（通用版）2020版高考数学复习专题四数列4.2数列解答题练习理.docx

4.2数列解答题命题角度1等差、等比数列的判定与证明高考真题体验对方向1.(2019全国19)已知数列an和bn满足a1=1,b1=0,4an+1=3an-bn+4,4bn+1=3bn-an-4.(1)证明:an+bn是等比数列,an-bn是等差数列;(2)求an和bn的通项公式.(1)证明由题设得4(an+1+bn+1)=2(an+bn),即an+1+bn+1=12(an+bn).又因为a1+b1=1,所以an+bn是首项为1,公比为12的等比数列.由题设得4(an+1-bn+1)=4(an-bn)+8,即an+1-bn+1=an-bn+2.又因为a1-b1=1,所以an-bn是首项为1,公差为2的等差数列.(2)解由(1)知,an+bn=12n-1,an-bn=2n-1.所以an=12(an+bn)+(an-bn)=12n+n-12,bn=12(an+bn)-(an-bn)=12n-n+12.2.(2014全国17)已知数列an满足a1=1,an+1=3an+1.(1)证明an+12是等比数列,并求an的通项公式;(2)证明1a1+1a2+1an32.解(1)由an+1=3an+1得an+1+12=3an+12.又a1+12=32,所以an+12是首项为32,公比为3的等比数列.an+12=3n2,因此an的通项公式为an=3n-12.(2)由(1)知1an=23n-1.因为当n1时,3n-123n-1,所以13n-1123n-1.于是1a1+1a2+1an1+13+13n-1=321-13n32.所以1a1+1a2+1an32.典题演练提能刷高分1.(2019江西临川第一中学高三下学期考前模拟)已知数列an中,a1=m,且an+1=3an+2n-1,bn=an+n(nN*).(1)判断数列bn是否为等比数列,并说明理由;(2)当m=2时,求数列(-1)nan的前2 020项和S2 020.解(1)an+1=3an+2n-1,bn+1=an+1+n+1=3an+2n-1+n+1=3(an+n)=3bn.当m=-1时,b1=0,数列bn不是等比数列;当m-1时,数列bn是等比数列,其首项为b1=m+10,公比为3.(2)由(1)且当m-1时,有bn=an+n=33n-1=3n,即an=3n-n,(-1)nan=(-3)n-(-1)nn.S2020=-31-(-3)20201-(-3)-(-1+2)+(-3+4)+(-2019+2020)=-3+320214-1010=32021-40434.2.(2019重庆一中高三下学期5月月考)已知数列an满足:an1,an+1=2-1an(nN*),数列bn中,bn=1an-1,且b1,b2,b4成等比数列.(1)求证:数列bn是等差数列;(2)若Sn是数列bn的前n项和,求数列1Sn的前n项和Tn.(1)证明bn+1-bn=1an+1-1-1an-1=12-1an-1-1an-1=anan-1-1an-1=1,所以数列bn是公差为1的等差数列.(2)解由题意可得b22=b1b4,即(b1+1)2=b1(b1+3),b1=1.故bn=n.所以Sn=n(n+1)2.所以1Sn=2n(n+1)=21n-1n+1.所以Tn=21-12+12-13+1n-1n+1=21-1n+1=2nn+1.3.已知数列an的前n项和为Sn,且Sn=2an-1.(1)证明数列an是等比数列;(2)设bn=(2n-1)an,求数列bn的前n项和Tn.解(1)当n=1时,a1=S1=2a1-1,所以a1=1,当n2时,an=Sn-Sn-1=(2an-1)-(2an-1-1),所以an=2an-1,所以数列an是以a1=1为首项,以2为公比的等比数列.(2)由(1)知,an=2n-1,所以bn=(2n-1)2n-1,所以Tn=1+32+522+(2n-3)2n-2+(2n-1)2n-1,2Tn=12+322+(2n-3)2n-1+(2n-1)2n,-得-Tn=1+2(21+22+2n-1)-(2n-1)2n=1+22-2n-121-2-(2n-1)2n=(3-2n)2n-3,所以Tn=(2n-3)2n+3.4.设a1=2,a2=4,数列bn满足:bn+1=2bn+2且an+1-an=bn.(1)求证:数列bn+2是等比数列;(2)求数列an的通项公式.解(1)由题知bn+1+2bn+2=2bn+2+2bn+2=2,又b1=a2-a1=4-2=2,b1+2=4,bn+2是以4为首项,以2为公比的等比数列.(2)由(1)可得bn+2=42n-1,故bn=2n+1-2.an+1-an=bn,a2-a1=b1,a3-a2=b2,a4-a3=b3,an-an-1=bn-1.累加得an-a1=b1+b2+b3+bn-1,an=2+(22-2)+(23-2)+(24-2)+(2n-2)=2+22(1-2n-1)1-2-2(n-1)=2n+1-2n,即an=2n+1-2n(n2).而a1=2=21+1-21,an=2n+1-2n(nN*).命题角度2等差、等比数列的通项公式与前n项和公式的应用高考真题体验对方向1.(2019天津19)设an是等差数列,bn是等比数列.已知a1=4,b1=6,b2=2a2-2,b3=2a3+4.(1)求an和bn的通项公式;(2)设数列cn满足c1=1,cn=1,2kn35成立的n的最小值.解(1)设等差数列an的公差为d,d0.因为a2,a3,a6成等比数列,所以a32=a2a6,即(1+d)2=1+4d,解得d=2或d=0(舍去).所以an的通项公式为an=a2+(n-2)d=2n-3.(2)因为an=2n-3,所以Sn=n(a1+an)2=n(a2+an-1)2=n2-2n.依题意有n2-2n35,解得n7.故使Sn35成立的n的最小值为8.2.(2019云南昆明高三1月复习诊断)已知数列an是等比数列,公比q1,前n项和为Sn,若a2=2,S3=7.(1)求an的通项公式;(2)设mZ,若Snm恒成立,求m的最小值.解(1)由a2=2,S3=7,得a1q=2,a1+a1q+a1q2=7,解得a1=4,q=12或a1=1,q=2(舍).所以an=412n-1=12n-3.(2)由(1)可知:Sn=a1(1-qn)1-q=4(1-12n)1-12=81-12n0,所以Sn单调递增.所以要使Snm恒成立,需m8.又因为mZ,故m的最小值为8.3.(2019北京大学附属中学高三下)设an是首项为1,公比为3的等比数列.(1)求an的通项公式及前n项和Sn;(2)已知bn是等差数列,Tn为前n项和,且b1=a2,b3=a1+a2+a3,求T20.解(1)由题意可得an=3n-1,Sn=1-3n1-3=3n-12.(2)b1=a2,b3=a1+a2+a3,b3-b1=a1+a3=30+32=10=2d.d=5.T20=203+201925=1010.4.已知等差数列an的公差d为1,且a1,a3,a4成等比数列.(1)求数列an的通项公式;(2)设数列bn=2an+5+n,求数列bn的前n项和Sn.解(1)在等差数列an中,因为a1,a3,a4成等比数列,所以a32=a1a4,即(a1+2d)2=a12+3a1d,解得a1d+4d2=0.因为d=1,所以a1=-4,所以数列an的通项公式an=n-5.(2)由(1)知an=n-5,所以bn=2an+5+n=2n+n.Sn=b1+b2+b3+bn=(2+22+23+2n)+(1+2+3+n)=2(1-2n)1-2+n(1+n)2=2n+1+n(n+1)2-2.5.已知数列an是等差数列,其前n项和为Sn,a2=37,S4=152.(1)求数列an的通项公式;(2)求数列|an-2n|的前n项和Tn.解(1)设数列an的首项为a1,公差为d,则a1+d=37,4a1+6d=152,解得a1=35,d=2,所以数列an的通项公式为an=2n+33(nN*).(2)由(1)知,|an-2n|=|2n+33-2n|=2n+33-2n(0n5),2n-(2n+33)(n6),当0n5时,|2n+33-2n|=2n+33-2n,有Tn=(35+2n+33)n2-2(1-2n)1-2=n2+34n-2n+1+2.当n6时,T5=133,|2n+33-2n|=2n-(2n+33),Tn-T5=64(1-2n-5)1-2-(45+2n+33)(n-5)2=2n+1-n2-34n+131,Tn=2n+1-n2-34n+264.综上所述:Tn=n2+34n-2n+1+2(0n5,nN*),2n+1-n2-34n+264(n6,nN*).6.已知an为等差数列,且a2=3,an前4项的和为16,数列bn满足b1=4,b4=88,且数列bn-an为等比数列.(1)求数列an和bn-an的通项公式;(2)求数列bn的前n项和Sn.解(1)设an的公差为d,因为a2=3,an前4项的和为16,所以a1+d=3,4a1+432d=16,解得a1=1,d=2,所以an=1+(n-1)2=2n-1.设bn-an的公比为q,则b4-a4=(b1-a1)q3,所以q3=b4-a4b1-a1=88-74-1=27,得q=3,所以bn-an=(4-1)3n-1=3n.(2)由(1)得bn=3n+2n-1,所以Sn=(3+32+33+3n)+(1+3+5+2n-1)=3(1-3n)1-3+n(1+2n-1)2=32(3n-1)+n2=3n+12+n2-32.命题角度3一般数列的通项公式与前n项和的求解高考真题体验对方向1.(2019浙江20)设等差数列an的前n项和为Sn,a3=4,a4=S3.数列bn满足:对每个nN*,Sn+bn,Sn+1+bn,Sn+2+bn成等比数列.(1)求数列an,bn的通项公式;(2)记cn=an2bn,nN*,证明:c1+c2+cn2n,nN*.解(1)设数列an的公差为d,由题意得a1+2d=4,a1+3d=3a1+3d,解得a1=0,d=2.从而an=2n-2,nN*.所以Sn=n2-n,nN*.由Sn+bn,Sn+1+bn,Sn+2+bn成等比数列得(Sn+1+bn)2=(Sn+bn)(Sn+2+bn).解得bn=1d(Sn+12-SnSn+2).所以bn=n2+n,nN*.(2)cn=an2bn=2n-22n(n+1)=n-1n(n+1),nN*.我们用数学归纳法证明.当n=1时,c1=02,不等式成立;假设n=k(kN*)时不等式成立,即c1+c2+ck2k.那么,当n=k+1时,c1+c2+ck+ck+12k+k(k+1)(k+2)2k+1k+12k+2k+1+k=2k+2(k+1-k)=2k+1,即当n=k+1时不等式也成立.根据和,不等式c1+c2+cn0,解得q=2.所以,bn=2n.由b3=a4-2a1,可得3d-a1=8.由S11=11b4,可得a1+5d=16,联立,解得a1=1,d=3,由此可得an=3n-2.所以,数列an的通项公式为an=3n-2,数列bn的通项公式为bn=2n.(2)设数列a2nb2n-1的前n项和为Tn,由a2n=6n-2,b2n-1=24n-1,有a2nb2n-1=(3n-1)4n,故Tn=24+542+843+(3n-1)4n,4Tn=242+543+844+(3n-4)4n+(3n-1)4n+1,上述两式相减,得-3Tn=24+342+343+34n-(3n-1)4n+1=12(1-4n)1-4-4-(3n-1)4n+1=-(3n-2)4n+1-8.得Tn=3n-234n+1+83.所以,数列a2nb2n-1的前n项和为3n-234n+1+83.3.(2015全国17)Sn为数列an的前n项和.已知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.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是首项为3,公差为2的等差数列,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).典题演练提能刷高分1.(2019山东烟台高三3月诊断性测试)已知等差数列an的公差是1,且a1,a3,a9成等比数列.(1)求数列an的通项公式;(2)求数列an2an的前n项和Tn.解(1)因为an是公差为1的等差数列,且a1,a3,a9成等比数列,所以a32=a1a9,即(a1+2)2=a1(a1+8),解得a1=1.所以an=a1+(n-1)d=1+(n-1)1=n.(2)Tn=1121+2122+3123+n12n,12Tn=1122+2123+(n-1)12n+n12n+1,两式相减,得12Tn=121+122+123+12n-n12n+1,所以12Tn=12-12n+11-12-n12n+1=1-12n-n2n+1.所以Tn=2-2+n2n.2.(2019江西景德镇高三第二次质检)已知首项为1的等差数列an的前n项和为Sn,S3为a4与a5的等差中项,数列bn满足bn=2Sn+n2n.(1)求数列an与bn的通项公式;(2)求数列anbn的前n项和Tn.解(1)设等差数列an的公差为d.因为S3为a4与a5的等差中项,所以2S3=a4+a5,即2(3+3d)=(1+3d)+(1+4d),解得d=4.an=a1+(n-1)d=1+(n-1)4=4n-3.Sn=na1+n(n-1)2d=n1+n(n-1)24=2n2-n.bn=2Sn+n2n=2n.(2)anbn=(4n-3)2n,Tn=121+522+923+(4n-3)2n,2Tn=122+523+(4n-7)2n+(4n-3)2n+1,下式减上式,得Tn=(4n-3)2n+1-4(22+23+2n)-2=(4n-3)2n+1-44(1-2n-1)1-2-2=(4n-7)2n+1+14.3.已知等比数列an的前n项和为Sn,满足S4=2a4-1,S3=2a3-1.(1)求an的通项公式;(2)记bn=log2(anan+1),数列bn的前n项和为Tn,求证:1T1+1T2+1Tn2.解(1)设an的公比为q,由S4-S3=a4得2a4-2a3=a4,所以a4a3=2,所以q=2.又因为S3=2a3-1,所以a1+2a1+4a1=8a1-1,所以a1=1.所以an=2n-1.(2)由(1)知bn=log2(an+1an)=log2(2n2n-1)=2n-1,所以Tn=1+(2n-1)2n=n2,所以1T1+1T2+1Tn=112+122+1n21+112+123+1(n-1)n=1+1-12+12-13+1n-1-1n=2-1n0,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n项和Tn.(1)证明由题意可知S1=2a1-,即a1=;当n2时,an=Sn-Sn-1=(2an-)-(2an-1-)=2an-2an-1,即an=2an-1.所以数列an是首项为,公比为2的等比数列,所以an=2n-1.(2)解由(1)可知当=4时,an=2n+1,从而bn=2n+1,n是奇数,n+1,n是偶数.当n为偶数时,Tn=4(1-4n2)1-4+n2(3+n+1)2=4(