高一数学同步数列概念2

数列的概念2一、选择题1.已知数列{an}的通项公式为nnan22,那么101是它的（）A.第4项B.第5项C.第6项D.第7项解析：设101是{an}中的第n项,则10122nn.解得n＝4或n＝-5(舍去),故选A.答案：A2.已知数列{an}的前n项和21nnSn(n∈N*),则a4等于（）A.301B.341C.201D.321解析：由已知,得a4＝S4-S3＝3015465.答案：A3.已知Sn为数列{an}的前n项和,若Sn＝2an-1,则a5的值为（）A.-16B.16C.32D.-32解析：由Sn＝2an-1,得Sn-1＝2an-1-1,∴an＝2(an-an-1),即an＝2an-1.∴21nnaa.又S1＝2a1-1a1＝1,∴a5＝a1·q4＝1·24＝16.答案：B4.在数列{an}中,a1＝2,nan+1＝(n+1)an+2(n∈N*),则a10为（）A.34B.36C.38D.40解析：先取n＝1,2,3,可求得a2＝6,a3＝10,可推出an＝4n-2,从而求出a10＝38.答案：C5.数列{-2n2+29n+3}中的最大项是（）A.107B.108C.81108D.109解析：-2n2+29n+3＝3829)429(222n,当n＝7时最大,最大项为108.选B.答案：B6.由1,3,5,…,2n-1,…构成数列{an},数列{bn}满足b1＝2,当n≥2时,1nbnab,则b5等于（）A.17B.15C.33D.63解析：根据题意,得b2＝1ba＝a2＝3b3＝2ba＝a3＝5b4＝3ba＝a5＝9b5＝4ba＝a9＝17.答案：A7.数列{an}中,an＝(-1)n+1(4n-3),其前n项和为Sn,则S22-S11等于（）A.-85B.85C.-65D.65解析：S22＝1-5+9-13+17-21+…-85＝-44,S11＝1-5+9-13+…+33-37+41＝21,∴S22-S11＝-65.或S22-S11＝a12+a13+…+a22＝a12+(a13+a14)+(a15+a16)+…+(a21+a22)＝-65.答案：C8.已知数列{an}满足a1＝0,1331nnnaaa(n∈N*),则a20等于（）A.0B.3C.3D.23解析：∵a1＝0,1331nnnaaa,∴32a,33a,a4＝0,….从而可知{an}是以3为周期的数列,3220aa.答案：B9.(2009北京崇文高三第一学期期末练习,理6）若正项数列{an}满足a1＝2,an+12-3an+1an-4an2＝0,则数列{an}的通项an等于()A.22n-1B.2nC.22n+1D.22n-3解析:由an+12-3an+1an-4an2＝0,得043)(121nnnnaaaa,解得1411nnnnaaaa或(舍去,因为an＞0),故数列{an}是首项a1＝2,公比q＝4的等比数列.所以an＝2·4n-1＝22n-1.答案:A10.如果f(a+b)＝f(a)·f(b)且f(1)＝2,则)2003()2004()2()3()1()2(ffffff等于（）A.2003B.1001C.2004D.4006解析：令a＝n,b＝1,f(n+1)＝f(n)·f(1),∴2)1()()1(fnfnf.∴)2003()2004()2()3()1()2(ffffff＝2×2003＝4006.答案：D二、填空题11.已知数列{an}的首项a1＝1,并且对任意n∈N*都有an＞0.设其前n项和为Sn,若以(an,Sn)(n∈N*)为坐标的点在曲线)1(21xxy上运动,则数列{an}的通项公式为______.解析：由题意,得Sn＝21an(an+1),∴Sn-1＝21an-1(an-1+1).作差,得an＝21(an2-an-12+an-an-1),即(an+an-1)(an-an-1-1)＝0.∵an＞0,∴an-an-1-1＝0,即an-an-1＝1(n≥2).∴数列{an}为首项a1＝1,公差d＝1的等差数列.∴an＝n(n∈N*).答案：an＝n(n∈N*)12.数列{an}满足,121,12,210,21nnnnnaaaaa若761a,则a2＝________,a24＝______.解析：∵761a,且21＜76＜1,∴7517622a,7317523a.又∵73＜21,∴a4＝2a3＝76.由此推得此数列的周期为3.∴a24＝a3＝73.答案：757313.数列{an}中,Sn是其前n项和,若a1＝1,nnSa311(n≥1),则an＝_________.解析：∵3an+1＝Sn,∴3an＝Sn-1.两式相减,得3(an+1-an)＝Sn-Sn-1＝an(n≥2)341nnaan≥2时,数列{an}是以34为公比,以a2为首项的等比数列,∴n≥2时,an＝a2(34)n-2.令n＝1,由3an+1＝Sn,得3a2＝a1,又a1＝1a2＝31,∴2)34(31nna(n≥2),故.2,)34(31,1,12nnann答案：2,)34(31,1,12nnn14.已知a1＝1,111nnaa(n≥2),则a5＝____________.解析：2111112aa,232111123aa,353211134aa,585311145aa.答案：58三、解答题15.数列{an}满足211a,a1+a2+…+an＝n2an,求an.解：∵Sn＝n2an,∴当n≥2时,an＝Sn-Sn-1＝n2an-(n-1)2an-1.∴(n2-1)an＝(n-1)2an-1.∴111)1(221nnnnaann.∴11,2,,64,53,42,3112145342312nnaannaaaaaaaaaannnn.上式相乘,得)1(21nnaan(n≥2).又211a,∴)1(1nnan(n≥2).∵当n＝1时,211a,∴)1(1nnan.16.数列{an}中,前n项和Sn＝an2+bn,其中a、b是常数,且a＞0,a+b＞1,n∈N*.(1)求{an}的通项公式an,并证明an+1＞an＞1(n∈N*);(2)令1lognnaanc,试判断数列{cn}中任意相邻两项的大小.解：(1)a1＝S1＝a+b＞1.an＝Sn-Sn-1(当n≥2时)＝(an2+bn)-［a(n-1)2+b(n-1)］＝2an-a+b(n＝2,3,4,…).当n＝1时也能满足上式,∴an＝2an-a+b(n∈N*).an+1-an＝2a(n+1)-a+b-(2an-a+b)＝2a＞0,∴an+1＞an＞1(n∈N*).(2)由(1)及对数的性质可得数列{an}中各项皆为正值,nnnnnnnaaaaaanannacc12111loglogloglog212)2loglog(121nnnnaaaa22)]([log411nnaaan222])2([log411nnaaan(an+2≠an)1])([log412211naan.来高考资源网(又∵an＞1,∴0log1nnaanc.∴cn+1＜cn(n∈N*).教学参考例题志鸿优化系列丛书【例题】有一数列{an}，a1＝a,由递推公式nnnaaa121写出这个数列的前4项，并根据前4项观察规律，写出该数列的一个通项公式.解:∵a1＝a,nnnaaa121,∴aaa122,aaaaaaaaa3141211412223,aaaaaaaaa718314131812334.观察规律，得通项公式为yaxaan1的形式,其中x与n的关系可由n＝1,2,3,4得出，即x＝2n-1,而y比x小1,∴aaannn)12(1211.