2025高考數(shù)學(xué)一輪復(fù)習(xí)-6.4-數(shù)列求和-專項(xiàng)訓(xùn)練【含答案】

2025高考數(shù)學(xué)一輪復(fù)習(xí)-6.4-數(shù)列求和-專項(xiàng)訓(xùn)練【A級(jí)基礎(chǔ)鞏固】1.數(shù)列112,314,518,7116,…,(2n-1)+A.n2+1-12n B.2n2C.n2+1-12n-1 2.數(shù)列{an}的前n項(xiàng)和為Sn,已知Sn=1-2+3-4+…+(-1)n-1·n,則S17等于()A.9 B.8 C.17 D.163.數(shù)列{an}的通項(xiàng)公式是an=1nA.9 B.99 C.10 D.1004.已知數(shù)列{an}的前n項(xiàng)和Sn滿足Sn=n2+n,則數(shù)列{4aA.67 B.78 C.895.已知函數(shù)f(n)=n2,n為奇數(shù),-n2,na100等于()A.0 B.100C.-100 D.102006.有窮數(shù)列1,1+2,1+2+4,…,1+2+4+…+2n-1所有項(xiàng)的和為.7.已知數(shù)列{an}滿足a1=1,且an+1+an=n-1009(n∈N*),則其前2023項(xiàng)之和S2023=.8.已知單調(diào)遞增的等差數(shù)列{an}的前n項(xiàng)和為Sn,且S4=20,a2,a4,a8成等比數(shù)列.(1)求數(shù)列{an}的通項(xiàng)公式;(2)若bn=2an+1-3n+2,求數(shù)列{bn}的前n項(xiàng)和Tn.INCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【B級(jí)能力提升】9.若數(shù)列{an}滿足an=(-1)n-1(12n-1+A.Sn<1 B.Sn>1C.Sn有最小值 D.Sn無(wú)最大值10.數(shù)列22+122-A.1755 B.11C.1143132 D.1111.已知數(shù)列{nan}的前n項(xiàng)和為Sn,且an=2n,則使得Sn-nan+1+50<0的最小正整數(shù)n的值為.12.已知等差數(shù)列{an}的公差d>0,其前n項(xiàng)和為Sn,若S3=6,且a1,a2,1+a3成等比數(shù)列.(1)求數(shù)列{an}的通項(xiàng)公式;(2)若bn=an+2-an,求數(shù)列{bn13.已知等差數(shù)列{an}中,a3=3,a6=6,且bn=a(1)求數(shù)列{bn}的通項(xiàng)公式及前20項(xiàng)和;(2)若cn=b2n-1b2n,記數(shù)列{cn}的前n項(xiàng)和為Sn,求Sn.INCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【C級(jí)應(yīng)用創(chuàng)新練】14.已知數(shù)列{an}的前n項(xiàng)和為Sn,滿足S3=3a2+2,且2an=Sn+a1.(1)求{an}的通項(xiàng)公式;(2)數(shù)列{bn}滿足1b1+2b2+3b3+…+nb參考答案【A級(jí)基礎(chǔ)鞏固】1.解析:該數(shù)列的通項(xiàng)公式為an=(2n-1)+12則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.故選A.3.解析:因?yàn)閍n=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:當(dāng)n≥2時(shí),an=Sn-Sn-1=2n,當(dāng)n=1時(shí),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:由題意,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.故選B.6.解析:由題意知所求數(shù)列的通項(xiàng)為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)設(shè)數(shù)列{an}的公差為d(d>0),由題意得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組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【B級(jí)能力提升】9.解析:數(shù)列{an}滿足an=(-1)n-1(12n-當(dāng)n為偶數(shù)時(shí),Sn=1+13-13-15+15+17-…-(1當(dāng)n為奇數(shù)時(shí),Sn=1+13-13-15+15+17-…+(1故當(dāng)n=2時(shí),Sn最小為45;當(dāng)n=1時(shí),Sn最大為410.解析:因?yàn)?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,則2Sn=1×22+2×23+…+n·2n+1,兩式相減得-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,依題意52-2n+1<0,故最小正整數(shù)n的值為5.答案:512.解:(1)依題意,得a即a1(a1+2d+1)=(a1+1+(n-1)=n,即數(shù)列{an}的通項(xiàng)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)設(shè)等差數(shù)列{an}的公差為d,則d=a6-a36-3=1,所以an=a3+(n-3)d=n,從而bn=n+1,n為奇數(shù),(22+24+…+220)=10×(2+20)2+4×((2)因?yàn)閏n=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,相減得,-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組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【C級(jí)應(yīng)用創(chuàng)新練】14.解:(1)由2an=Sn+a1得Sn=2an-a1,當(dāng)n≥2時(shí),Sn-1=2an-1-a1,故Sn-Sn-1=2(an-an-1),則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