《(江蘇專版)2018年高考數(shù)學(xué)二輪復(fù)習(xí) 6個(gè)解答題專項(xiàng)強(qiáng)化練(四)數(shù)列》由會(huì)員分享,可在線閱讀,更多相關(guān)《(江蘇專版)2018年高考數(shù)學(xué)二輪復(fù)習(xí) 6個(gè)解答題專項(xiàng)強(qiáng)化練(四)數(shù)列(9頁(yè)珍藏版)》請(qǐng)?jiān)谘b配圖網(wǎng)上搜索。
1、6個(gè)解答題專項(xiàng)強(qiáng)化練(四)數(shù)列1已知an為等差數(shù)列,前n項(xiàng)和為Sn(nN*),bn是首項(xiàng)為2的等比數(shù)列,且公比大于0,b2b312,b3a42a1,S1111b4.(1)求an和bn的通項(xiàng)公式;(2)求數(shù)列a2nb2n1的前n項(xiàng)和(nN*)解:(1)設(shè)等差數(shù)列an的公差為d,等比數(shù)列bn的公比為q.由已知b2b312,得b1(qq2)12,而b12,所以q2q60.又因?yàn)閝0,解得q2.所以bn2n.由b3a42a1,可得3da18.由S1111b4,可得a15d16.由,解得a11,d3,由此可得an3n2.所以數(shù)列an的通項(xiàng)公式為an3n2,數(shù)列bn的通項(xiàng)公式為bn2n.(2)設(shè)數(shù)列a2n
2、b2n1的前n項(xiàng)和為Tn,由a2n6n2,b2n124n1,得a2nb2n1(3n1)4n,故Tn24542843(3n1)4n,4Tn242543844(3n4)4n(3n1)4n1,上述兩式相減,得3Tn2434234334n(3n1)4n14(3n1)4n1(3n2)4n18.故Tn4n1.所以數(shù)列a2nb2n1的前n項(xiàng)和為4n1.2已知數(shù)列an滿足:a1,an1anp3n1nq,nN*,p,qR.(1)若q0,且數(shù)列an為等比數(shù)列,求p的值;(2)若p1,且a4為數(shù)列an的最小項(xiàng),求q的取值范圍解:(1)q0,an1anp3n1,a2a1pp,a3a23p4p,由數(shù)列an為等比數(shù)列,得
3、2,解得p0或p1.當(dāng)p0時(shí),an1an,an,符合題意;當(dāng)p1時(shí),an1an3n1,ana1(a2a1)(a3a2)(anan1)(133n2)3n1,3.符合題意p的值為0或1.(2)法一:若p1,則an1an3n1nq,ana1(a2a1)(a3a2)(anan1)(133n2)12(n1)q3n1n(n1)q數(shù)列an的最小項(xiàng)為a4,對(duì)任意的nN*,有3n1n(n1)qa4(2712q)恒成立,即3n127(n2n12)q對(duì)任意的nN*恒成立當(dāng)n1時(shí),有2612q,q;當(dāng)n2時(shí),有2410q,q;當(dāng)n3時(shí),有186q,q3;當(dāng)n4時(shí),有00,qR;當(dāng)n5時(shí),n2n120,所以有q恒成立,
4、令cn(n5,nN*),則cn1cn0,即數(shù)列cn為遞增數(shù)列,qc5.綜上所述,q的取值范圍為.法二:p1,an1an3n1nq,又a4為數(shù)列an的最小項(xiàng),即3q.此時(shí)a2a11q0,a3a232qa2a3a4.當(dāng)n4時(shí),令bnan1an,bn1bn23n1q23410,bn1bn,0b4b5b6,即a4a5a6a7.綜上所述,當(dāng)3q時(shí),a4為數(shù)列an的最小項(xiàng),即q的取值范圍為.3數(shù)列an的前n項(xiàng)和為Sn,a12,Snan(rR,nN*)(1)求r的值及數(shù)列an的通項(xiàng)公式;(2)設(shè)bn(nN*),記bn的前n項(xiàng)和為Tn.當(dāng)nN*時(shí),0.Bn1Bn,Bn單調(diào)遞增,故(Bn)minB1,0,即(1
5、)n.當(dāng)n為大于等于4的偶數(shù)時(shí),有恒成立,又隨著n的增大而增大,此時(shí)min,即,故的取值范圍為.當(dāng)n為大于等于3的奇數(shù)時(shí),有恒成立,此時(shí)min,即0,得0,所以g(x)g(1)0,即2ln xx,用替代x可得ln x,x(1,2,所以f(x)Tn. 所以數(shù)列an為可拆分?jǐn)?shù)列(2)設(shè)數(shù)列bn,cn的公差分別為d1,d2.由an5n,得b1(n1)d1c1(n1)d2(d1d2)nb1c1d1d25n對(duì)任意的nN*都成立所以即 由SnTn,得nb1d1nc1d2,則n2n0.由n1,得n0對(duì)任意的nN*成立則0且0即d1d2且b1c1.由數(shù)列bn,cn各項(xiàng)均為正整數(shù),則b1,c1,d1,d2均為正整數(shù),當(dāng)d1d2時(shí),由d1d25,得d1d2N*,不符合題意,所以d1d2. 聯(lián)立,可得或或或所以或或或(3)證明:設(shè)ana1qn1,a1N*,q0,q1,則q2.當(dāng)q為無(wú)理數(shù)時(shí),a2a1q為無(wú)理數(shù),與anN*矛盾故q為有理數(shù),設(shè)q(a,b為正整數(shù),且a,b互質(zhì))此時(shí)ana1.則對(duì)任意的nN*,an1均為a1的約數(shù),則an11,即a1,故qbN*,所以qN*,q2. 所以ana1qn1(a11)qn1qn1,令bn(a11)qn1,cnqn1.則bn,cn各項(xiàng)均為正整數(shù)因?yàn)閍13,所以a1121,則SnTn,所以數(shù)列an為可拆分?jǐn)?shù)列9