2020版广西高考人教版数学（文）一轮复习考点规范练：31 数列求和 Word版含解析.docx

考点规范练31数列求和考点规范练A册第22页 一、基础巩固1.数列112,314,518,7116,(2n-1)+12n,的前n项和Sn的值等于()A.n2+1-12nB.2n2-n+1-12nC.n2+1-12n-1D.n2-n+1-12n答案A解析该数列的通项公式为an=(2n-1)+12n,则Sn=1+3+5+(2n-1)+12+122+12n=n2+1-12n.2.已知数列an满足a1=1,且对任意的nN*都有an+1=a1+an+n,则1an的前100项和为()A.100101B.99100C.101100D.200101答案D解析an+1=a1+an+n,a1=1,an+1-an=1+n.an-an-1=n(n2).an=(an-an-1)+(an-1-an-2)+(a2-a1)+a1=n+(n-1)+2+1=n(n+1)2.1an=2n(n+1)=21n-1n+1.1an的前100项和为21-12+12-13+1100-1101=21-1101=200101.故选D.3.已知数列an满足an+1-an=2,a1=-5,则|a1|+|a2|+|a6|=()A.9B.15C.18D.30答案C解析an+1-an=2,a1=-5,数列an是首项为-5,公差为2的等差数列.an=-5+2(n-1)=2n-7.数列an的前n项和Sn=n(-5+2n-7)2=n2-6n.令an=2n-70,解得n72.当n3时,|an|=-an;当n4时,|an|=an.|a1|+|a2|+|a6|=-a1-a2-a3+a4+a5+a6=S6-2S3=62-6×6-2(32-6×3)=18.4.已知函数f(x)=xa的图象过点(4,2),令an=1f(n+1)+f(n),nN*.记数列an的前n项和为Sn,则S2 018等于()A.2 018-1B.2 018+1C.2 019-1D.2 019+1答案C解析由f(4)=2,可得4a=2,解得a=12,则f(x)=x12.an=1f(n+1)+f(n)=1n+1+n=n+1-n,S2 018=a1+a2+a3+a2 018=(2-1)+(3-2)+(4-3)+(2 019-2 018)=2 019-1.5.已知数列an满足an+1+(-1)nan=2n-1,则an的前60项和为()A.3 690B.3 660C.1 845D.1 830答案D解析an+1+(-1)nan=2n-1,当n=2k(kN*)时,a2k+1+a2k=4k-1,当n=2k+1(kN*)时,a2k+2-a2k+1=4k+1,+得:a2k+a2k+2=8k.则a2+a4+a6+a8+a60=(a2+a4)+(a6+a8)+(a58+a60)=8(1+3+29)=8×15×(1+29)2=1 800.由得a2k+1=a2k+2-(4k+1),a1+a3+a5+a59=a2+a4+a60-4×(0+1+2+29)+30=1 800-4×30×292+30=30,a1+a2+a60=1 800+30=1 830.6.已知等差数列an,a5=2.若函数f(x)=sin 2x+1,记yn=f(an),则数列yn的前9项和为. 答案9解析由题意,得yn=sin 2an+1,所以数列yn的前9项和为sin 2a1+sin 2a2+sin 2a3+sin 2a8+sin 2a9+9.由a5=2,得sin 2a5=0.a1+a9=2a5=,2a1+2a9=4a5=2,2a1=2-2a9,sin 2a1=sin2-2a9=-sin 2a9.由倒序相加可得12(sin 2a1+sin 2a2+sin 2a3+sin 2a8+sin 2a9+sin 2a1+sin 2a2+sin 2a3+sin 2a8+sin 2a9)=0,y1+y2+y3+y8+y9=9.7.已知数列an满足:a3=15,an-an+1=2anan+1,则数列anan+1前10项的和为. 答案1021解析an-an+1=2anan+1,an-an+1anan+1=2,即1an+1-1an=2.数列1an是以2为公差的等差数列.1a3=5,1an=5+2(n-3)=2n-1.an=12n-1.anan+1=1(2n-1)(2n+1)=1212n-1-12n+1.数列anan+1前10项的和为121-13+13-15+12×10-1-12×10+1=12×1-121=12×2021=1021.8.(2018云南昆明第二次统考)在数列an中,a1=3,an的前n项和Sn满足Sn+1=an+n2.(1)求数列an的通项公式;(2)设数列bn满足bn=(-1)n+2an,求数列bn的前n项和Tn.解(1)由Sn+1=an+n2,得Sn+1+1=an+1+(n+1)2,-,得an=2n+1.a1=3满足上式,所以数列an的通项公式为an=2n+1.(2)由(1)得bn=(-1)n+22n+1,所以Tn=b1+b2+bn=(-1)+(-1)2+(-1)n+(23+25+22n+1)=(-1)×1-(-1)n1-(-1)+23×(1-4n)1-4=(-1)n-12+83(4n-1).9.设等差数列an的公差为d,前n项和为Sn,等比数列bn的公比为q,已知b1=a1,b2=2,q=d,S10=100.(1)求数列an,bn的通项公式;(2)当d>1时,记cn=anbn,求数列cn的前n项和Tn.解(1)由题意,得10a1+45d=100,a1d=2,即2a1+9d=20,a1d=2,解得a1=1,d=2或a1=9,d=29.故an=2n-1,bn=2n-1或an=19(2n+79),bn=9·29n-1.(2)由d>1,知an=2n-1,bn=2n-1,故cn=2n-12n-1,于是Tn=1+32+522+723+924+2n-12n-1,12Tn=12+322+523+724+925+2n-12n.-可得12Tn=2+12+122+12n-2-2n-12n=3-2n+32n,故Tn=6-2n+32n-1.10.已知Sn为数列an的前n项和,an>0,an2+2an=4Sn+3.(1)求an的通项公式;(2)设bn=1anan+1,求数列bn的前n项和.解(1)由an2+2an=4Sn+3,可知an+12+2an+1=4Sn+1+3.两式相减可得an+12-an2+2(an+1-an)=4an+1,即2(an+1+an)=an+12-an2=(an+1+an)·(an+1-an).由于an>0,可得an+1-an=2.又a12+2a1=4a1+3,解得a1=-1(舍去),a1=3.所以an是首项为3,公差为2的等差数列,故an的通项公式为an=2n+1.(2)由an=2n+1可知bn=1anan+1=1(2n+1)(2n+3)=1212n+1-12n+3.设数列bn的前n项和为Tn,则Tn=b1+b2+bn=1213-15+15-17+12n+1-12n+3=n3(2n+3).11.已知各项均为正数的数列an的前n项和为Sn,满足an+12=2Sn+n+4,a2-1,a3,a7恰为等比数列bn的前3项.(1)求数列an,bn的通项公式;(2)若cn=(-1)nlog2bn-1anan+1,求数列cn的前n项和Tn.解(1)因为an+12=2Sn+n+4,所以an2=2Sn-1+n-1+4(n2).两式相减,得an+12-an2=2an+1,所以an+12=an2+2an+1=(an+1)2.因为an是各项均为正数的数列,所以an+1=an+1,即an+1-an=1.又a32=(a2-1)a7,所以(a2+1)2=(a2-1)(a2+5),解得a2=3,a1=2,所以an是以2为首项,1为公差的等差数列,所以an=n+1.由题意知b1=2,b2=4,b3=8,故bn=2n.(2)由(1)得cn=(-1)nlog22n-1(n+1)(n+2)=(-1)nn-1(n+1)(n+2),故Tn=c1+c2+cn=-1+2-3+(-1)nn-12×3+13×4+1(n+1)×(n+2).设Fn=-1+2-3+(-1)nn,则当n为偶数时,Fn=(-1+2)+(-3+4)+-(n-1)+n=n2;当n为奇数时,Fn=Fn-1+(-n)=n-12-n=-(n+1)2.设Gn=12×3+13×4+1(n+1)×(n+2),则Gn=12-13+13-14+1n+1-1n+2=12-1n+2.所以Tn=n-12+1n+2,n为偶数,-n+22+1n+2,n为奇数.二、能力提升12.在数列an中,a1=1,且an+1=an2an+1.若bn=anan+1,则数列bn的前n项和Sn为()A.2n2n+1B.n2n+1C.2n2n-1D.2n-12n+1答案B解析由an+1=an2an+1,得1an+1=1an+2,数列1an是以1为首项,2为公差的等差数列,1an=2n-1,又bn=anan+1,bn=1(2n-1)(2n+1)=1212n-1-12n+1,Sn=1211-13+13-15+12n-1-12n+1=n2n+1,故选B.13.(2018福建宁德期末)今要在一个圆周上标出一些数,第一次先把圆周二等分,在这两个分点处分别标上1,如图(1)所示;第二次把两段半圆弧二等分,在这两个分点处分别标上2,如图(2)所示;第三次把四段圆弧二等分,并在这4个分点处分别标上3,如图(3)所示.如此继续下去,当第n次标完数以后,这个圆周上所有已标出的数的总和是. 答案(n-1)×2n+2解析由题意可得,第n次标完后,圆周上所有已标出的数的总和为Tn=1+1+2×2+3×22+n×2n-1.设S=1+2×2+3×22+n×2n-1,则2S=2+2×22+(n-1)×2n-1+n×2n,两式相减可得-S=1+2+22+2n-1-n×2n=(1-n)×2n-1,则S=(n-1)×2n+1,故Tn=(n-1)×2n+2.14.已知首项为32的等比数列an不是递减数列,其前n项和为Sn(nN*),且S3+a3,S5+a5,S4+a4成等差数列.(1)求数列an的通项公式;(2)设bn=(-1)n+1·n(nN*),求数列an·bn的前n项和 Tn.解(1)设等比数列an的公比为q.由S3+a3,S5+a5,S4+a4成等差数列,可得2(S5+a5)=S3+a3+S4+a4,即2(S3+a4+2a5)=2S3+a3+2a4,即4a5=a3,则q2=a5a3=14,解得q=±12.由等比数列an不是递减数列,可得q=-12,故an=32·-12n-1=(-1)n-1·32n.(2)由bn=(-1)n+1·n,可得an·bn=(-1)n-1·32n·(-1)n+1·n=3n·12n.故前n项和Tn=31·12+2·122+n·12n,则12Tn=31·122+2·123+n·12n+1,两式相减可得,12Tn=312+122+12n-n·12n+1=3121-12n1-12-n·12n+1,化简可得Tn=61-n+22n+1.15.(2018湖南长沙雅礼中学模拟)若数列an的前n项和Sn满足Sn=2an-(>0,nN*).(1)证明:数列an为等比数列,并求an;(2)若=4,bn=an,n是奇数,log2an,n是偶数(nN*),求数列bn的前2n项和T2n.(1)证明Sn=2an-,当n=1时,得a1=,当n2时,Sn-1=2an-1-,则Sn-Sn-1=2an-2an-1,即an=2an-2an-1,an=2an-1,数列an是以为首项,2为公比的等比数列,an=·2n-1.(2)解=4,an=4·2n-1=2n+1,bn=2n+1,n是奇数,n+1,n是偶数.T2n=22+3+24+5+26+7+22n+2n+1=(22+24+26+22n)+(3+5+2n+1)=4-22n·41-4+n(3+2n+1)2=4n+1-43+n(n+2),T2n=4n+13+n2+2n-43.三、高考预测16.已知数列an的前n项和为Sn,且a1=2,Sn=2an+k,等差数列bn的前n项和为Tn,且Tn=n2.(1)求k和Sn;(2)若cn=an·bn,求数列cn的前n项和Mn.解(1)Sn=2an+k,当n=1时,S1=2a1+k.a1=-k=2,即k=-2.Sn=2an-2.当n2时,Sn-1=2an-1-2.an=Sn-Sn-1=2an-2an-1.an=2an-1.数列an是以2为首项,2为公比的等比数列.即an=2n.Sn=2n+1-2.(2)等差数列bn的前n项和为Tn,且Tn=n2,当n2时,bn=Tn-Tn-1=2n-1.又b1=T1=1符合bn=2n-1,bn=2n-1.cn=an·bn=(2n-1)2n.数列cn的前n项和Mn=1×2+3×22+5×23+(2n-3)×2n-1+(2n-1)×2n,2Mn=1×22+3×23+5×24+(2n-3)×2n+(2n-1)×2n+1,由-,得-Mn=2+2×22+2×23+2×24+2×2n-(2n-1)×2n+1=2+2×22-2n+11-2-(2n-1)×2n+1,即Mn=6+(2n-3)2n+1.