2018届高三数学（理）一轮复习考点规范练：第六章数列32Word版含解析.doc

考点规范练32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.(2016山西太原三模)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.(2016山西太原五中4月模拟)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 016等于()A.-1B.+1C.-1D.+15.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导学号372704586.32-1+42-2+52-3+(n+2)2-n=.导学号372704597.(2016河南商丘三模)已知数列an满足:a3=,an-an+1=2anan+1,则数列anan+1前10项的和为.导学号372704608.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.导学号372704619.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.导学号3727046210.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.导学号3727046311.(2016全国高考预测模拟一)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.导学号37270464能力提升12.(2016河南商丘二模)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.导学号3727046513.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.导学号37270467高考预测14.已知在等差数列an中,公差d0,a10=19,且a1,a2,a5成等比数列.(1)求an;(2)设bn=an2n,求数列bn的前n项和Sn.导学号37270468参考答案考点规范练32数列求和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 016=a1+a2+a3+a2 016=()+()+()+()=-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,有即解得故(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.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.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=613.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+1.14.解 (1)a1,a2,a5成等比数列,=a1a5