2021版高考数学（文）导学大一轮人教A广西专用考点规范练31　数列求和 WORD版含解析.docx

1、考点规范练31数列求和考点规范练A册第22页基础巩固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

2、=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

3、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=62-66-2(32-63)=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 020等于()A.2 020-1B.2 020+1C.2 021-1D.2 021+1答案:C解析:由f(4)=2,可得4a=2,解得a=12,则f(x)=x12.故an=1f(n+1)+f(n)=1n+1+n=n+1-n,S2 020=a1+a2+a3+a2 020=(

4、2-1)+(3-2)+(4-3)+(2 021-2 020) =2 021-1.5.(2019辽宁沈阳质量监测)已知数列an满足an+1+(-1)n+1an=2,则其前100项和为()A.250B.200C.150D.100答案:D解析:当n=2k-1时,a2k+a2k-1=2,an的前100项和为(a1+a2)+(a3+a4)+(a99+a100)=502=100,故选D.6.(2019广东深圳月考)已知数列an的前n项和为Sn,且满足a1=1,a2=2,Sn+1=an+2-an+1(nN*),则Sn=.答案:2n-1解析:Sn+1=an+2-an+1(nN*),Sn+1=Sn+2-Sn+1

5、-(Sn+1-Sn),则Sn+2+1=2(Sn+1+1).由a1=1,a2=2,可得S2+1=2(S1+1),Sn+1+1=2(Sn+1)对任意的nN*都成立,数列Sn+1是首项为2,公比为2的等比数列,Sn+1=2n,即Sn=2n-1.7.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

6、-1-12n+1.数列anan+1前10项的和为121-13+13-15+1210-1-1210+1=121-121=122021=1021.8.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

7、)=(-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d1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=929n-1.(2)由d1,知an=2n-1,bn=2n-1,故cn=2n-12n-1

8、,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.(2019云南玉溪五调)若数列an的前n项和为Sn,首项a10,且2Sn=an2+an(nN*).(1)求数列an的通项公式.(2)若an0,令bn=4an(an+2),数列bn的前n项和为Tn.若Tn0,得an=n,bn=4n(n+2)=21n-1n+2,Tn=21-13+12-14+13-15+1n-1n+2=3-4n+6(n+1)(n+2)3,又mZ,

9、故mmin=3.11.(2019广东珠海高三上学期期末)已知Sn为等差数列an的前n项和,公差d=-2,且a1,a3,a4成等比数列.(1)求an,Sn;(2)设Tn=|a1|+|a2|+|an|,求Tn.解:(1)由a1,a3,a4成等比数列可知,a32=a1a4,从而(a1+2d)2=a1(a1+3d),又d=-2,解得a1=8.所以an=-2n+10.故Sn=na1+n(n-1)2d=-n2+9n.(2)由an=-2n+10,知当n0;当n=5时,an=0;当n5时,an5时,Tn=S5+|a6|+|an|=2S5-Sn=n2-9n+40.能力提升12.在数列an中,a1=1,且an+1

10、=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.(2019新疆乌鲁木齐模拟)把函数f(x)=sin x(x0)所有的零点按从小到大的顺序排列,构成数列an,数列bn满足bn=3nan,则数列bn的前n项和Tn=.答

11、案:(2n-1)3n+1+34解析:令sin x=0(x0),解得x=k,kN*,即an=n,nN*,则bn=3nan=n3n.则数列bn的前n项和Tn=(3+232+333+n3n),3Tn=32+233+(n-1)3n+n3n+1,-,得-2Tn=(3+32+3n-n3n+1)=3(1-3n)1-3-n3n+1,可得Tn=(2n-1)3n+1+34.14.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

12、:(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=

13、-1+2-3+(-1)nn-123+134+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=123+134+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.15.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

14、(nN*),求数列bn的前2n项和T2n.答案:(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=42n-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)=4-22n41-4+n(3+2n+1)2=4n+1-43+n(n+2),T2n=4n+13+n2+2n-43.高考预测16.已知数列an

15、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=anbn,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-=anbn=(2n-1)2n.数列cn的前n项和Mn=12+322+523+(2n-3)2n-1+(2n-1)2n,2Mn=122+323+524+(2n-3)2n+(2n-1)2n+1,由-,得-Mn=2+222+223+224+22n-(2n-1)2n+1=2+222-2n+11-2-(2n-1)2n+1,即Mn=6+(2n-3)2n+1.