2019高考数学二轮复习第7讲等差数列、等比数列练习理.docx

第7讲等差数列、等比数列1.(2018全国课标,4,5分)记Sn为等差数列an的前n项和.若3S3=S2+S4,a1=2,则a5=() A.-12B.-10C.10D.122.已知数列an的前n项和Sn=an2+bn(a,bR),且a2=3,a6=11,则S7=()A.13B.49C.35D.633.已知Sn为等比数列an的前n项和,a2+a3+a4=42,a3+a4+a5=84,则S3=()A.12B.21C.36D.484.已知数列an满足:an+1an+1+1=12,且a2=2,则a4=()A.-12B.23C.12D.115.我国古代数学著作九章算术中有如下问题:“今有金棰,长五尺,斩本一尺,重四斤,斩末一尺,重二斤,问次一尺各重几何?”意思是“现有一根金杖,长5尺,一头粗,一头细,在粗的一端截下1尺,重4斤;在细的一端截下1尺,重2斤.问:依次每一尺各重多少斤?”设该金杖由粗到细是均匀变化的,其质量为M.现将该金杖截成长度相等的10段,记第i段的质量为ai(i=1,2,10),且a1a2S9S7,则满足SnSn+10,a1+a2=4,a3-a2=6.(1)求数列an的通项公式;(2)若对任意的nN*,kan,Sn,-1都成等差数列,求实数k的值.12.已知数列an中,an2+2an-n2+2n=0.(1)求数列an的通项公式;(2)求数列an的前n项和Sn.13.已知数列an满足a1=1,且nan+1-(n+1)an=2n2+2n.(1)求a2,a3;(2)证明:数列ann是等差数列,并求an的通项公式.14.(2018北京,15,12分)设an是等差数列,且a1=ln 2,a2+a3=5ln 2.(1)求an的通项公式;(2)求ea1+ea2+ean.答案全解全析1.B设等差数列an的公差为d,则3(3a1+3d)=2a1+d+4a1+6d,即d=-32a1.又a1=2,d=-3.a5=a1+4d=-10.故选B.2.B由Sn=an2+bn(a,bR)可知,数列an是等差数列.依题意,得d=a6-a26-2=11-34=2,则an=a2+(n-2)d=2n-1.所以a1=1,a7=13.所以S7=a1+a727=1+1327=49.故选B.3.B设an的公比为q(q0).由a2+a3+a4=42,a3+a4+a5=84,易得a3+a4+a5a2+a3+a4=q(a2+a3+a4)a2+a3+a4=q=2.又a2+a3+a4=42,所以a1q+a1q2+a1q3=a1(2+4+8)=42.解得a1=3.所以S3=3+6+12=21.故选B.4.D因为数列an满足:an+1an+1+1=12,所以an+1+1=2(an+1),即数列an+1是等比数列,公比为2.所以a4+1=22(a2+1)=12.所以a4=11.5.C由题意知,由细到粗每段的质量成等差数列,记为an.设其公差为d,则a1+a2=2,a9+a10=4,即2a1+d=2,2a1+17d=4,解得a1=1516,d=18.所以该金杖的总质量M=101516+109218=15.因为48ai=5M,所以481516+(i-1)18=75,解得i=6.故选C.6.C若a1=-2,a2=4,a3=8,满足an2=4n,nN*,但an不是等比数列,故A错;若an=0,满足anan+2=an+12,nN*,但an不是等比数列,故B错;若an=0,满足anan+3=an+1an+2,nN*,但an不是等比数列,故D错;若aman=2m+n,m,nN*,则有aman+1aman=an+1an=2m+n+12m+n=2,故an是等比数列.7.答案1解析由等比数列的性质,得a42=a2a6=16,a4=4.又a4+a8=8,a4=4,a8=4或a4=-4,a8=12.a62=a4a80,a4=4,a8=4,公比q满足q4=1,q2=1.a20a10=q10=1.8.答案225解析因为数列an满足a1=0,数列bn为等差数列,且an+1=an+bn,b15+b16=15,所以an+1=b1+b2+b3+bn.所以a31=b1+b2+b3+b30=302(b1+b30)=15(b15+b16)=1515=225.9.答案16解析由S8S9S7,得S8-S90,S9-S70.a90.an是等差数列,S16=16(a1+a16)2=16(a8+a9)20,S17=17(a1+a17)2=172a92=19a90.满足SnSn+10,q=3,a1=1.an=13n-1=3n-1.故数列an的通项公式为an=3n-1.(2)由(1)知,an=3n-1,Sn=1(1-3n)1-3=3n-12,kan,Sn,-1成等差数列,2Sn=kan-1,即23n-12=k3n-1-1.解得k=3.12.解析(1)由an2+2an-n2+2n=0,得(an-n+2)(an+n)=0.an=n-2或an=-n.an的通项公式为an=n-2或an=-n.(2)当an=n-2时,易知an为等差数列,且a1=-1,Sn=n(a1+an)2=n(-1+n-2)2=n(n-3)2.当an=-n时,易知an也为等差数列,且a1=-1,Sn=n(a1+an)2=n(-1-n)2=-n(n+1)2.13.解析(1)由已知,得a2-2a1=4,则a2=2a1+4.又a1=1,所以a2=6.由2a3-3a2=12,得2a3=12+3a2,所以a3=15.(2)证明:由已知nan+1-(n+1)an=2n(n+1),得nan+1-(n+1)ann(n+1)=2,即an+1n+1-ann=2.所以数列ann是首项为1,公差为2的等差数列.所以ann=1+2(n-1)=2n-1.所以an=2n2-n.14.解析(1)设an的公差为d.因为a2+a3=5ln 2,所以2a1+3d=5ln 2.又a1=ln