（新教材適用）高中數(shù)學(xué)第四章數(shù)列習(xí)題課數(shù)列求和課后習(xí)題新人教A版選擇性

習(xí)題課——數(shù)列求和課后訓(xùn)練鞏固提升A組1.已知數(shù)列{an}的通項(xiàng)公式an=2n-12n,其前n項(xiàng)和Sn=321A.13 B.10 C.9 D.6解析:∵an=2n-12∴Sn=n121-12n1-1∴n=6.答案:D2.已知Sn=12+34+…+(1)n1n,則S17+S33+S50等于()A.0 B.1 C.1 D.2解析:因?yàn)镾17=(12)+(34)+…+(1516)+17=9,S33=(12)+(34)+…+(3132)+33=17,S50=(12)+(34)+…+(4950)=25,所以S17+S33+S50=1.答案:B3.已知數(shù)列{an}:12,13+23,14+24A.41-1n+1C.11n+1 D解析:∵an=1+2+3+…∴bn=1anan∴數(shù)列{bn}的前n項(xiàng)和Sn=41-12+答案:A4.數(shù)列112,214,318,4116,…的前A.12(n2+n+2)B.12n(n+1)+1C.12(n2n+2)D.12n(n+1)+2解析:112+214+318+…+n+12n=(1+2+…+n)+12+14+…+1答案:A5.數(shù)列12,24,38,…,解析:Sn=12+222+12Sn=122+22①②,得12Sn=12+122+1故Sn=212答案:Sn=216.已知an=ln1+1n(n∈N*),則數(shù)列{an}的前n項(xiàng)和Sn=解析:∵an=lnn+1n=ln(n+1)ln∴Sn=(ln2ln1)+(ln3ln2)+(ln4ln3)+…+[ln(n+1)lnn]=ln(n+1)ln1=ln(n+1).答案:ln(n+1)7.在100以內(nèi)所有能被3整除但不能被7整除的正整數(shù)之和是.

解析:100以內(nèi)所有能被3整除的正整數(shù)的和為S1=3+6+…+99=33×(3+99100以內(nèi)所有能被21整除的正整數(shù)的和為S2=21+42+63+84=210.故100以內(nèi)能被3整除不能被7整除的所有正整數(shù)之和為S1S2=1683210=1473.答案:14738.已知正項(xiàng)數(shù)列{an}滿足an2(2n1)an2n=(1)求數(shù)列{an}的通項(xiàng)公式;(2)令bn=1(n+1)an,求數(shù)列{bn解:(1)由an2(2n1)an2得(an2n)(an+1)=0.因?yàn)閧an}是正項(xiàng)數(shù)列,所以an=2n.(2)由an=2n,bn=1(則bn=12Tn=121-9.已知數(shù)列{an}的前n項(xiàng)和為Sn,且Sn=2n2+n,n∈N*,數(shù)列{bn}滿足an=4log2bn+3,n∈N*.(1)求an,bn;(2)求數(shù)列{anbn}的前n項(xiàng)和Tn.解:(1)由Sn=2n2+n,得當(dāng)n=1時(shí),a1=S1=3;當(dāng)n≥2時(shí),an=SnSn-1=4當(dāng)n=1時(shí)也適合上式,所以an=4n1,n∈N*.由4n1=an=4log2bn+3,得bn=2n1,n∈N*.(2)由(1)知anbn=(4n1)·2n1,n∈N*.所以Tn=3+7×2+11×22+…+(4n1)·2n1,2Tn=3×2+7×22+…+(4n5)·2n1+(4n1)·2n,所以2TnTn=(4n1)·2n[3+4(2+22+…+2n1)]=(4n5)·2n+5.故Tn=(4n5)·2n+5,n∈N*.B組1.若數(shù)列{an}的通項(xiàng)公式為an=2n+2n1,則數(shù)列{an}的前n項(xiàng)和為()A.2n+n21B.2n+1+n21C.2n+1+n22D.2n+n2解析:Sn=(2+22+…+2n)+(1+3+5+…+2n1)=2(1-2n)1-答案:C2.已知數(shù)列{an}的通項(xiàng)公式an=ncosnπ2,其前n項(xiàng)和為Sn,則S2022等于(A.1012 B.2022 C.506 D.0解析:∵函數(shù)y=cosnπ2的周期T=2∴可分四組求和:a1+a5+…+a2021=0,a2+a6+…+a2022=26…2022=506×1012,a3+a7+…+a2019=0,a4+a8+…+a2020=4+8+…+2020=505×1012.故S2022=0506×1012+0+505×1012=1012.答案:A3.已知函數(shù)f(n)=n2,n為奇數(shù),-n2,n為偶數(shù),且an=f(n)+f(n+1),則a1+aA.0 B.100C.100 D.10200解析:由題意,得a1+a2+a3+…+a100=122222+32+324242+52+…+99210021002+1012=(1+2)+(3+2)(4+3)+…(99+100)+(101+100)=(1+2+…+99+100)+(2+3+…+100+101)=50×101+50×103=100.故選B.答案:B4.(多選題)已知數(shù)列{an}滿足a1=1,an+an+1=2n+1,n∈N*,Sn是數(shù)列1an的前n項(xiàng)和,則下列結(jié)論正確的是(A.S2n1=(2n1)·1B.S2n=12SC.S2n≥32-D.S2n≥Sn+1解析:易知a2=2,由an+an+1=2n+1,an+1+an+2=2n+3,兩式相減,得an+2an=2,即此數(shù)列每隔一項(xiàng)成等差數(shù)列,由a1=1,可得數(shù)列{an}的奇數(shù)項(xiàng)為1,3,5,…,由a2=2,可得其偶數(shù)項(xiàng)為2,4,6,…,故an=n.A.令n=2,S2n1=S3=116,(2n1)·1an=32,S2n1≠B.令n=1,S2n=S2=1+12=32,12Sn=12S1=12C.∵S2n12Sn=1+13+15+…+12n-1∴1+13+15+…+12n-1D.S2nSn=1n+1+1n+2+…+12n,設(shè)f(n則f(n+1)f(n)=12n∴f(n+1)>f(n),∴f(n)單調(diào)遞增,∴f(n)≥f(1)=12,∴S2nSn≥1∴S2n≥Sn+12,故D正確答案:CD5.已知等比數(shù)列{an}的各項(xiàng)都為正數(shù),且當(dāng)n≥3時(shí),a4·a2n4=102n,則數(shù)列l(wèi)ga1,2lga2,22lga3,23lga4,…,2n1lgan的前n項(xiàng)和Sn=.

解析:∵{an}是等比數(shù)列,∴a4a2n4=an2=102n,又?jǐn)?shù)列{a∴an=10n,∴2n1lgan=n·2n1.利用錯(cuò)位相減法,求得Sn=1+(n1)·2n.答案:1+(n1)·2n6.設(shè)數(shù)列{an}滿足a1=2,an+1an=3·22n1.(1)求數(shù)列{an}的通項(xiàng)公式;(2)令bn=nan,求數(shù)列{bn}的前n項(xiàng)和Sn.解:(1)由已知,當(dāng)n≥1時(shí),an+1=[(an+1an)+(anan1)+…+(a2a1)]+a1=3(22n1+22n3+…+2)+2=22(n+1)1.而a1=2符合上式,故數(shù)列{an}的通項(xiàng)公式為an=22n1.(2)由bn=nan=n·22n1,知Sn=1×2+2×23+3×25+…+n·22n1,①?gòu)亩?2·Sn=1×23+2×25+3×27+…+n·22n+1.②①②,得(122)Sn=2+23+25+…+22n1n·22n+1,即Sn=19[(3n1)22n+1+2]7.已知等差數(shù)列{an}滿足a3=7,a5+a7=26,{an}的前n項(xiàng)和為Sn.(1)求an及Sn;(2)令bn=1an2-1(