1、高考资源网() 您身边的高考专家第2讲数列的求和及综合应用一、选择题1已知数列1,3,5,7,则其前n项和Sn为()An21 Bn22Cn21 Dn22解析因为an2n1,则Snnn21.另解:用特值验证答案A2(2015青岛模拟)数列an满足a12,an,其前n 项积为Tn,则T2 015()A. B C3 D3解析由anan1,所以a23,a3,a4,a52,因此可推知数列an的项具有周期性,且一个周期内的四项之积为1.因为2 01545033,且a2 013a12,a2 014a23,a2 015a3.则T2 0152(3)3.答案C3(2015日照模拟)在数列an中,a12,an1an
2、ln,则an()A2ln n B2(n1)ln nC2nln n D1nln n解析an(anan1)(an1an2)(a2a1)a1ln nln(n1)ln(n1)ln(n2)ln 2ln 122ln n.答案A4(2015临汾模拟)的值为()A. B.C. D.解析,.答案C5各项均为正数的数列an的前n项和为Sn,且3Snanan1,则a2k()A. B.C. D.解析当n1时,3S1a1a2,即3a1a1a2,a23,当n2时,由3Snanan1,可得3Sn1an1an,两式相减得:3anan(an1an1)an0,an1an13,a2n为一个以3为首项,3为公差的等差数列,a2ka2
3、a4a6a2n3n3,选B.答案B二、填空题6(2015全国卷)设Sn是数列an的前n项和,且a11,an1SnSn1,则Sn_解析由题意,得S1a11,又由an1SnSn1,得Sn1SnSnSn1,所以Sn0,所以1,即1,故数列是以1为首项,1为公差的等差数列,得1(n1)n,所以Sn.答案7(2015江苏卷)设数列an满足a11,且an1ann1(nN*),则数列前10项的和为_解析a11,an1ann1,a2a12,a3a23,anan1n,将以上n1个式子相加得ana123n,即an,令bn,故bn2,故S10b1b2b102.答案8设Sn为数列an的前n项和,Sn(1)nan,nN
4、*,则(1)a3_;(2)S1S2S100_解析(1)当n1时,S1(1)a1,得a1.当n2时,Sn(1)n(SnSn1).当n为偶数时,Sn1,当n为奇数时,SnSn1,从而S1,S3,又由S3S2,得S20,则S3S2a3a3.(2)由(1)得S1S3S5S99,S101,又S2S4S6S1002S32S52S72S1010,故S1S2S100.答案(1)(2)三、解答题9(2015湖北卷)设等差数列an的公差为d,前n项和为Sn,等比数列bn的公比为q,已知b1a1,b22,qd,S10100.(1) 求数列an,bn的通项公式;(2) 当d1时,记cn,求数列cn的前n项和Tn.解(
5、1)由题意有即解得或故或(2)由d1,知an2n1,bn2n1,故cn,于是Tn1,Tn.可得Tn23,故Tn6.10(2015四川卷)设数列an(n1,2,3,)的前n项和Sn满足Sn2ana1,且a1,a21,a3成等差数列(1)求数列an的通项公式;(2)记数列的前n项和为Tn,求使得|Tn1|成立的n的最小值解(1)由已知Sn2ana1,有anSnSn12an2an1(n2),即an2an1(n2),所以q2.从而a22a1,a32a24a1,又因为a1,a21,a3成等差数列,即a1a32(a21),所以a14a12(2a11),解得a12,所以,数列an是首项为2,公比为2的等比数
6、列,故an2n.(2)由(1)得,所以Tn1.由|Tn1|,得,即2n1 000,因为295121 0001 024210,所以n10,于是,使|Tn1|成立的n的最小值为10.11(2015威海模拟)已知数列an前n项和为Sn,首项为a1,且,an,Sn成等差数列(1)求数列an的通项公式;(2)数列bn满足bn(log2a2n1)(log2a2n3),求数列的前n项和解(1),an,Sn成等差数列,2anSn,当n1时,2a1S1,a1,当n2时,Sn2an,Sn12an1,两式相减得:anSnSn12an2an1,2,所以数列an是首项为,公比为2的等比数列,即an2n12n2.(2)bn(log2a2n1)(log2a2n3)(log222n12)(log222n32)(2n1)(2n1),数列的前n项和Tn.- 7 - 版权所有高考资源网