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1、 跟蹤強化訓(xùn)練(十九)1(20xx沈陽質(zhì)檢)已知數(shù)列an是公差不為0的等差數(shù)列,首項a11,且a1,a2,a4成等比數(shù)列(1)求數(shù)列an的通項公式;(2)設(shè)數(shù)列bn滿足bnan,求數(shù)列bn的前n項和Tn.解(1)設(shè)數(shù)列an的公差為d,由已知得,aa1a4,即(1d)213d,解得d0或d1.又d0,d1,可得ann.(2)由(1)得bnn2n,Tn(121)(222)(323)(n2n)(123n)(222232n)2n12. 解(1)由題意得,解得當n2時,Sn1(n1)an(n1)n,所以annan1n(n1)(n1)an(n1)n,即an1an2.又a2a12,因而數(shù)列an是首項為1,公
2、差為2的等差數(shù)列,從而an2n1.Tn121322523(2n3)2n1(2n1)2n,2Tn122323524(2n3)2n(2n1)2n1.兩式相減得Tn12122222322n(2n1)2n122(2122232n)(2n1)2n122(2n1)2n122n24(2n1)2n16(2n3)2n1.所以Tn6(2n3)2n1.3數(shù)列an的前n項和為Sn,且首項a13,an1Sn3n(nN*)(1)求證:Sn3n是等比數(shù)列;(2)若an為遞增數(shù)列,求a1的取值范圍解(1)證明:an1Sn3n,(nN*)Sn12Sn3n,Sn13n12(Sn3n),a13.2,數(shù)列Sn3n是公比為2,首項為a
3、13的等比數(shù)列(2)由(1)得Sn3n(a13)2n1,Sn(a13)2n13n,當n2時,anSnSn1(a13)2n223n1,an為遞增數(shù)列,n2時,(a13)2n123n(a13)2n223n1,n2時,2n20,可得n2時,a1312n2,又當n2時,312n2有最大值為9,a19,又a2a13滿足a2a1,a1的取值范圍是(9,)4(20xx昆明模擬)設(shè)數(shù)列an的前n項和為Sn,a11,當n2時,an2anSn2S.(1)求數(shù)列an的通項公式;(2)是否存在正數(shù)k,使(1S1)(1S2)(1Sn)k對一切正整數(shù)n都成立?若存在,求k的取值范圍;若不存在,請說明理由解(1)當n2時,anSnSn1,an2anSn2S,SnSn12(SnSn1)Sn2S.Sn1Sn2SnSn1.2.數(shù)列是首項為1,公差為2的等差數(shù)列,即1(n1)22n1.Sn.當n2時,anSnSn1.數(shù)列an的通項公式為an(2)設(shè)bn,則bn1.由(1)知Sn,Sn1, 1.又bn0,數(shù)列bn是單調(diào)遞增數(shù)列由(1S1)(1S2)(1Sn)k,得bnk.kb1.存在正數(shù)k,使(1S1)(1S2)(1Sn)k對一切正整數(shù)n都成立,且k的取值范圍為.