2022年高考数学一轮复习 考点规范练31 数列求和（含解析）新人教A版（文）.docx

1、考点规范练31数列求和基础巩固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-

2、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

3、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.2020-1B.2020+1C.2021-1D.2021+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,S2020=a1+a2+a3+a2020=(2-1)+(3-2)+(4-3)+(

4、2021-2020)=2021-1.5.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)=815(1+29)2=1800.由得a2k+1=a2k+2-(4k+1),a1+a3+a5+a59=a2+a4+a60-4(

5、0+1+2+29)+30=1800-430292+30=30,a1+a2+a60=1800+30=1830.6.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=sin2an+1,所以数列yn的前9项和为sin2a1+sin2a2+sin2a3+sin2a8+sin2a9+9.由a5=2,得sin2a5=0.a1+a9=2a5=,2a1+2a9=4a5=2,2a1=2-2a9,sin2a1=sin2-2a9=-sin2a9.由倒序相加可得12(sin2a1+sin2a2+sin2a3+sin2a8+sin2a

6、9+sin2a1+sin2a2+sin2a3+sin2a8+sin2a9)=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+1210-1-1210+1=121-121=122

7、021=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)=(-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

8、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,于是Tn=1+32+522+723+924+2n-12n-1,12Tn=12+322+523+724+925+2n-12n

9、.-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,an0,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an0,可得an+1-an=2.又a12+2a1=4a1+3,解得a1=-1(舍去),a1=3.所以an是

10、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.(2020山东,18)已知公比大于1的等比数列an满足a2+a4=20,a3=8.(1)求an的通项公式;(2)记bm为an在区间(0,m(mN*)中的项的个数,求数列bm的前100项和S100.解:(1)设an的公比为q.由题设得a1q+a1q3=20,a1q2=8.解得q=12(舍去),q=2.因

11、为a1q2=8,所以a1=2.所以an的通项公式为an=2n.(2)由题设及(1)知b1=0,且当2nm0,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.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

12、+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.