高考专题突破三

,高考专题突破三高考中的数列问题,第五章数列,内容索引,考点自测快速解答自查自纠,题型分类对接高考深度剖析,练出高分,考点自测,1.设等差数列an和等比数列bn首项都是1，公差与公比都是2，则等于()A.54B.56C.58D.57解析由题意得，an12(n1)2n1，bn12n12n1，a1a2a4a8a16137153157.,D,考点自测,1,解析答案,1,2,3,4,5,2.已知等比数列的各项都为正数，且当n3时，a4a2n4102n，则数列lga1,2lga2,22lga3,23lga4，2n1lgan，的前n项和Sn等于()A.n2nB.(n1)2n11C.(n1)2n1D.2n1解析等比数列an的各项都为正数，且当n3时，a4a2n4102n，a102n，即an10n，2n1lgan2n1lg10nn2n1，Sn122322n2n1，2Sn12222323n2n，得Sn12222n1n2n2n1n2n(1n)2n1，Sn(n1)2n1.,C,解析答案,1,2,3,4,5,解析答案,1,2,3,4,5,解析因为an1a1ann1ann，所以an1ann1.用累加法：ana1(a2a1)(anan1),解析答案,答案B,解析答案,1,2,3,4,5,解析设等差数列an的首项为a1，公差为d.a55，S515，,ana1(n1)dn.,解析答案,答案A,5.设数列an的前n项和为Sn.已知a10，an1Sn3n，nN*.(1)Sn_；,解析答案,1,2,3,4,5,解析由an1Sn3n，得anSn13n1(n2，且nN*)，两式相减得an1anan23n1，an12an23n1.,又a2S13a133，,解析答案,故an23n132n2.则an123n32n1Sn3n，,Sn3n32n1(n2，且nN*).S1a10也适合上式，Sn3n32n1(nN*).答案3n32n1,解析答案,返回,2k23|k|.即k23|k|20.当k0时，k23k20，解得1k2；当k0，可得an1an2.,所以an是首项为3，公差为2的等差数列，通项公式为an2n1.,解析答案,1,2,3,4,5,解由an2n1可知,设数列bn的前n项和为Tn，则Tnb1b2bn,解析答案,3.已知数列an的前n项和Sn满足Sn2an(1)n(nN*).(1)求数列an的前三项a1，a2，a3；,1,2,3,4,5,解析答案,(2)求证：数列an23(1)n为等比数列，并求出an的通项公式.,1,2,3,4,5,解析答案,解由Sn2an(1)n(nN*)得：Sn12an1(1)n1(n2)，两式相减得：an2an12(1)n(n2)，,1,2,3,4,5,解析答案,1,2,3,4,5,4.(2015湖南)设数列an的前n项和为Sn，已知a11，a22，且an23SnSn13,nN*.(1)证明：an23an；证明由条件，对任意nN*，有an23SnSn13，因而对任意nN*，n2，有an13Sn1Sn3.两式相减，得an2an13anan1，即an23an，n2.又a11，a22，所以a33S1S233a1(a1a2)33a1，故对一切nN*，an23an.,1,2,3,4,5,解析答案,1,2,3,4,5,(2)求Sn.,解析答案,1,2,3,4,5,数列a2n是首项a22，公比为3的等比数列.因此a2n13n1，a2n23n1.于是S2na1a2a2n(a1a3a2n1)(a2a4a2n)(133n1)2(133n1),解析答案,1,2,3,4,5,3(133n1),1,2,3,4,5,解析答案,1,2,3,4,5,(2)设annf(n)，nN*，求证：a1a2a3an2；,解析答案,1,2,3,4,5,证明设Tn为an的前n项和，,解析答案,1,2,3,4,5,即a1a2