（新教材适用）高中数学第四章数列习题课数列求和课后习题新人教A版选择性

习题课——数列求和课后训练巩固提升A组1.已知数列{an}的通项公式an=2n-12n,其前n项和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,则S17+S33+S50等于()A.0 B.1 C.1 D.2解析:因为S17=(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.已知数列{an}:12,13+23,14+24A.41-1n+1C.11n+1 D解析:∵an=1+2+3+…∴bn=1anan∴数列{bn}的前n项和Sn=41-12+答案:A4.数列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.数列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*),则数列{an}的前n项和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以内所有能被3整除但不能被7整除的正整数之和是.

解析:100以内所有能被3整除的正整数的和为S1=3+6+…+99=33×(3+99100以内所有能被21整除的正整数的和为S2=21+42+63+84=210.故100以内能被3整除不能被7整除的所有正整数之和为S1S2=1683210=1473.答案:14738.已知正项数列{an}满足an2(2n1)an2n=(1)求数列{an}的通项公式;(2)令bn=1(n+1)an,求数列{bn解:(1)由an2(2n1)an2得(an2n)(an+1)=0.因为{an}是正项数列,所以an=2n.(2)由an=2n,bn=1(则bn=12Tn=121-9.已知数列{an}的前n项和为Sn,且Sn=2n2+n,n∈N*,数列{bn}满足an=4log2bn+3,n∈N*.(1)求an,bn;(2)求数列{anbn}的前n项和Tn.解:(1)由Sn=2n2+n,得当n=1时,a1=S1=3;当n≥2时,an=SnSn-1=4当n=1时也适合上式,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组1.若数列{an}的通项公式为an=2n+2n1,则数列{an}的前n项和为()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.已知数列{an}的通项公式an=ncosnπ2,其前n项和为Sn,则S2022等于(A.1012 B.2022 C.506 D.0解析:∵函数y=cosnπ2的周期T=2∴可分四组求和: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.已知函数f(n)=n2,n为奇数,-n2,n为偶数,且an=f(n)+f(n+1),则a1+aA.0 B.100C.100 D.10200解析:由题意,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.故选B.答案:B4.(多选题)已知数列{an}满足a1=1,an+an+1=2n+1,n∈N*,Sn是数列1an的前n项和,则下列结论正确的是(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+3,两式相减,得an+2an=2,即此数列每隔一项成等差数列,由a1=1,可得数列{an}的奇数项为1,3,5,…,由a2=2,可得其偶数项为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,设f(n则f(n+1)f(n)=12n∴f(n+1)>f(n),∴f(n)单调递增,∴f(n)≥f(1)=12,∴S2nSn≥1∴S2n≥Sn+12,故D正确答案:CD5.已知等比数列{an}的各项都为正数,且当n≥3时,a4·a2n4=102n,则数列lga1,2lga2,22lga3,23lga4,…,2n1lgan的前n项和Sn=.

解析:∵{an}是等比数列,∴a4a2n4=an2=102n,又数列{a∴an=10n,∴2n1lgan=n·2n1.利用错位相减法,求得Sn=1+(n1)·2n.答案:1+(n1)·2n6.设数列{an}满足a1=2,an+1an=3·22n1.(1)求数列{an}的通项公式;(2)令bn=nan,求数列{bn}的前n项和Sn.解:(1)由已知,当n≥1时,an+1=[(an+1an)+(anan1)+…+(a2a1)]+a1=3(22n1+22n3+…+2)+2=22(n+1)1.而a1=2符合上式,故数列{an}的通项公式为an=22n1.(2)由bn=nan=n·22n1,知Sn=1×2+2×23+3×25+…+n·22n1,①从而22·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.已知等差数列{an}满足a3=7,a5+a7=26,{an}的前n项和为Sn.(1)求an及Sn;(2)令bn=1an2-1(