2025高考数学一轮复习-6.4-数列求和-专项训练【含答案】

2025高考数学一轮复习-6.4-数列求和-专项训练【A级基础巩固】1.数列112,314,518,7116,…,(2n-1)+A.n2+1-12n B.2n2C.n2+1-12n-1 2.数列{an}的前n项和为Sn,已知Sn=1-2+3-4+…+(-1)n-1·n,则S17等于()A.9 B.8 C.17 D.163.数列{an}的通项公式是an=1nA.9 B.99 C.10 D.1004.已知数列{an}的前n项和Sn满足Sn=n2+n,则数列{4aA.67 B.78 C.895.已知函数f(n)=n2,n为奇数,-n2,na100等于()A.0 B.100C.-100 D.102006.有穷数列1,1+2,1+2+4,…,1+2+4+…+2n-1所有项的和为.7.已知数列{an}满足a1=1,且an+1+an=n-1009(n∈N*),则其前2023项之和S2023=.8.已知单调递增的等差数列{an}的前n项和为Sn,且S4=20,a2,a4,a8成等比数列.(1)求数列{an}的通项公式;(2)若bn=2an+1-3n+2,求数列{bn}的前n项和Tn.INCLUDEPICTURE"B组.TIF"INCLUDEPICTURE"E:\\大样\\人教数学\\B组.TIF"INET【B级能力提升】9.若数列{an}满足an=(-1)n-1(12n-1+A.Sn<1 B.Sn>1C.Sn有最小值 D.Sn无最大值10.数列22+122-A.1755 B.11C.1143132 D.1111.已知数列{nan}的前n项和为Sn,且an=2n,则使得Sn-nan+1+50<0的最小正整数n的值为.12.已知等差数列{an}的公差d>0,其前n项和为Sn,若S3=6,且a1,a2,1+a3成等比数列.(1)求数列{an}的通项公式;(2)若bn=an+2-an,求数列{bn13.已知等差数列{an}中,a3=3,a6=6,且bn=a(1)求数列{bn}的通项公式及前20项和;(2)若cn=b2n-1b2n,记数列{cn}的前n项和为Sn,求Sn.INCLUDEPICTURE"B组.TIF"INCLUDEPICTURE"E:\\大样\\人教数学\\B组.TIF"INET【C级应用创新练】14.已知数列{an}的前n项和为Sn,满足S3=3a2+2,且2an=Sn+a1.(1)求{an}的通项公式;(2)数列{bn}满足1b1+2b2+3b3+…+nb参考答案【A级基础巩固】1.解析:该数列的通项公式为an=(2n-1)+12则Sn=[1+3+5+…+(2n-1)]+(12+122+…+12n2.解析:S17=1-2+3-4+5-6+…+15-16+17=1+(-2+3)+(-4+5)+(-6+7)+…+(-14+15)+(-16+17)=1+1+1+…+1=9.故选A.3.解析:因为an=1n+n+1=所以Sn=a1+a2+…+an=(2-1)+(3-2)+…+(n-n-(n+1-n)=n+1-1,令4.解析:当n≥2时,an=Sn-Sn-1=2n,当n=1时,a1=2也符合上式,所以an=2n(n∈N*),所以4anan+1=42n(2n+2)=1n(n+1)=1n-5.解析:由题意,得a1+a2+a3+…+a100=12-22-22+32+32-42-42+52+…+992-1002-1002+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.6.解析:由题意知所求数列的通项为1-2n1-2=2答案:2n+1-2-n7.解析:S2023=a1+(a2+a3)+(a4+a5)+…+(a2022+a2023),又an+1+an=n-1009(n∈N*),且a1=1,所以S2023=1+(2-1009)+(4-1009)+…+(2022-1009)=1+(2+4+6+…+2022)-1009×1011=1+2+20222×答案:30348.解:(1)设数列{an}的公差为d(d>0),由题意得S即4解得a所以an=2+(n-1)·2=2n.(2)由(1)得,an=2n,所以bn=4(n+1)-3n+2,所以Tn=4×2-33+4×3-34+…+4(n+1)-3n+2=4[2+3+…+(n+1)]-(33+34+…+3n+2)=4n·2+n+12-27(1-3INCLUDEPICTURE"B组.TIF"INCLUDEPICTURE"E:\\大样\\人教数学\\B组.TIF"INET【B级能力提升】9.解析:数列{an}满足an=(-1)n-1(12n-当n为偶数时,Sn=1+13-13-15+15+17-…-(1当n为奇数时,Sn=1+13-13-15+15+17-…+(1故当n=2时,Sn最小为45;当n=1时,Sn最大为410.解析:因为(n+1)2+1(n+1)(1-13)+(12-14)+(13-15)+…+(19-111)+(110-11211.解析:Sn=1×21+2×22+…+n·2n,则2Sn=1×22+2×23+…+n·2n+1,两式相减得-Sn=2+22+…+2n-n·2n+1=2(1-2n故Sn=2+(n-1)·2n+1.又an=2n,所以Sn-nan+1+50=2+(n-1)·2n+1-n·2n+1+50=52-2n+1,依题意52-2n+1<0,故最小正整数n的值为5.答案:512.解:(1)依题意,得a即a1(a1+2d+1)=(a1+1+(n-1)=n,即数列{an}的通项公式an=n.(2)bn=an+2-an=n+2-n=n+(12)n,Tn=b1+b2+…+bn=1+12+2+(12=(1+2+…+n)+[12+(12)2+(12)3+…+(12)n=n(n+1)213.解:(1)设等差数列{an}的公差为d,则d=a6-a36-3=1,所以an=a3+(n-3)d=n,从而bn=n+1,n为奇数,(22+24+…+220)=10×(2+20)2+4×((2)因为cn=b2n-1b2n=2n·22n=2n·4n,所以Sn=2×41+4×42+6×43+…+2n·4n,4Sn=2×42+4×43+6×44+…+2(n-1)·4n+2n·4n+1,相减得,-3Sn=2×41+2×42+2×43+…+2×4n-2n·4n+1,所以-3Sn=8(1-4n(23-2n)4n+1-8即Sn=(23n-29)4n+1+INCLUDEPICTURE"B组.TIF"INCLUDEPICTURE"E:\\大样\\人教数学\\B组.TIF"INET【C级应用创新练】14.解:(1)由2an=Sn+a1得Sn=2an-a1,当n≥2时,Sn-1=2an-1-a1,故Sn-Sn-1=2(an-an-1),则an=2an-2an-1,即an=2an-1,所以anan由S3=3a2+2得a1+a3=2a2+2,即a1+a1q2=2a1q+2,所以a1=2,故an=a1qn