2019届高考数学复习数列考点规范练31数列求和文新人教A版.docx

考点规范练31数列求和基础巩固1.数列1,3,5,7,(2n-1)+,的前n项和Sn的值等于()A.n2+1-B.2n2-n+1-C.n2+1-D.n2-n+1-2.已知数列an满足a1=1,且对任意的nN*都有an+1=a1+an+n,则的前100项和为()A.B.C.D.3.已知函数f(n)=n2cos(n),且an=f(n)+f(n+1),则a1+a2+a3+a100=()A.0B.-100C.100D.10 2004.已知函数f(x)=xa的图象过点(4,2),令an=,nN*.记数列an的前n项和为Sn,则S2 018等于()A.-1B.+1C.-1D.+15.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 8306.32-1+42-2+52-3+(n+2)2-n=.7.已知数列an满足:a3=,an-an+1=2anan+1,则数列anan+1前10项的和为.8.已知an是等差数列,bn是等比数列,且b2=3,b3=9,a1=b1,a14=b4.(1)求an的通项公式;(2)设cn=an+bn,求数列cn的前n项和.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=,求数列cn的前n项和Tn.10.已知Sn为数列an的前n项和,an0,+2an=4Sn+3.(1)求an的通项公式;(2)设bn=,求数列bn的前n项和.11.已知各项均为正数的数列an的前n项和为Sn,满足=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-,求数列cn的前n项和Tn.能力提升12.(2017山东烟台模拟)已知在数列an中,a1=1,且an+1=,若bn=anan+1,则数列bn的前n项和Sn为()A.B.C.D.13.(2017福建龙岩一模)已知Sn为数列an的前n项和,对nN*都有Sn=1-an,若bn=log2an,则+=.14.已知首项为的等比数列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+1n(nN*),求数列anbn的前n项和Tn.15.已知数列an的前n项和为Sn,且a1=0,对任意nN*,都有nan+1=Sn+n(n+1).(1)求数列an的通项公式;(2)若数列bn满足an+log2n=log2bn,求数列bn的前n项和Tn.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.答案：1.A解析:该数列的通项公式为an=(2n-1)+,则Sn=1+3+5+(2n-1)+=n2+1-.2.D解析:an+1=a1+an+n,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=.=2.的前100项和为2=2.故选D.3.B解析:f(n)=n2cos(n)=(-1)nn2,an=f(n)+f(n+1)=(-1)nn2+(-1)n+1(n+1)2=(-1)nn2-(n+1)2=(-1)n+1(2n+1).a1+a2+a3+a100=3+(-5)+7+(-9)+199+(-201)=50(-2)=-100.故选B.4.C解析:由f(4)=2,可得4a=2,解得a=,则f(x)=.an=,S2 018=a1+a2+a3+a2 018=()+()+()+()=-1.5.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=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-=30,a1+a2+a60=1 800+30=1 830.6.4-解析:设Sn=32-1+42-2+52-3+(n+2)2-n,则Sn=32-2+42-3+(n+1)2-n+(n+2)2-n-1.-,得Sn=32-1+2-2+2-3+2-n-(n+2)2-n-1=22-1+2-1+2-2+2-3+2-n-(n+2)2-n-1=1+-(n+2)2-n-1=2-(n+4)2-n-1.故Sn=4-.7.解析:an-an+1=2anan+1,=2,即=2.数列是以2为公差的等差数列.=5,=5+2(n-3)=2n-1.an=.anan+1=.数列anan+1前10项的和为=.8.解:(1)因为等比数列bn的公比q=3,所以b1=1,b4=b3q=27.设等差数列an的公差为d.因为a1=b1=1,a14=b4=27,所以1+13d=27,即d=2.所以an=2n-1.(2)由(1)知,an=2n-1,bn=3n-1.因此cn=an+bn=2n-1+3n-1.从而数列cn的前n项和Sn=1+3+(2n-1)+1+3+3n-1=n2+.9.解:(1)由题意,有解得故(2)由d1,知an=2n-1,bn=2n-1,故cn=,于是Tn=1+,Tn=+.-可得Tn=2+=3-,故Tn=6-.10.解:(1)由+2an=4Sn+3,可知+2an+1=4Sn+1+3.两式相减可得+2(an+1-an)=4an+1,即2(an+1+an)=(an+1+an)(an+1-an).由于an0,可得an+1-an=2.又+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=.设数列bn的前n项和为Tn,则Tn=b1+b2+bn=.11.解:(1)因为=2Sn+n+4,所以=2Sn-1+n-1+4(n2).两式相减,得=2an+1,所以+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=(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)nn-,故Tn=c1+c2+cn=-1+2-3+(-1)nn-.设Fn=-1+2-3+(-1)nn,则当n为偶数时,Fn=(-1+2)+(-3+4)+-(n-1)+n=;当n为奇数时,Fn=Fn-1+(-n)=-n=.设Gn=+,则Gn=+.所以Tn=12.B解析:由an+1=,得+2,数列是以1为首项,2为公差的等差数列,=2n-1,又bn=anan+1,bn=,Sn=,故选B.13.解析:对nN*都有Sn=1-an,当n=1时,a1=1-a1,解得a1=.当n2时,an=Sn-Sn-1=1-an-(1-an-1),化为an=an-1.数列an是等比数列,公比为,首项为.an=.bn=log2an=-n.则+=1-.14.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=,解得q=.由等比数列an不是递减数列,可得q=-,故an=(-1)n-1.(2)由bn=(-1)n+1n,可得anbn=(-1)n-1(-1)n+1n=3n.故前n项和Tn=3,则Tn=3,两式相减可得,Tn=3=3,化简可得Tn=6.15.解:(1)(方法一)nan+1=Sn+n(n+1),当n2时,(n-1)an=Sn-1+n(n-1),两式相减,得nan+1-(n-1)an=Sn-Sn-1+n(n+1)-n(n-1),即nan+1-(n-1)an=an+2n,得an+1-an=2.当n=1时,1a2=S1+12,即a2-a1=2.数列an是以0为首项,2为公差的等差数列.an=2(n-1)=2n-2.(方法二)由nan+1=Sn+n(n+1),得n(Sn+1-Sn)=Sn+n(n+1),整理,得nSn+1=(n+1)Sn+n(n+1),两边同除以n(n+1)得,=1.数列是以=0为首项,1为公差的等差数列.=0+n-1=n-1.Sn=n(n-1).当n2时,an=Sn-Sn-1=n(n-1)-(n-1)(n-2)=2n-2.又a1=0适合上式,数列an的通项公式为an=2n-2.(2)an+log2n=log2bn,bn=n=n22n-2=n4n-1.Tn=b1+b2+b3+bn-1+bn=40+241+342+(n-1)4n-2+n4n-1,4Tn=41+242+343+(n-1)4n-1+n4n,-,得-3Tn=40+41+42+4n-1-n4n=-n4n=.Tn=(3n-1)4n+1.16.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=n