,大一轮复习讲义,高考专题突破四高考中的数列问题,第,七,章数列与数学归纳法,大一轮复习讲义高考专题突破四高考中的数列问题第七章数列与,1,NEIRONGSUOYIN,内容索引,题型分类 深度剖析,课时作业,NEIRONGSUOYIN内容索引题型分类 深度剖析课,2,题型分类深度剖析,1,PART ONE,题型分类深度剖析1PART ONE,3,题型一等差数列、等比数列的基本问题,师生共研,题型一等差数列、等比数列的基本问题师生共研,2020版高考数学新增分大一轮浙江专用版：第七章-数列与数学归纳法高考专题突破四课件,(2),求正整数,t,的最小值,.,(2)求正整数t的最小值.,等差数列、等比数列综合问题的解题策略,(1),分析已知条件和求解目标，为最终解决问题设置中间问题，例如求和需要先求出通项、求通项需要先求出首项和公差,(,公比,),等，确定解题的顺序,.,(2),注意细节：在等差数列与等比数列综合问题中，如果等比数列的公比不能确定，则要看其是否有等于,1,的可能，在数列的通项问题中第一项和后面的项能否用同一个公式表示等，这些细节对解题的影响也是巨大的,.,思维升华,等差数列、等比数列综合问题的解题策略思维升华,跟踪训练,1,(2018,浙江名校联盟联考,),已知等差数列,a,n,的前,n,项和为,S,n,，等比数列,b,n,的公比是,q,(,q,1),，且满足：,a,1,2,，,b,1,1,，,S,2,3,b,2,，,a,2,b,3,.,(1),求,a,n,与,b,n,；,跟踪训练1(2018浙江名校联盟联考)已知等差数列an,(2),设,c,n,2,b,n,，若数列,c,n,是递减数列，求实数,的取值范围,.,解,由,(1),可知,c,n,2,n,3,n,，,若,c,n,是递减数列，则,c,n,1,c,n,，,即,2,n,1,3,n,1,2,n,3,n,，,(2)设cn2bn ，若数列cn是递减数列,题型二数列的通项与求和,师生共研,(1),求数列,a,n,的通项公式；,题型二数列的通项与求和师生共研(1)求数列an的通项公,所以,S,n,2,n,2,n,.,当,n,1,时，,a,1,S,1,1,；,当,n,2,时，,a,n,S,n,S,n,1,(2,n,2,n,),2(,n,1),2,(,n,1),4,n,3,，,当,n,1,时，,a,1,1,也符合上式,.,所以数列,a,n,的通项公式为,a,n,4,n,3(,n,N,*,).,所以Sn2n2n.,2020版高考数学新增分大一轮浙江专用版：第七章-数列与数学归纳法高考专题突破四课件,(1),可以利用数列的递推关系探求数列的通项，利用递推关系构造数列或证明数列的有关结论,.,(2),根据数列的特点选择合适的求和方法，常用的求和方法有错位相减法、分组转化法、裂项相消法等,.,思维升华,(1)可以利用数列的递推关系探求数列的通项，利用递推关系构造,(1),求数列,a,n,的通项公式；,解,由题意知,S,n,S,n,1,S,n,1,S,n,2,2,n,1,(,n,3),，,即,a,n,a,n,1,2,n,1,(,n,3),，,所以,a,n,(,a,n,a,n,1,),(,a,n,1,a,n,2,),(,a,3,a,2,),a,2,2,n,1,2,n,2,2,2,5,2,n,1,2,n,2,2,2,2,1,2,2,n,1(,n,3),，,检验知,n,1,2,时，结论也成立，故,a,n,2,n,1.,(1)求数列an的通项公式；解由题意知SnSn1,故,T,n,b,1,f,(1),b,2,f,(2),b,n,f,(,n,),故Tnb1f(1)b2f(2)bnf(n),题型三数列与不等式的交汇,师生共研,a,n,1,a,n,，,n,N,*,.,题型三数列与不等式的交汇师生共研an1,a,n,；,求证：(1)an1an；,2020版高考数学新增分大一轮浙江专用版：第七章-数列与数学归纳法高考专题突破四课件,课时作业,2,PART TWO,课时作业2PART TWO,24,基础保分练,1,2,3,4,5,6,1.(2018,绍兴市上虞区调研,),已知数列,a,n,满足,a,1,511,4,a,n,a,n,1,3(,n,2).,(1),求证：,a,n,1,是等比数列；,a,1,1,512,0,，,基础保分练1234561.(2018绍兴市上虞区调研)已知,1,2,3,4,5,6,(2),令,b,n,|log,2,(,a,n,1)|,，求,b,n,的前,n,项和,S,n,.,则,log,2,(,a,n,1),11,2,n,.,b,n,|11,2,n,|,，,令,c,n,11,2,n,，当,n,5,时，,c,n,0,；,当,n,6,时，,c,n,0,，,设,c,n,的前,n,项和为,T,n,，则,T,n,10,n,n,2,，,当,n,5,时，,S,n,T,n,10,n,n,2,；,当,n,6,时，,S,n,2,T,5,T,n,n,2,10,n,50.,123456(2)令bn|log2(an1)|，求bn,1,2,3,4,5,6,(1),求数列,a,n,的通项公式；,解,当,n,1,时，可得,a,2,4,，,当,n,2,时，,4,S,n,a,n,a,n,1,4,S,n,1,a,n,a,n,1,，,两式相减，得,4,a,n,a,n,(,a,n,1,a,n,1,),，,a,n,0,，,a,n,1,a,n,1,4,，,a,n,的奇数项和偶数项分别成以,4,为公差的等差数列，,当,n,2,k,1,，,k,N,*,时，,a,n,2,n,；,当,n,2,k,，,k,N,*,时，,a,n,2,n,.,a,n,2,n,(,n,N,*,).,123456(1)求数列an的通项公式；解当n1时，,1,2,3,4,5,6,(2),设,(,n,N,*,),，求,c,n,的前,n,项和,T,n,.,123456(2)设,1,2,3,4,5,6,123456,1,2,3,4,5,6,123456,1,2,3,4,5,6,(1),求数列,a,n,的通项公式；,则,a,n,S,n,S,n,1,n,2,(,n,1),2,2,n,1(,n,2),，,当,n,1,时，,a,1,1,，适合上式，因此,a,n,2,n,1(,n,N,*,).,123456(1)求数列an的通项公式；则anSnS,1,2,3,4,5,6,123456,1,2,3,4,5,6,解,2,k,a,n,2,2,k,，,2,k,2,n,12,2,k,，,2,k,1,1,n,2,2,k,1,，则,b,k,2,2,k,1,(2,k,1,1),1,2,2,k,1,2,k,1,，,k,N,*,.,123456解2kan22k，2k2n12,1,2,3,4,5,6,则,4,2,m,(4,m,)2,k,1,4,，,即有,08,2,m,(4,m,)2,k,1,，因此,m,4,，对于,m,N,*,，则当,m,1,时，正整数,k,不存在，,m,2,时，正整数,k,不存在，,m,3,时，,k,3,，,因此存在符合条件的,k,，,m,，且,m,3,，,k,3.,123456则42m(4m)2k14，,1,2,3,4,5,6,4.(2018,浙江名校协作体联考,),已知数列,a,n,中，,a,1,1,，且点,P,(,a,n,，,a,n,1,),(,n,N,*,),在直线,x,y,1,0,上,.,(1),求数列,a,n,的通项公式；,解,因为,a,n,a,n,1,1,0,，所以,a,n,1,a,n,1,，,因此数列,a,n,是首项为,1,，公差为,1,的等差数列，,则,a,n,1,(,n,1),1,n,.,1234564.(2018浙江名校协作体联考)已知数列a,1,2,3,4,5,6,123456,1,2,3,4,5,6,123456,1,2,3,4,5,6,所以,S,1,S,2,S,n,1,123456所以S1S2Sn1,因此,g,(,n,),n,.,故存在关于,n,的整式,g,(,n,),n,，使得对于一切不小于,2,的自然数恒成立,.,故,nS,n,(,n,1),S,n,1,S,n,1,1,，,(,n,1),S,n,1,(,n,2),S,n,2,S,n,2,1,，,，,2,S,2,S,1,S,1,1,，,以上式子相加得,nS,n,S,1,S,1,S,2,S,n,1,(,n,1),，,则有,S,1,S,2,S,n,1,nS,n,n,n,(,S,n,1)(,n,2),，因此,g,(,n,),n,，,故存在关于,n,的整式,g,(,n,),n,，使得对于一切不小于,2,的自然数恒成立,.,1,2,3,4,5,6,因此g(n)n.故nSn(n1)Sn1Sn11,1,2,3,4,5,6,技能提升练,123456技能提升练,a,n,1,(,a,n,1,a,n,),(,a,2,a,1,),a,1,n,2.,1,2,3,4,5,6,原不等式得证,.,an1(an1an)(a2a1)a1n,1,2,3,4,5,6,123456,1,2,3,4,5,6,原不等式得证,.,123456原不等式得证.,1,2,3,4,5,6,拓展冲刺练,123456拓展冲刺练,假设当,n,k,(,k,N,*,),时不等式成立，即,0,a,k,1,，,那么当,n,k,1,时，,1,2,3,4,5,6,0,a,k,1,1.,即当,n,k,1,时不等式也成立,.,综合,可知，,0,a,n,1,对任意,n,N,*,成立,.,假设当nk(kN*)时不等式成立，即0ak1，12,1,2,3,4,5,6,123456,1,2,3,4,5,6,即,a,n,1,a,n,，,数列,a,n,为递增数列,.,123456即an1an，数列an为递增数列.,1,2,3,4,5,6,当,n,2,时，,当,n,2,时，,123456当n2时，当n2时，,