《(全国通用版)2019高考数学二轮复习 中档大题规范练(二)数列 理.doc》由会员分享,可在线阅读,更多相关《(全国通用版)2019高考数学二轮复习 中档大题规范练(二)数列 理.doc(5页珍藏版)》请在得力文库 - 分享文档赚钱的网站上搜索。
1、1( (二二) )数数 列列1(2018三明质检)已知正项数列an的前n项和为Sn,a11,且(t1)Sna3an2(tR R)2n(1)求数列an的通项公式;(2)若数列bn满足b11,bn1bnan1,求数列的前n项和Tn.1 2bn7n解 (1)因为a11,且(t1)Sna3an2,2n所以(t1)S1a3a12,所以t5.2 1所以 6Sna3an2.2n当n2 时,有 6Sn1a3an12,2n1得 6ana3ana3an1,2n2n1所以(anan1)(anan13)0,因为an0,所以anan13,又因为a11,所以an是首项a11,公差d3 的等差数列,所以an3n2(nN N
2、*)(2)因为bn1bnan1,b11,所以bnbn1an(n2,nN N*),所以当n2 时,bn(bnbn1)(bn1bn2)(b2b1)b1anan1a2b1.3n2n 2又b11 也适合上式,所以bn(nN N*)3n2n 2所以1 2bn7n1 3n2n7n ,1 31nn21 6(1 n1 n2)所以Tn 1 6(11 31 21 41 n1 n2) ,1 6(3 21 n11 n2).3n25n12n1n222(2018葫芦岛模拟)设等差数列an的前n项和为Sn,且S3, ,S4成等差数列,S5 2a53a22a12.(1)求数列an的通项公式;(2)设bn2n1,求数列的前n项
3、和Tn.an bn解 (1)设等差数列an的首项为a1,公差为d,由S3, ,S4成等差数列,S5 2可知S3S4S5,得 2a1d0,由a53a22a12,得 4a1d20,由,解得a11,d2,因此,an2n1(nN N*)(2)令cn(2n1)n1,an bn(1 2)则Tnc1c2cn,Tn113 52(2n1)n1,1 2(1 2)(1 2)Tn1 3253(2n1)n,1 21 2(1 2)(1 2)(1 2),得Tn12(2n1)n1 21 2(1 2)2(1 2)n1(1 2)12 (2n1)n 3,1(1 2)n1(1 2)2n3 2nTn6(nN N*)2n3 2n13(2
4、018厦门质检)已知等差数列an满足(n1)an2n2nk,kR R.(1)求数列an的通项公式;(2)设bn,求数列bn的前n项和Sn.4n2 anan1解 (1)方法一 由(n1)an2n2nk,令n1,2,3,得到a1,a2,a3,3k 210k 321k 4an是等差数列,2a2a1a3,即,202k 33k 221k 43解得k1.由于(n1)an2n2n1(2n1)(n1),又n10,an2n1(nN N*)方法二 an是等差数列,设公差为d,则ana1d(n1)dn(a1d),(n1)an(n1)(dna1d)dn2a1na1d,dn2a1na1d2n2nk对于nN N*均成立,
5、则Error!解得k1,an2n1(nN N*)(2)由bn4n2 anan14n22n12n114n2 4n211 4n2111,12n12n11 2(1 2n11 2n1)得Snb1b2b3bn11111 2(11 3)1 2(1 31 5)1 2(1 51 7)1 2(1 2n11 2n1)n1 2(11 31 31 51 51 71 2n11 2n1)n1 2(11 2n1)n(nN N*)n 2n12n22n 2n14(2018天津河东区模拟)已知等比数列an满足条件a2a43(a1a3),a2n3a,nN N*.2n(1)求数列an的通项公式;(2)数列bn满足n2,nN N*,求
6、bn的前n项和Tn.b1 a1b2 a2bn an解 (1)设an的通项公式为ana1qn1(nN N*),由已知a2a43(a1a3),得a1qa1q33(a1a1q2),所以q3.又由已知a2n3a,2n得a1q2n13a q2n2,所以q3a1,2 1所以a11,所以an的通项公式为an3n1(nN N*)(2)当n1 时,1,b11,b1 a14当n2 时,n2,b1 a1b2 a2bn an所以(n1)2,b1 a1b2 a2bn1 an1由得2n1,bn an所以bn(2n1)3n1,b11 也符合,综上,bn(2n1)3n1(nN N*)所以Tn130331(2n3)3n2(2n
7、1)3n1,3Tn131332(2n3)3n1(2n1)3n,由得2Tn1302(31323n1)(2n1)3n13023(2n1)3n3n11 3113n3(2n1)3n(22n)3n2,所以Tn1(n1)3n(nN N*)5(2018宿州模拟)已知数列an的前n项和为Sn,数列Sn的前n项和为Tn,满足Tn2Snn2.(1)证明数列an2是等比数列,并求出数列an的通项公式;(2)设bnnan,求数列bn的前n项和Kn.解 (1)由Tn2Snn2,得a1S1T12S11,解得a1S11,由S1S22S24,解得a24.当n2 时,SnTnTn1 2Snn22Sn1(n1)2,即Sn2Sn12n1,Sn12Sn2n1,由得an12an2,an122(an2),又a222(a12),数列an2是以a123 为首项,2 为公比的等比数列,an232n1,即an32n12(nN N*)(2)bn3n2n12n,Kn3(120221n2n1)2(12n)3(120221n2n1)n2n.5记Rn120221n2n1,2Rn121222(n1)2n1n2n,由,得Rn2021222n1n2nn2n (1n)2n1,12n 12Rn(n1)2n1.Kn3(n1)2nn2n3(nN*)
限制150内