2019届北京专用高考数学一轮复习第六章数列第一节数列的概念及简单表示法讲义理.ppt

,总纲目录,教材研读,考点突破,栏目索引,第一节数列的概念及简单表示法,总纲目录,教材研读,1.,数列的定义,考点突破,2.,数列的分类,3.,数列的表示方法,考点二由递推关系式求数列的通项公式,考点一由an与Sn的关系求通项an,4.,数列的通项公式,考点三数列的性质,5.,已知数列an的前n项和Sn,教材研读,1.数列的定义,按照,一定顺序,排列的一列数叫做数列,数列中的每一个数叫做这,个数列的,项,.,2.数列的分类,分类原则,类型,满足条件,按项数分类,有穷数列,项数,有限,无穷数列,项数,无限,按项与项,间的大小,关系分类,递增数列,a,n,+1,a,n,其中,n,N,*,递减数列,a,n,+1,a,n,常数列,a,n,+1,=,a,n,按其他,标准分类,有界数列,存在正数,M,使对于任意的,n,N,*,都有|,a,n,|,M,摆动数列,从第二项起,有些项大于它的前一项,有些项小于它的前一项,3.数列的表示方法,数列有三种表示方法,它们分别是,列表法,、,图象法,和,解析式法,.,4.数列的通项公式,如果数列,a,n,的第,n,项与,序号,n,之间的关系可以用一个式子来表,示,那么这个式子叫做这个数列的通项公式.,5.已知数列,a,n,的前,n,项和,S,n,则,a,n,=,1.数列,a,n,的前,n,项和为,S,n,若,S,n,-,S,n,-1,=2,n,-1(,n,2),且,S,2,=3,则,a,1,+,a,3,的值为,(),A.1B.3C.5D.6,C,答案,C由题意知,S,2,-,S,1,=3-,a,1,=3,a,1,=0,a,3,=,S,3,-,S,2,=2,3-1=5,a,1,+,a,3,=5,故,选C.,2.(2018北京海淀期中,4)已知数列,a,n,满足,a,1,+,a,2,+,+,a,n,=2,a,2,(,n,=1,2,3,),则,(),A.,a,1,0C.,a,1,a,2,D.,a,2,=0,D,答案,D,a,1,=2,a,2,a,1,+,a,2,=2,a,2,a,1,=,a,1,+,a,2,a,2,=0,故选D.,3.(2017北京西城一模,7)若数列,a,n,的通项公式为,a,n,=|,n,-,c,|(,n,N,*,),则“,c,1”是“,a,n,为递增数列”的,(),A.充分不必要条件B.必要不充分条件,C.充要条件D.既不充分也不必要条件,A,答案,A充分性:当,c,1时,a,n,=|,n,-,c,|=,n,-,c,(,n,N,*,),易知,a,n,为递增数列,所以充分性成立;,必要性:当,c,=1.1时,易验证,a,n,为递增数列,所以必要性不成立.,综上,“,c,1”是“,a,n,为递增数列”的充分不必要条件.,4.已知数列,a,n,的通项公式为,a,n,=2,n,(3,n,-13),则数列,a,n,的前,n,项和,S,n,的最,小值是,(),A.,S,3,B.,S,4,C.,S,5,D.,S,6,B,答案,B易知数列,a,n,是递增数列,且,a,1,a,2,a,3,a,4,的值为负数,从第5项开,始以后各项均为正数,所以数列,a,n,的前,n,项和,S,n,的最小值是,S,4,选B.,5.已知数列,a,n,的前,n,项和,S,n,=2,n,-3,则数列,a,n,的通项公式是,a,n,=,.,答案,a,n,=,解析,当,n,=1时,a,1,=,S,1,=2-3=-1,当,n,2时,a,n,=,S,n,-,S,n,-1,=(2,n,-3)-(2,n,-1,-3)=2,n,-2,n,-1,=2,n,-1,.,又,a,1,=-1不适合上式,故,a,n,=,6.已知函数,f,(,x,)的对应关系如下表所示,数列,a,n,满足,a,1,=3,a,n,+1,=,f,(,a,n,),则,a,4,=,1,a,2 015,=,3,.,x,1,2,3,f,(,x,),3,2,1,答案,1;3,解析,a,1,=3,a,2,=,f,(,a,1,)=,f,(3)=1,a,3,=,f,(,a,2,)=,f,(1)=3,a,4,=,f,(,a,3,)=,f,(3)=1,可,知数列,a,n,是以2为周期的周期数列,a,2 015,=,a,1,=3.,考点一由,a,n,与,S,n,的关系求通项,a,n,考点突破,典例1,已知数列,a,n,的前,n,项和,S,n,求,a,n,的通项公式.,(1),S,n,=2,n,2,-3,n,;(2),S,n,=3,n,+,b,.,解析,(1)当,n,=1时,a,1,=,S,1,=-1;当,n,2时,a,n,=,S,n,-,S,n,-1,=(2,n,2,-3,n,)-2(,n,-1),2,-3(,n,-1),=4,n,-5,又,a,1,=-1也适合上式,因此,a,n,=4,n,-5(,n,N,*,).,(2)当,n,=1时,a,1,=,S,1,=3+,b,当,n,2时,a,n,=,S,n,-,S,n,-1,=(3,n,+,b,)-(3,n,-1,+,b,)=23,n,-1,.,当,b,=-1时,a,1,适合上式;当,b,-1时,a,1,不适合上式.,当,b,=-1时,a,n,=23,n,-1,(,n,N,*,);,当,b,-1时,a,n,=,方法技巧,已知,S,n,求,a,n,的三个步骤,(1)先利用,a,1,=,S,1,求出,a,1,;,(2)用,n,-1替换,S,n,中的,n,得出,S,n,-1,利用,a,n,=,S,n,-,S,n,-1,(,n,2)便可求出当,n,2时,a,n,的,表达式;,(3)看,a,1,是否符合,n,2时,a,n,的表达式,如果符合,则可以把数列的通项公式,合写,如果不符合,则应该分,n,=1与,n,2两段来写.,1-1,已知数列,a,n,的前,n,项和,S,n,=3-3,2,n,n,N,*,则,a,n,=,-3,2,n,-1,.,答案,-3,2,n,-1,解析,分情况讨论:,当,n,=1时,a,1,=,S,1,=3-3,2,1,=-3;,当,n,2时,a,n,=,S,n,-,S,n,-1,=(3-3,2,n,)-(3-3,2,n,-1,)=-3,2,n,-1,.,综合,得,a,n,=-3,2,n,-1,.(经检验,a,1,=-3符合该式),典例2,(1)已知数列,a,n,满足,a,1,=,a,n,+1,=,a,n,+,则,a,n,=,;,(2)若数列,a,n,满足,a,1,=,a,n,+1,=,a,n,则通项,a,n,=,;,(3)若数列,a,n,满足,a,1,=1,a,n,+1,=2,a,n,+3,则,a,n,=,.,考点二由递推关系式求数列的通项公式,答案,(1),-,(2),(3)2,n,+1,-3,解析,(1)由条件知,a,n,+1,-,a,n,=,=,=,-,则,n,2时,(,a,2,-,a,1,)+(,a,3,-,a,2,)+(,a,4,-,a,3,)+,+(,a,n,-,a,n,-1,),=,+,+,+,+,即,n,2时,a,n,-,a,1,=1-,又,a,1,=,n,2时,a,n,=1-,+,=,-,又,n,=1时,a,1,=,符合上式,a,n,=,-,.,(2)由,a,n,+1,=,a,n,(,a,n,0),得,=,故,n,2时,a,n,=,a,1,=,=,又,n,=1时,a,1,=,符,合上式,a,n,=,.,(3)设递推关系式,a,n,+1,=2,a,n,+3可以转化为,a,n,+1,-,t,=2(,a,n,-,t,),则,a,n,+1,=2,a,n,-,t,则,t,=-3.,故,a,n,+1,+3=2(,a,n,+3).,令,b,n,=,a,n,+3,则,b,1,=,a,1,+3=4,b,n,0,且,=,=2.,所以,b,n,是以4为首项,2为公比的等比数列.,所以,b,n,=4,2,n,-1,=2,n,+1,即,a,n,=2,n,+1,-3.,方法技巧,由数列的递推关系求通项公式的常用方法,递推式,方法,转化过程,a,n,+1,=,a,n,+,f,(,n,),累加法,(,a,2,-,a,1,)+(,a,3,-,a,2,)+,+(,a,n,-,a,n,-1,)=,a,n,-,a,1,(,n,2),=,f,(,n,),累乘法,=,(,n,2),a,n,+1,=,pa,n,+,q,(,p,0,1,q,0),转化法,化为,a,n,+1,+,m,=,p,(,a,n,+,m,),以此构造等比数列,a,n,+,m,2-1,数列,a,n,的首项,a,1,=2,且(,n,+1),a,n,=,na,n,+1,则,a,3,的值为,(),A.5B.6C.7D.8,答案,B由题意知,=,由累乘法得,=,n,2,a,n,=,na,1,n,2,又,a,1,=2,a,n,=2,n,n,2,又,a,1,=2适合上式,a,n,=2,n,a,3,=6,故选B.,B,2-2,数列,a,n,中,a,1,=2,a,n,+1,=,a,n,+,n,+1,则,a,n,=,.,答案,解析,由条件知,a,n,+1,-,a,n,=,n,+1,则,n,2时,a,n,=(,a,2,-,a,1,)+(,a,3,-,a,2,)+(,a,4,-,a,3,)+,+(,a,n,-,a,n,-1,)+,a,1,=(2+3+4+,+,n,)+2=,又当,n,=1时,a,1,=2符合上式,a,n,=,.,典例3,(1)已知数列,a,n,的首项为2,且数列,a,n,满足,a,n,+1,=,数列,a,n,的前,n,项和为,S,n,则,S,2 016,为(),A.504B.588C.-588D.-504,(2)设等比数列,a,n,满足,a,1,+,a,3,=10,a,2,+,a,4,=5,则,a,1,a,2,a,n,的最大值为,.,考点三数列的性质,答案,(1)C(2)64,解析,(1),a,1,=2,a,n,+1,=,a,2,=,a,3,=-,a,4,=-3,a,5,=2,数列,a,n,是以4为周期的数列,且,a,1,+,a,2,+,a,3,+,a,4,=-,2 016,4=504,S,2 016,=504,=-588,故选C.,(2)设,a,n,的公比为,q,于是,a,1,(1+,q,2,)=10,a,1,(,q,+,q,3,)=5,联立解得,a,1,=8,q,=,a,n,=2,4-,n,a,1,a,2,a,n,=2,3+2+1+,+(4-,n,),=,=,2,6,=64.,a,1,a,2,a,n,的最大值为64.,方法技巧,(1)解决数列的单调性问题可用以下三种方法:,用作差比较法,根据,a,n,+1,-,a,n,的符号进行判断.,用作商比较法,根据,(,a,n,中各项都同号)与1的大小关系进行判断.,结合相应函数的图象直观判断.,(2)解决数列周期性问题的方法:,先根据已知条件求出数列的前几项,确定数列的周期,再根据周期求值.,(3)数列的最值问题可以利用数列的单调性或求函数最值的方法求解.,3-1,已知数列,a,n,的前,n,项和,S,n,=1-5+9-13+17-21+,+(-1),n,-1,(4,n,-3),则,S,11,=,(),A.-21B.-19C.19D.21,答案,D,S,11,=-4,5+(-1),11-1,(4,11-3)=-20+41=21.,D,