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1、跟踪强化训练(十九)1(2017沈阳质检)已知数列an是公差不为0的等差数列,首项a11,且a1,a2,a4成等比数列(1)求数列an的通项公式;(2)设数列bn满足bnan,求数列bn的前n项和Tn.解(1)设数列an的公差为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)由题意得,解得当n2时,Sn1(n1)an(n1)n,所以annan1n(n1)(n1)an(n1)n,即an1an2.又a2a12,因而数列an是首项为1,公差
2、为2的等差数列,从而an2n1.Tn121322523(2n3)2n1(2n1)2n,2Tn122323524(2n3)2n(2n1)2n1.两式相减得Tn12122222322n(2n1)2n122(2122232n)(2n1)2n122(2n1)2n122n24(2n1)2n16(2n3)2n1.所以Tn6(2n3)2n1.3数列an的前n项和为Sn,且首项a13,an1Sn3n(nN*)(1)求证:Sn3n是等比数列;(2)若an为递增数列,求a1的取值范围解(1)证明:an1Sn3n,(nN*)Sn12Sn3n,Sn13n12(Sn3n),a13.2,数列Sn3n是公比为2,首项为a1
3、3的等比数列(2)由(1)得Sn3n(a13)2n1,Sn(a13)2n13n,当n2时,anSnSn1(a13)2n223n1,an为递增数列,n2时,(a13)2n123n(a13)2n223n1,n2时,2n20,可得n2时,a1312n2,又当n2时,312n2有最大值为9,a19,又a2a13满足a2a1,a1的取值范围是(9,)4(2017昆明模拟)设数列an的前n项和为Sn,a11,当n2时,an2anSn2S.(1)求数列an的通项公式;(2)是否存在正数k,使(1S1)(1S2)(1Sn)k对一切正整数n都成立?若存在,求k的取值范围;若不存在,请说明理由解(1)当n2时,anSnSn1,an2anSn2S,SnSn12(SnSn1)Sn2S.Sn1Sn2SnSn1.2.数列是首项为1,公差为2的等差数列,即1(n1)22n1.Sn.当n2时,anSnSn1.数列an的通项公式为an(2)设bn,则bn1.由(1)知Sn,Sn1, 1.又bn0,数列bn是单调递增数列由(1S1)(1S2)(1Sn)k,得bnk.kb1.存在正数k,使(1S1)(1S2)(1Sn)k对一切正整数n都成立,且k的取值范围为.5