高考数学复习专题四数列推理与证明第1讲等差数列与等比数列文市赛课公开课一等奖省名师优质课获奖PPT课.pptx

,第,1,讲等差数列与等比数列,专题四数列、推理与证实,1/43,热点分类突破,真题押题精练,2/43,热点分类突破,3/43,热点一等差数列、等比数列运算,1.,通项公式,等差数列：,a,n,a,1,(,n,1),d,；,等比数列：,a,n,a,1,q,n,1,.,2.,求和公式,4/43,3.,性质,若,m,n,p,q,，,在等差数列中,a,m,a,n,a,p,a,q,；,在等比数列中,a,m,a,n,a,p,a,q,.,5/43,答案,解析,6/43,解析,由题设可得,log,2,a,9,log,2,a,2 009,2,，,即,b,9,b,2 009,2,，,由等差数列通项性质，可得,b,9,b,2 009,b,1,b,2 017,2,，,7/43,答案,解析,解析,因为,a,3,a,5,a,7,a,3,a,3,q,2,a,3,q,4,6(,q,4,q,2,1),78,，得,q,4,q,2,12,0,，得,q,2,3,或,q,2,4(,舍去,),，则,a,5,a,3,q,2,6,3,18,，故选,B.,思维升华,(2)(,届四川省成城市诊疗性检测,),在等比数列,a,n,中，已知,a,3,6,a,3,a,5,a,7,78,，则,a,5,等于,A.12 B.18C.24 D.36,8/43,思维升华,在进行等差,(,比,),数列项与和运算时，若条件和结论间联络不显著，则均可化成关于,a,1,和,d,(,q,),方程组求解，但要注意消元法及整体计算，以降低计算量,.,9/43,答案,解析,跟踪演练,1,(1)(,河北省曲周县第一中学模拟,),设等差数列,a,n,前,n,项和为,S,n,，若,S,4,4,，,S,6,6,，则,S,5,等于,A.0 B.,2C.4 D.1,10/43,答案,解析,ln 2,11/43,热点二等差数列、等比数列判定与证实,数列,a,n,是等差数列或等比数列证实方法,(1),证实数列,a,n,是等差数列两种基本方法：,利用定义，证实,a,n,1,a,n,(,n,N,*,),为一常数；,利用等差中项，即证实,2,a,n,a,n,1,a,n,1,(,n,2).,12/43,(2),证实,a,n,是等比数列两种基本方法,13/43,例,2,(,届东北三省三校联考,),已知数列,a,n,满足,a,1,3,，,a,n,1,2,a,n,n,1,，数列,b,n,满足,b,1,2,，,b,n,1,b,n,a,n,n,.,(1),证实：,a,n,n,为等比数列；,证实,a,n,1,2,a,n,n,1,，,a,n,1,(,n,1),2(,a,n,n,),，,又,a,1,1,2,，,a,n,n,是以,2,为首项，,2,为公比等比数列,.,证实,思维升华,14/43,思维升华,判断一个数列是等差,(,比,),数列，也能够利用通项公式及前,n,项和公式，但不能作为证实方法,.,15/43,解答,思维升华,16/43,解,由,(1),知,a,n,n,(,a,1,1)2,n,1,2,n,，,b,n,1,b,n,a,n,n,，,b,n,1,b,n,2,n,，,当,n,1,时，,b,1,2,，,b,n,2,n,，,17/43,18/43,跟踪演练,2,(,届吉林省长白山市模拟,),在数列,a,n,中，设,f,(,n,),a,n,，且,f,(,n,),满足,f,(,n,1),2,f,(,n,),2,n,(,n,N,*,),，且,a,1,1.,证实,证实,由已知得,a,n,1,2,a,n,2,n,，,b,n,1,b,n,1,，,又,a,1,1,，,b,1,1,，,b,n,是首项为,1,，公差为,1,等差数列,.,19/43,解答,(2),求数列,a,n,前,n,项和,S,n,.,解,由,(1),知，,b,n,n,，,a,n,n,2,n,1,.,S,n,1,22,1,32,2,n,2,n,1,，,两边乘以,2,，得,2,S,n,12,1,22,2,(,n,1)2,n,1,n,2,n,，,两式相减得,S,n,1,2,1,2,2,2,n,1,n,2,n,2,n,1,n,2,n,(1,n,)2,n,1,，,S,n,(,n,1)2,n,1.,20/43,热点三等差数列、等比数列综合问题,处理等差数列、等比数列综合问题，要从两个数列特征入手，理清它们关系；数列与不等式、函数、方程交汇问题，能够结合数列单调性、最值求解,.,21/43,例,3,已知等差数列,a,n,公差为,1,，且,a,2,a,7,a,12,6.,(1),求数列,a,n,通项公式,a,n,与前,n,项和,S,n,；,解,由,a,2,a,7,a,12,6,，得,a,7,2,，,a,1,4,，,解答,22/43,(2),将数列,a,n,前,4,项抽去其中一项后，剩下三项按原来次序恰为等比数列,b,n,前,3,项，记,b,n,前,n,项和为,T,n,，若存在,m,N,*,，使对任意,n,N,*,，总有,S,n,T,m,恒成立，求实数,取值范围,.,解答,思维升华,23/43,解,由题意知,b,1,4,，,b,2,2,，,b,3,1,，设等比数列,b,n,公比为,q,，,T,m,为递增数列，得,4,T,m,8.,24/43,故,(,S,n,),max,S,4,S,5,10,，,若存在,m,N,*,，使对任意,n,N,*,总有,S,n,T,m,，,则,102.,即实数,取值范围为,(2,，,).,25/43,思维升华,(1),等差数列与等比数列交汇问题，惯用,“,基本量法,”,求解，但有时灵活地利用性质，可使运算简便,.,(2),数列项或前,n,项和能够看作关于,n,函数，然后利用函数性质求解数列问题,.,(3),数列中恒成立问题能够经过分离参数，经过求数列值域求解,.,26/43,跟踪演练,3,(,北京,),已知等差数列,a,n,和等比数列,b,n,满足,a,1,b,1,1,，,a,2,a,4,10,，,b,2,b,4,a,5,.,(1),求,a,n,通项公式；,解,设等差数列,a,n,公差为,d,.,因为,a,2,a,4,10,，所以,2,a,1,4,d,10,，,解得,d,2,，所以,a,n,2,n,1.,解答,27/43,解答,(2),求和：,b,1,b,3,b,5,b,2,n,1,.,解,设等比数列,b,n,公比为,q,，,所以,b,2,n,1,b,1,q,2,n,2,3,n,1,.,28/43,真题押题精练,29/43,真题体验,1.(,全国,改编,),记,S,n,为等差数列,a,n,前,n,项和,.,若,a,4,a,5,24,，,S,6,48,，则,a,n,公差为,_.,4,答案,解析,1,2,3,4,解析,设,a,n,公差为,d,，,解得,d,4.,30/43,2.(,浙江改编,),已知等差数列,a,n,公差为,d,，前,n,项和为,S,n,，则,“,d,0,”,是,“,S,4,S,6,2,S,5,”,_,条件,.,充要,答案,解析,1,2,3,4,31/43,解析,方法一,数列,a,n,是公差为,d,等差数列，,S,4,4,a,1,6,d,，,S,5,5,a,1,10,d,，,S,6,6,a,1,15,d,，,S,4,S,6,10,a,1,21,d,2,S,5,10,a,1,20,d,.,若,d,0,，则,21,d,20,d,10,a,1,21,d,10,a,1,20,d,，,即,S,4,S,6,2,S,5,.,若,S,4,S,6,2,S,5,，则,10,a,1,21,d,10,a,1,20,d,，,即,21,d,20,d,，,d,0.,“,d,0,”,是,“,S,4,S,6,2,S,5,”,充要条件,.,1,2,3,4,32/43,方法二,S,4,S,6,2,S,5,S,4,S,4,a,5,a,6,2(,S,4,a,5,),a,6,a,5,a,5,d,a,5,d,0.,“,d,0,”,是,“,S,4,S,6,2,S,5,”,充要条件,.,1,2,3,4,33/43,3.(,北京,),若等差数列,a,n,和等比数列,b,n,满足,a,1,b,1,1,，,a,4,b,4,8,，则,_.,1,答案,解析,1,2,3,4,解析,设等差数列,a,n,公差为,d,，等比数列,b,n,公比为,q,，则由,a,4,a,1,3,d,，,q,2.,34/43,32,解析,设,a,n,首项为,a,1,，公比为,q,，,1,2,3,4,答案,解析,35/43,押题预测,答案,解析,押题依据,等差数列性质和前,n,项和是数列最基本知识点，也是高考热点，能够考查学生灵活变换能力,.,1,2,3,4,1.,设等差数列,a,n,前,n,项和为,S,n,，且,a,1,0,，,a,3,a,10,0,，,a,6,a,7,0,最大自然数,n,值为,A.6 B.7C.12 D.13,押题依据,36/43,1,2,3,解析,a,1,0,，,a,6,a,7,0,，,a,7,0,，,a,1,a,13,2,a,7,0,，,S,13,0,最大自然数,n,值为,12.,4,37/43,2.(,安庆模拟,),等比数列,a,n,中，,a,3,3,a,2,2,，且,5,a,4,为,12,a,3,和,2,a,5,等差中项，则,a,n,公比等于,A.3 B.2,或,3,C.2 D.6,答案,解析,押题依据,等差数列、等比数列综合问题可反应知识利用综合性和灵活性，是高考出题重点,.,1,2,3,4,押题依据,38/43,解析,设公比为,q,5,a,4,为,12,a,3,和,2,a,5,等差中项，,可得,10,a,4,12,a,3,2,a,5,10,a,3,q,12,a,3,2,a,3,q,2,，得,10,q,12,2,q,2,，解得,q,2,或,3.,又,a,3,3,a,2,2,，所以有,a,2,q,3,a,2,2,，所以有,q,2,，故选,C.,1,2,3,4,39/43,答案,解析,押题依据,本题在数列、方程、不等式交汇处命题，综合考查学生应用数学能力，是高考命题方向,.,1,2,3,4,押题依据,40/43,解析,由,a,7,a,6,2,a,5,，得,a,1,q,6,a,1,q,5,2,a,1,q,4,，整理得,q,2,q,2,0,，,解得,q,2,或,q,1(,不合题意，舍去,),，,1,2,3,4,41/43,押题依据,先定义一个新数列，然后要求依据定义条件推断这个新数列一些性质或者判断一个数列是否属于这类数列问题是近年来高考中逐步兴起一类问题，这类问题普通形式新奇，难度不大，常给人耳目一新感觉,.,4.,定义在,(,，,0),(0,，,),上函数,f,(,x,),，假如对于任意给定等比数列,a,n,，,f,(,a,n,),仍是等比数列，则称,f,(,x,),为,“,保等比数列函数,”.,现有定义在,(,，,0),(0,，,),上以下函数：,f,(,x,),x,2,；,f,(,x,),2,x,；,f,(,x,),；,f,(,x,),ln|,x,|.,则其中是,“,保等比数列函数,”,f,(,x,),序号为,A.,B.,C.,D.,解析,1,2,3,4,答案,押题依据,42/43,1,2,3,4,f,(,a,n,),f,(,a,n,2,),f,2,(,a,n,1,),；,f,(,a,n,),f,(,a,n,2,),ln|,a,n,|ln|,a,n,2,|,(ln|,a,n,1,|),2,f,2,(,a,n,1,).,故选,C.,43/43,