=6,S4=10,S5=S6=9,當(dāng)n>6(n∈N*)時(shí),Sn>Sn+1,所以當(dāng)n=4時(shí),Sn取得最大值,且S4=10.17.(1)解因?yàn)榈缺葦?shù)列{an}的前n項(xiàng)和為Sn,公比q>0,S2=2a2-2,S3=a4-2,所以S3-S2=a4-2a2=a3,整理得a2q2-2a2=a2q.又a2≠0,所以q2-q-2=0.由于q>0,解得q=2.由于a1+a2=2a2-2,解得a1=2,所以an=2n.(2)證明數(shù)列{an}滿足a2=4b1,解得b1=1.由于nbn+1-(n+1)bn=n2+n,所以bn+1所