高考数学 59数列概念考点专项复习课件 新人教A版 课件.ppt

单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,*,*,原创,2011,届高考数学考点专项复习课件,59,数列概念,数列的概念,一、,数列的概念,1.,定义,按一定次序排列的一列数叫做数列,.,2.,数列是特殊的函数,从函数的观点看数列,对于定义域为正整数集,N*,(,或它的有限子集,1,2,3,n,),的函数来说,数列就是这个函数当自变量从小到大依次取值时对应的一系列函数值,其图象是无限个或有限个孤立的点,.,注,:,依据此观点可以用函数的思想方法来解决有关数列的问题,.,二、,数列的表示,1.,列举法,2.,图象法,3.,通项公式法,若数列的每一项,a,n,与项数,n,之间的函数关系可以用一个公式来表达,即,a,n,=,f,(,n,),则,a,n,=,f,(,n,),叫做数列的,通项公式,.,4.,递推公式法,如果已知数列的第一项,(,或前几项,),且任一项与它的前一项,(,或前几项,),的关系可以用一个公式来表示,这个公式就叫做数列的,递推公式,.,注,:,递推公式有两要素,:,递推关系与初始条件,.,三、,数列的分类,1.,按项数,：,有穷数列和无穷数列,；,2.,按,a,n,的增减性：递增、递减、常数、摆动数列,；,3.,按,|,a,n,|,是否有界：有界数列和无界数列,.,四、数列的前,n,项和,S,n,=,a,1,+,a,2,+,+,a,n,=,a,k,;,n,k,=1,a,n,=,S,1,(,n,=1),S,n,-,S,n,-,1,(,n,2).,五、数列的单调性,设,D,是由连续的正整数构成的集合,若对于,D,中的每一个,n,都有,a,n,+1,a,n,(,或,a,n,+1,a,n,),则称数列,a,n,在,D,内单调递增,(,或单调递减,),.,方法,：作差、作商、函数求导,.,六、重要变换,a,n,=,a,1,+(,a,2,-,a,1,)+(,a,3,-,a,2,)+,+(,a,n,-,a,n,-,1,);,a,n,=,a,1,.,a,n,a,n,-,1,a,2,a,1,a,3,a,2,典型例题,1.,若数列,a,n,满足,a,1,=1,a,n,=,a,1,+2,a,2,+3,a,3,+,+(,n,-,1),a,n,-,1,(,n,2),则当,n,2,时,a,n,的通项,a,n,=,.,2.,定义“等和数列”,:,在一个数列中,如果每一项与它的后一项的和都为同一个常数,那么这个数列叫做等和数列,这个常数叫做该数列的公和,.,已知数列,a,n,是等和数列,且,a,1,=2,公和为,5,那么,a,18,的值为,这个数列的前,n,项和,S,n,的计算公式为,.,3.,设数列,a,n,的前,n,项和为,S,n,S,n,=,(,对于所有,n,1,),且,a,4,=54,则,a,1,的数值为,.,a,1,(3,n,-,1),2,4.,在数列,a,n,中,a,1,=,a,n,+1,-,a,n,=,求数列,a,n,的通项公式,.,1,2,4,n,2,-,1,1,n,!,2,a,n,=,3,2,4,n,-,2,4,n,-,3,a,n,=,n,为奇数时,S,n,=,n,-,;,n,为偶数时,S,n,=,n,.,1,2,5,2,5,2,5.,已知数列,a,n,的前,n,项和,S,n,满足,:log,2,(1+,S,n,)=,n,+1,求数列,a,n,的通项公式,.,6.,设数列,a,n,的前,n,项和,S,n,=2,a,n,-,1(,n,=1,2,3,);,数列,b,n,满足,:,b,1,=3,b,k,+1,=,a,k,+,b,k,(,k,=1,2,3,).,求数列,a,n,、,b,n,的通项公式,.,3,n,=1,2,n,n,2.,a,n,=,a,n,=2,n,-,1,b,n,=2,n,-,1,+2,7.,设数列,a,n,的前,n,项和,S,n,=3,n,2,-,65,n,求数列,|,a,n,|,的前,n,项和,T,n,.,-,3,n,2,+65,n,n,11,3,n,2,-,65,n,+704,n,12.,T,n,=,8.,已知数列,a,n,的通项,a,n,=(,n,+1),(),问是否存在正整数,M,使得对任意正整数,n,都有,a,n,a,M,?,n,10,9,当,n,a,n,a,n,单调递增；,当,n,8,时,a,n,+1,a,n,a,n,单调递减,.,而,a,8,=,a,9,即,a,1,a,2,a,10,a,11,a,8,与,a,9,是数列,a,n,的最大项,.,故存在,M,=8,或,9,使得,a,n,a,M,对,n,N,+,恒成立,.,解,:,a,n,+1,-,a,n,=(,n,+2)(,),n,+1,-,(,n,+1)(,),n,11,9,11,9,=(,),n,.,11,9,10,8,-,n,9.,求使得不等式,+,+,2,a,-,5,对,n,N,*,恒成立的正整数,a,的最大值,.,1,3,n,+1,1,n,+1,1,n,+2,1,n,+3,解,:,记,f,(,n,)=+,+,考察,f,(,n,),的单调性,.,1,3,n,+1,1,n,+1,1,n,+2,1,n,+3,f,(,n,+1,),f,(,n,),f,(,n,+1,),-,f,(,n,)=+,+,-,1,3,n,+2,1,3,n,+3,1,3,n,+4,1,n,+1,=+,-,1,3,n,+2,1,3,n,+4,2,3,n,+3,=0,2,(3,n,+2)(3,n,+3)(3,n,+4),评析,数列的单调性是探索数列的最大项、最小项及解决其它许多数列问题的重要途径,因此要熟练掌握求数列单调性的程序,.,当,n,=1,时,f,(,n,),有最小值,f,(1)=+=.,1,2,1,3,1,4,12,13,要使题中不等式对,n,N,*,恒成立,只须,2,a,-,5 .,12,13,正整数,a,的最大值是,3.,解得,a,.,24,73,课后练习,1.,根据下列数列的前几项的值,写出数列的一个,通项公式,:,(1),-,1,-,-,;,3,4,3,6,3,2,1,3,1,5,(2),5,55,555,.,a,n,=(,-,1),n,2+(,-,1),n,n,a,n,=555,5=(999,9)=(10,n,-,1),n,个,5,9,n,个,5,9,(3),-,1,7,-,13,19,;,(4),7,77,777,7777,;,(5),;,2,3,63,8,99,10,15,4,35,6,(6),5,0,-,5,0,5,0,-,5,0,.,a,n,=(,-,1),n,(6,n,-,5),a,n,=,(10,n,-,1),7,9,a,n,=,2,n,(2,n,-,1,)(2,n,+1),a,n,=5sin,2,n,2.,已知下面各数列,a,n,的前,n,项和,S,n,的公式,求,a,n,的通项公式,:(1),S,n,=2,n,2,-,3,n,;(2),S,n,=3,n,2,+,n,+1;(3),S,n,=3,n,-,2.,解,:,(1),当,n,=1,时,a,1,=,S,1,=,-,1;,当,n,2,时,a,n,=,S,n,-,S,n,-,1,=4,n,-,5,故,a,n,=4,n,-,5(,n,N,*,).,(2),当,n,=1,时,a,1,=,S,1,=5;,当,n,2,时,a,n,=,S,n,-,S,n,-,1,=6,n,-,2,故,a,n,=,5,n,=1,6,n,-,2,n,2.,(3),当,n,=1,时,a,1,=,S,1,=1;,当,n,2,时,a,n,=,S,n,-,S,n,-,1,=2,3,n,-,1,故,a,n,=,1,n,=1,2,3,n,-,1,n,2.,3.,已知数列,a,n,满足,a,1,=1,a,n,=3,n,-,1,+,a,n,-,1,(,n,2).(1),求,a,2,a,3,;(2),证明,:,a,n,=.,3,n,-,1,2,(1),解,:,a,1,=1,a,n,=3,n,-,1,+,a,n,-,1,(,n,2),a,2,=3,2,-,1,+,a,1,=3+1=4,a,3,=3,3,-,1,+,a,2,=9+4=13.,故,a,2,a,3,的值分别为,4,13.,(2),证,:,a,1,=1,a,n,=3,n,-,1,+,a,n,-,1,a,n,-,a,n,-,1,=3,n,-,1,.,a,n,=,a,1,+(,a,2,-,a,1,)+(,a,3,-,a,2,)+,+,(,a,n,-,a,n,-,1,),=1+3+3,2,+,+,3,n,-,1,3,n,-,1,2,故,a,n,=.,3,n,-,1,2,3,n,-,1,3,-,1,3,-,1,=.,4.,设函数,f,(,x,)=log,2,x,-,log,x,2(0,x,1),数列,a,n,满足,f,(2,a,n,)=2,n,n,=1,2,3,.(1),求数列,a,n,的通项公式,;(2),判断数列,a,n,的单调性,.,解,:,(1),由已知,log,2,2,a,n,-,=2,n,log,2,2,a,n,1,a,n,-,=2,n,1,a,n,即,a,n,2,-,2,na,n,-,1=0.,解得,a,n,=,n,n,2,+1.,故,a,n,=,n,-,n,2,+1(,n,N,*,).,0,x,1,即,02,a,n,1,a,n,0.,(2),=,a,n,+1,a,n,(,n,+1),-,(,n,+1),2,+1,n,-,n,2,+1,(,n,+1)+,(,n,+1),2,+1,n,+,n,2,+1,=,1.,而,a,n,a,n,.,故,数列,a,n,是递增数列,.,5.,已知数列,a,n,的通项,a,n,=(,n,+1)(),n,(,n,N,*,),试问该数列,a,n,有没有最大项,?,若有,求出最大项和最大项的项数,;,若没有,说明理由,.,11,10,当,n,0,即,a,n,+1,a,n,;,当,n,9,时,a,n,+1,-,a,n,0,即,a,n,+1,0),则有,:,a,24,=,a,14,q,=(,a,11,+3,d,),q,a,32,=,a,12,q,2,=(,a,11,+,d,),q,2,1,2,(+3,d,),q,=1,(+,d,),q,2,=,1,2,1,4,即,:,解得,:,q,=,d,=.,1,2,1,2,故公比,q,的值为,.,1,2,1,2,(2),a,1k,=,a,11,+(,k,-,1),d,=,+(,k,-,1),=.,k,2,n,2,1,2,(3)A,1,=,a,11,+,a,12,+,a,13,+,+,a,1,n,=,(+)=.,n,2,n,(,n,+1),4,A,k,=,a,k,1,+,a,k,2,+,a,k,3,+,+,a,kn,=,q,k,-,1,A,1,=(),k,-,1,=.,1,2,n,(,n,+1),4,n,(,n,+1),2,k,+1,7.,已知数列,2,n,-,1,a,n,的前,n,项和,S,n,=9,-,6,n,.(1),求数列,a,n,的通项公式,;(2),设,b,n,=,n,(3,-,log,2,),求数列,的前,n,项和,.,|,a,n,|,3,b,n,1,解,:,(1),当,n,=1,时,2,0,a,1,=,S,1,=9,-,6=3,a,1,=3;,当,n,2,时,2,n,-,1,a,n,=,S,n,-,S,n,-,1,=,-,6,故,a,n,=,-,n,2.,3,n,=1,2,n,-,2,3,a,n,=,-,.,2,n,-,2,3,(2),当,n,=1,时,b,1,=3,-,log,2,1=3,=;,b,1,1,1,3,当,n,2,时,b,n,=,n,(3,-,log,2,)=,n,(,n,+1),3,2,n,-,2,3,b,n,1,=,-,.,n,1,n,+1,1,5,6,=,-,.,n,+1,1,+,+,=+(,-,)+,+(,-,),b,1,1,b,2,1,b,n,1,1,3,n,1,1,2,1,3,n,+1,1,(2),当,b,n,是等比数列时,甲同学说,:,a,n,一定是等比数列,乙同学说,:,a,n,一定不是等比数列,.,你认为他们的说法是否正确,?,为什么,?,解,:,甲,乙两个同学的说法均不正确,理由如下,:,设,b,n,的公比为,q,则,:,b,n,+1,b,n,a,n,a,n,+1,a,n,+1,a,n,+2,=,q,且,a,0.,a,n,+2,a,n,又,a,1,=1,a,2,=,a,a,1,a,3,a,5,a,2,n,-,1,是以,1,为首项,q,为公比的等比数列,.,a,2,a,4,a,6,a,2,n,是以,a,为首项,q,为公比的等比数列,.,即,a,n,为,:1,a,q,aq,q,2,aq,2,.,当,q,=,a,2,时,a,n,是等比数列,当,q,a,2,时,a,n,不,是等比数列,.,法二,:,举例说明,a,n,可能是等比数列,也可能不是,:,设,b,n,的公比为,q,取,a,=,q,=1,则,:,a,n,=1(,n,N,*,).,此时,b,n,=1,a,n,与,b,n,都,是等比数列,;,取,a,=2,q,=1,则,:,a,n,=,b,n,=2.,1(,n,为奇数,),2(,n,为偶数,),此时,b,n,是等比数列,而,a,n,不,是等比数列,.,