高中数学第二章数列章末复习课省公开课一等奖新名师优质课获奖PPT课件.pptx

,第,2,章 数列,章末复习课,1/54,1.,整合知识结构，梳理知识网络，深入巩固、深化所学知识,.,2.,提升处理等差数列、等比数列问题能力，培养综合利用,知识处理问题能力,.,学习目标,2/54,题型探究,知识梳理,内容索引,当堂训练,3/54,知识梳理,4/54,知识点一对比归纳等差数列和等比数列基本概念和公式,等差数列,等比数列,定义,假如一个数列从第二项起，每一项减去它前一项所得差都等于同一个常数，那么这个数列就叫做等差数列，这个常数叫做等差数列公差，公差通惯用字母d表示.,假如一个数列从第二项起，每一项与它前一项比都等于同一个常数，那么这个数列就叫做等比数列，这个常数叫做等比数列公比，公比通惯用字母q表示(q0).,5/54,6/54,7/54,8/54,9/54,知识点二数列中公式推导和解题过程中用到基本方法和思想,1.,在求等差数列和等比数列通项公式时，分别用到了,法和,法,.,2.,在求等差数列和等比数列前,n,项和时，分别用到了,法和,法,.,3.,等差数列和等比数列各自都包括,5,个量，已知其中任意,个求其余,个，用到了方程思想,.,4.,在研究等差数列和等比数列单调性，等差数列前,n,项和最值问题时，都用到了,思想,.,累加,累乘,倒序相加,错,位相减,三,两,函数,10/54,题型探究,11/54,例,1,设,a,n,是公比大于,1,等比数列，,S,n,为数列,a,n,前,n,项和,.,已知,S,3,7,，且,a,1,3,3,a,2,，,a,3,4,组成等差数列公式,.,(1),求数列,a,n,通项公式；,类型一方程思想求解数列问题,解答,12/54,故数列,a,n,通项公式为,a,n,2,n,1,.,13/54,(2),令,b,n,ln,a,3,n,1,，,n,1,2,，,，求数列,b,n,前,n,项和,T,n,.,解答,因为,b,n,ln,a,3,n,1,，,n,1,2,，,，,由,(1),得,a,3,n,1,2,3,n,，,b,n,ln 2,3,n,3,n,ln 2.,又,b,n,1,b,n,3ln 2,，,数列,b,n,是等差数列，,14/54,在等差数列和等比数列中，通项公式,a,n,和前,n,项和公式,S,n,共包括五个量：,a,1,，,a,n,，,n,，,q,(,d,),，,S,n,，其中首项,a,1,和公比,q,(,公差,d,),为基本量，,“,知三求二,”,是指将已知条件转换成关于,a,1,，,a,n,，,n,，,q,(,d,),，,S,n,方程组，经过方程思想解出需要量,.,反思与感悟,15/54,解答,16/54,17/54,类型二转化与化归思想求解数列问题,证实,18/54,由,S,n,1,4,a,n,2,，,则当,n,2,时，有,S,n,4,a,n,1,2.,得,a,n,1,4,a,n,4,a,n,1,.,方法一对,a,n,1,4,a,n,4,a,n,1,两边同除以,2,n,1,，得,即,c,n,1,c,n,1,2,c,n,，,数列,c,n,是等差数列,.,19/54,由,S,n,1,4,a,n,2,，得,a,1,a,2,4,a,1,2,，则,a,2,3,a,1,2,5,，,方法二,a,n,1,2,a,n,2,a,n,4,a,n,1,2(,a,n,2,a,n,1,),，,令,b,n,a,n,1,2,a,n,，,则,b,n,是以,a,2,2,a,1,4,a,1,2,a,1,2,a,1,3,为首项，,2,为公比等比数列，,b,n,32,n,1,，,20/54,21/54,解答,(2),求数列,a,n,通项公式及前,n,项和公式,.,22/54,即数列,a,n,通项公式是,a,n,(3,n,1)2,n,2,.,设,S,n,(3,1)2,1,(3,2,1)2,0,(3,n,1)2,n,2,，,2,S,n,(3,1)2,0,(3,2,1)2,1,(3,n,1)2,n,1,，,S,n,2,S,n,S,n,(3,1)2,1,3(2,0,2,1,2,n,2,),(3,n,1)2,n,1,23/54,1,3,(3,n,4)2,n,1,2,(3,n,4)2,n,1,.,数列,a,n,前,n,项和公式为,S,n,2,(3,n,4)2,n,1,.,24/54,反思与感悟,由递推公式求通项公式，要求掌握方法有两种，一个求法是先找出数列前几项，经过观察、归纳得出，然后证实；另一个是经过变形转化为等差数列或等比数列，再采取公式求出,.,25/54,a,1,2,a,2,3,a,3,na,n,(,n,1),S,n,2,n,(,n,N,*,),，,当,n,1,时，,a,1,2,1,2,；,当,n,2,时，,a,1,2,a,2,(,a,1,a,2,),4,，,a,2,4,；,当,n,3,时，,a,1,2,a,2,3,a,3,2(,a,1,a,2,a,3,),6,，,a,3,8.,跟踪训练,2,设数列,a,n,前,n,项和为,S,n,，已知,a,1,2,a,2,3,a,3,na,n,(,n,1),S,n,2,n,(,n,N,*,).,(1),求,a,2,，,a,3,值；,解答,26/54,(2),求证：数列,S,n,2,是等比数列,.,证实,27/54,a,1,2,a,2,3,a,3,na,n,(,n,1),S,n,2,n,(,n,N,*,),，,当,n,2,时，,a,1,2,a,2,3,a,3,(,n,1),a,n,1,(,n,2),S,n,1,2(,n,1).,得,na,n,(,n,1),S,n,(,n,2),S,n,1,2,n,(,S,n,S,n,1,),S,n,2,S,n,1,2,na,n,S,n,2,S,n,1,2.,S,n,2,S,n,1,2,0,，即,S,n,2,S,n,1,2,，,S,n,2,2(,S,n,1,2).,S,1,2,4,0,，,S,n,1,2,0,，,28/54,故,S,n,2,是以,4,为首项，,2,为公比等比数列,.,29/54,类型三函数思想求解数列问题,命题角度,1,借助函数性质解数列问题,例,3,已知等差数列,a,n,首项,a,1,1,，公差,d,0,，且第,2,项、第,5,项、第,14,项分别是一个等比数列第,2,项、第,3,项、第,4,项,.,(1),求数列,a,n,通项公式；,解答,由题意得,(,a,1,d,)(,a,1,13,d,),(,a,1,4,d,),2,，,整理得,2,a,1,d,d,2,.,d,0,，,d,2.,a,1,1.,a,n,2,n,1(,n,N,*,).,30/54,解答,31/54,32/54,33/54,反思与感悟,数列是一个特殊函数，在求解数列问题时，若包括参数取值范围、最值问题或单调性时，均可考虑采取函数性质及研究方法指导解题,.,值得注意是数列定义域是正整数集或,1,2,3,，,，,n,，这一特殊性对问题结果可能造成影响,.,34/54,解答,35/54,设等比数列,a,n,公比为,q,，,因为,S,3,a,3,，,S,5,a,5,，,S,4,a,4,成等差数列，,所以,S,5,a,5,S,3,a,3,S,4,a,4,S,5,a,5,，,36/54,解答,37/54,当,n,为奇数时，,S,n,随,n,增大而减小，,当,n,为偶数时，,S,n,随,n,增大而增大，,38/54,39/54,命题角度,2,以函数为载体给出数列,例,4,已知函数,f,(,x,),2,|,x,|,，无穷数列,a,n,满足,a,n,1,f,(,a,n,),，,n,N,*,.,(1),若,a,1,0,，求,a,2,，,a,3,，,a,4,；,解答,由,a,n,1,f,(,a,n,),a,n,1,2,|,a,n,|,，,a,1,0,a,2,2,，,a,3,0,，,a,4,2.,40/54,(2),若,a,1,0,，且,a,1,，,a,2,，,a,3,成等比数列，求,a,1,值,.,解答,41/54,42/54,反思与感悟,以函数为载体给出数列，只需代入函数式即可转化为数列问题,.,43/54,解答,44/54,45/54,解答,(2),令,T,n,a,1,a,2,a,2,a,3,a,3,a,4,a,4,a,5,a,2,n,a,2,n,1,，求,T,n,.,T,n,a,1,a,2,a,2,a,3,a,3,a,4,a,4,a,5,a,2,n,a,2,n,1,a,2,(,a,1,a,3,),a,4,(,a,3,a,5,),a,2,n,(,a,2,n,1,a,2,n,1,),46/54,当堂训练,47/54,1,2,3,4,答案,解析,a,n,36(2,n,1),48/54,设等差数列,a,n,公差为,d,，由前,n,项和概念及已知条件得，,1,2,3,4,4,a,1,6,d,4(2,a,1,d,).,解得,a,1,0(,舍去,),或,a,1,36.,所以,a,1,36,，,d,72,，,故数列,a,n,通项公式,a,n,36,(,n,1)72,72,n,36,36(2,n,1).,49/54,1,2,3,4,a,n,3,n,16,3,答案,解析,50/54,3.,设,a,n,为等比数列，,b,n,为等差数列，且,b,1,0,，,c,n,a,n,b,n,，若数列,c,n,是,1,1,2,，,，则数列,c,n,前,10,项和为,_.,由题意可得,a,1,1,，设数列,a,n,公比为,q,，数列,b,n,公差为,d,，,1,2,3,4,答案,解析,978,q,0,，,q,2,，,d,1,，,a,n,2,n,1,，,b,n,(,n,1)(,1),1,n,，,c,n,2,n,1,1,n,，,设数列,c,n,前,n,项和为,S,n,，,S,10,978.,51/54,1,2,3,4,4.,设等差数列,a,n,前,n,项和为,S,n,，公比是正数等比数列,b,n,前,n,项和为,T,n,，已知,a,1,1,，,b,1,3,，,a,3,b,3,17,，,T,3,S,3,12,，求,a,n,、,b,n,通项公式,.,解答,设数列,a,n,公差为,d,，数列,b,n,公比为,q,.,由,a,3,b,3,17,得,1,2,d,3,q,2,17,，,由,T,3,S,3,12,得,q,2,q,d,4.,由,、,及,q,0,，解得,q,2,，,d,2.,故所求通项公式为,a,n,2,n,1,，,b,n,32,n,1,.,52/54,规律与方法,1.,等差数列与等比数列是高中阶段学习两种最基本数列，也是高考中经常考查而且重点考查内容之一，这类问题多从数列本质入手，考查这两种基本数列概念、基本性质、简单运算、通项公式、求和公式等问题,.,2.,数列求和方法：普通数列求和，应从通项入手，若无通项，先求通项，然后经过对通项变形，转化为与特殊数列相关或具备某种方法适用特点形式，从而选择适当方法求和,.,53/54,本课结束,54/54,