,总纲目录,教材研读,考点突破,栏目索引,第一节数列的概念与简单表示法,(江苏专用)高考数学总复习第六章第一节数列的概念与简单表示法课件苏教版,1.数列的定义,2.数列的分类,3.数列与函数的关系,4.数列的通项公式,教材研读,1.数列的定义2.数列的分类3.数列与函数的关系4.数列的通,考点一 由an与Sn的关系求通项公式,考点二 由数列的递推公式求通项公式,考点突破,考点三 数列的性质,考点一 由an与Sn的关系求通项公式考点二 由数,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,.,教材研读,教材研读,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,n,N,*,恒成立,即(,n,+1),2,+,b,(,n,+1),n,2,+,bn,则,b,(-2,n,-1),max,=-3,则实数,b,的取值范围是(-3,+,).,5.设数列an的通项公式为an=n2+bn,若数列an,6.,(2018南京师大附中高三模拟)在数列,a,n,中,若,a,4,=1,a,12,=5,且任意连续,三项的和都是15,则,a,2 018,=,.,答案,9,解析,由任意连续三项的和都是15,得,a,n,+,a,n,+1,+,a,n,+2,=,a,n,+1,+,a,n,+2,+,a,n,+3,则,a,n,=,a,n,+3,则,a,12,=,a,3,=5,a,2,+,a,3,+,a,4,=15,则,a,2,=9,a,2 018,=,a,3,672+2,=,a,2,=9.,6.(2018南京师大附中高三模拟)在数列an中,若a4,考点一 由,a,n,与,S,n,的关系求通项公式,典例1,已知数列,a,n,的前,n,项和为,S,n,.,(1)若,S,n,=(-1),n,+1,n,求,a,5,+,a,6,及,a,n,;,(2)若,S,n,=3,n,+2,n,+1,求,a,n,.,考点突破,考点突破,解析,(1),a,5,+,a,6,=,S,6,-,S,4,=(-6)-(-4)=-2.,当,n,=1时,a,1,=,S,1,=1;,当,n,2时,a,n,=,S,n,-,S,n,-1,=(-1),n,+1,n,-(-1),n,(,n,-1),=(-1),n,+1,n,+(,n,-1),=(-1),n,+1,(2,n,-1),因为,a,1,也适合此式,所以,a,n,=(-1),n,+1,(2,n,-1)(,n,N,*,).,解析(1)a5+a6=S6-S4=(-6)-(-4)=-2,(2)当,n,=1时,a,1,=,S,1,=6;,当,n,2时,a,n,=,S,n,-,S,n,-1,=(3,n,+2,n,+1)-3,n,-1,+2(,n,-1)+1,=23,n,-1,+2,由于,a,1,不适合此式,所以,a,n,=,方法技巧,S,n,与,a,n,的关系问题的求解思路,(2)当n=1时,a1=S1=6;,根据所求结果的不同要求,将问题向不同的方向转化.,(1)利用,a,n,=,S,n,-,S,n,-1,(,n,2)转化为只含,S,n,S,n,-1,的关系式,再求解.,(2)利用,S,n,-,S,n,-1,=,a,n,(,n,2)转化为只含,a,n,a,n,-1,的关系式,再求解.,易错警示,注意检验,n,=1时的表达式是否可以与,n,2时的表达式合并.,根据所求结果的不同要求,将问题向不同的方向转化.,同类练,已知数列,a,n,的前,n,项和,S,n,=3+2,n,则数列,a,n,的通项公式为,.,答案,a,n,=,解析,当,n,=1时,a,1,=,S,1,=3+2=5;,当,n,2时,a,n,=,S,n,-,S,n,-1,=3+2,n,-(3+2,n,-1,)=2,n,-2,n,-1,=2,n,-1,.,因为当,n,=1时,不符合,a,n,=2,n,-1,所以数列,a,n,的通项公式为,a,n,=,同类练已知数列an的前n项和Sn=3+2n,则数列a,变式练,已知数列,a,n,的前,n,项和为,S,n,a,1,=1,S,n,=2,a,n,+1,则,S,n,=,.,答案,解析,S,n,=2,a,n,+1,当,n,2时,S,n,-1,=2,a,n,变式练已知数列an的前n项和为Sn,a1=1,Sn=2,a,n,=,S,n,-,S,n,-1,=2,a,n,+1,-2,a,n,(,n,2),3,a,n,=2,a,n,+1,(,n,2),又易知,a,2,=,a,n,0(,n,2),=,(,n,2).,a,n,=,(,n,2).,当,n,=1时,a,1,=1,=,an=Sn-Sn-1=2an+1-2an(n2),a,n,=,S,n,=2,a,n,+1,=2,=,.,an=,深化练1,设数列,a,n,满足,a,1,+3,a,2,+,+(2,n,-1),a,n,=2,n,则,a,n,=,.,答案,(,n,N,*,),深化练1设数列an满足a1+3a2+(2n-1)a,解析,因为,a,1,+3,a,2,+,+(2,n,-1),a,n,=2,n,抽以当,n,2时,a,1,+3,a,2,+,+(2,n,-3),a,n,-1,=2(,n,-1).,两式相减得(2,n,-1),a,n,=2.,所以,a,n,=,(,n,2).,又由题设可得,a,1,=2,满足上式,从而,a,n,的通项公式为,a,n,=,(,n,N,*,).,解析因为a1+3a2+(2n-1)an=2n,深化练2,若数列,a,n,满足,a,1,a,2,a,3,a,n,=,n,2,+3,n,+2,则数列,a,n,的通项公式,为,.,答案,a,n,=,深化练2若数列an满足a1a2a3an=n2,解析,a,1,a,2,a,3,a,n,=(,n,+1)(,n,+2),当,n,=1时,a,1,=6;,当,n,2时,a,1,a,2,a,3,a,n,-1,=,n,(,n,+1),两式相除得,a,n,=,(,n,2),经检验,a,1,=6不符合此式,所以,a,n,=,解析a1a2a3an=(n+1)(n+2),考点二 由数列的递推公式求通项公式,典例2,根据下列条件确定数列,a,n,的通项公式.,(1),a,1,=1,a,n,+1,=3,a,n,+2;,(2),a,1,=1,a,n,=,a,n,-1,(,n,2);,(3),a,1,=2,a,n,+1,=,a,n,+3,n,+2.,考点二 由数列的递推公式求通项公式,解析,(1),a,n,+1,=3,a,n,+2,a,n,+1,+1=3(,a,n,+1),即,=3.,数列,a,n,+1为等比数列,公比,q,=3.,又,a,1,+1=2,a,n,+1=2,3,n,-1,.,解析(1)an+1=3an+2,a,n,=2,3,n,-1,-1(,n,N,*,).,(2)由,a,n,=,a,n,-1,(,n,2),得,a,n,-1,=,a,n,-2,(,n,3),a,2,=,a,1,.,则,a,n,=,a,1,=,=,(,n,2).,又,a,1,=1符合上式,a,n,=,(,n,N,*,).,(3),a,n,+1,-,a,n,=3,n,+2,an=23n-1-1(nN*).,a,n,-,a,n,-1,=3,n,-1(,n,2),a,n,=(,a,n,-,a,n,-1,)+(,a,n,-1,-,a,n,-2,)+,+(,a,2,-,a,1,)+,a,1,=,(,n,2).,又,a,1,=2符合上式,a,n,=,n,2,+,(,n,N,*,).,an-an-1=3n-1(n2),方法技巧,由数列的递推关系求通项公式的常用方法,已知数列的递推关系,求数列的通项公式时,通常用累加、累乘、构造,法求解.当出现,a,n,=,a,n,-1,+,m,时,构造等差数列;当出现,a,n,=,xa,n,-1,+,y,时,构造等比,数列;当出现,a,n,=,a,n,-1,+,f,(,n,)时,用累加法求解;当出现,=,f,(,n,)时,用累乘法,求解.,方法技巧,2-1,(1)在数列,a,n,中,a,1,=2,a,n,+1,=,a,n,+,求数列,a,n,的通项公式.,(2)在数列,a,n,中,a,1,=1,a,n,+1,=,求数列,a,n,的通项公式.,2-1(1)在数列an中,a1=2,an+1=an+,解析,(1)由题意知,a,n,+1,-,a,n,=,=,-,a,n,=(,a,n,-,a,n,-1,)+(,a,n,-1,-,a,n,-2,)+,+(,a,2,-,a,1,)+,a,1,=,+,+,+,+,+2,=3-,(,n,N,*,).,(2),a,n,+1,=,a,1,=1,a,n,0,=,+,解析(1)由题意知,an+1-an=-,即,-,=,.,又,a,1,=1,=1,是以1为首项,为公差的等差数列,=,+(,n,-1),=,a,n,=,(,n,N,*,).,即-=.,考点三 数列的性质,角度一数列的周期性,典例3,(1)已知数列,a,n,的各项均为正整数,对于,n,N,*,有,a,n,+1,=,a,1,=11,则,a,65,=,.,(2)已知数列,a,n,满足,a,1,=2,a,n,+1,=,(,n,N,*,),则,a,1,a,2,a,3,a,2 017,=,.,答案,(1)31(2)2,考点三 数列的性质典例3(1)已知数列an的各项,解析,(1)由题设知,a,1,=11,a,2,=3,11+5=38,a,3,=,=19,a,4,=3,19+5=62,a,5,=,=31,a,6,=3,31+5=98,a,7,=,=49,a,8,=3,49+5=152,a,9,=,=19,所以数列,a,n,从第3项开始是周期为6的周期数列,所以,a,65,=,a,3+(6,10+2),=,a,5,=31.,(2)由,a,1,=2,a,n,+1,=,(,n,N,*,)得,a,2,=-3,a,3,=-,a,4,=,a,5,=2,所以数列,a,n,的,周期为4,且,a,1,a,2,a,3,a,4,=1,则,a,1,a,2,a,3,a,2017,=,a,1,=2.,解析(1)由题设知,a1=11,a2=311+5=38,探究1,若典例3(2)中的条件不变,则,a,1,+,a,2,+,a,3,+,+,a,2 017,=,.,答案,-586,解析,由典例3(2)的解析知数列,a,n,的周期为4,且,a,1,+,a,2,+,a,3,+,a,4,=-,则,a,1,+,a,2,+,a,3,+,+,a,2 017,=504(,a,1,+,a,2,+,a,3,+,a,4,)+,a,1,=-,504+2=-586.,探究1若典例3(2)中的条件不变,则a1+a2+a3+,探究2,若典例3(2)中的条件增加,a,1,+,a,2,+,a,3,+,+,a,k,=-589,k,N,*,则结论变,为,k,=,.,答案,2 018,解析,由探究1的解析可知,a,1,+,a,2,+,a,3,+,+,a,2 018,=504(,a,1,+,a,2,+,a,3,+,a,4,)+,a,1,+,a,2,=-,504+2-3=-589,所以,k,=2 018.,探究2若典例3(2)中的条件增加a1+a2+a3+ak,方法技巧,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,再根据周期性求值.,方法技巧,典例4,已知数列,a,n,的前,n,项和,S,n,=3,n,(,-,n,)-6,若数列,a,n,单调递减,则,的,取值范围是,.,角度二数列的单调性,答案,(-,2),解析,由题意可得,a,1,=,S,1,=3,-9,当,n,2时,a,n,=,S,n,-,S,n,-1,=3,n,-1,(2,-2,n,-1),因为,a,n,单调递减,所以,a,2,a,1,解得,2,当,n,2时,a,n,+1,-,a,n,=3,n,-1,(4,-