2019届全国通用高考数学二轮复习第一篇求准提速基础小题不失分第12练数列讲义文.ppt

第一篇求准提速,基础小题不失分,第,12,练数列,明考情,数列在高考中以,“,一大一小,”,的形式考查,.,“,一小,”,考查频率较高，难度为中档,.,知考向,1.,等差数列与等比数列,.,2.,数列的通项与求和,.,3.,等差、等比数列的综合应用,.,研透考点,核心考点突破练,栏目索引,明辨是非,易错易混专项练,演练模拟,高考押题冲刺练,研透考点,核心考点突破练,考点一等差数列与等比数列,要点重组,(1),在等差数列中，若,m,n,p,q,(,m,，,n,，,p,，,q,N,*,),，则,a,m,a,n,a,p,a,q,.,(2),若,a,n,是等差数列，则,也是等差数列,.,(3),在等差数列,a,n,中，,S,n,，,S,2,n,S,n,，,S,3,n,S,2,n,也成等差数列,.,(4),在等比数列中，若,m,n,p,q,(,m,，,n,，,p,，,q,N,*,),，则,a,m,a,n,a,p,a,q,.,(5),在等比数列中，,S,n,，,S,2,n,S,n,，,S,3,n,S,2,n,也成等比数列,(,n,为偶数且,q,1,除外,).,1.(2016,全国,),已知等差数列,a,n,的前,9,项和为,27,，,a,10,8,，则,a,100,等于,A.100 B.99 C.98 D.97,答案,解析,a,100,a,10,90,d,98,，故选,C.,1,2,3,4,5,6,2.,已知数列,1,a,n,是以,2,为公比的等比数列，且,a,1,1,，则,a,5,等于,A.31 B.24 C.21 D.7,解析,由题意可知，,1,a,n,22,n,1,2,n,，,则,a,n,2,n,1,，所以,a,5,31,，故选,A.,答案,解析,1,2,3,4,5,6,3.(2017,长春南关区校级模拟,),已知等差数列,a,n,满足：,a,2,2,，,S,n,S,n,3,54(,n,3),，,S,n,100,，则,n,等于,A.7 B.8 C.9 D.10,解析,等差数列,a,n,满足：,a,2,2,，,S,n,S,n,3,54(,n,3),，,S,n,100,，,a,n,a,n,1,a,n,2,54(,n,3).,又数列,a,n,为等差数列，,3,a,n,1,54(,n,2),，,a,n,1,18(,n,2).,又,a,2,2,，,S,n,100,，,n,10.,1,2,3,4,5,6,答案,解析,解析,设,S,2,k,，则,S,4,3,k,，,由数列,a,n,为等比数列,(,易知数列,a,n,的公比,q,1),，,得,S,2,，,S,4,S,2,，,S,6,S,4,为等比数列，,又,S,2,k,，,S,4,S,2,2,k,，,S,6,S,4,4,k,，,S,6,7,k,，,答案,解析,1,2,3,4,5,6,5.(2017,安徽蚌埠质检,),数列,a,n,是以,a,为首项，,q,(,q,1),为公比的等比数列，数列,b,n,满足,b,n,1,a,1,a,2,a,n,(,n,1,，,2,，,),，数列,c,n,满足,c,n,2,b,1,b,2,b,n,(,n,1,，,2,，,),，若,c,n,为等比数列，则,a,q,等于,答案,解析,1,2,3,4,5,6,解析,由题意知，,a,n,aq,n,1,，,1,2,3,4,5,6,6.,已知,a,n,为等差数列，,a,1,a,3,a,5,105,，,a,2,a,4,a,6,99,，以,S,n,表示,a,n,的前,n,项和，则使得,S,n,达到最大值的,n,是,_.,20,当,n,20,时，,S,n,取得最大值,.,1,2,3,4,5,6,答案,解析,考点二数列的通项与求和,方法技巧,(1),已知数列的递推关系，求数列的通项时，通常利用累加法、累乘法、构造法求解,.,7,8,9,10,11,12,答案,解析,又,a,1,1,，,7,8,9,10,11,12,答案,解析,7,8,9,10,11,12,7,8,9,10,11,12,A.4 B.16 C.32 D.64,所以,b,n,是以,2,为公比的等比数列，,所以,b,6,b,7,b,8,(,b,1,b,2,b,3,),2,5,2,2,5,64,，故选,D.,7,8,9,10,11,12,答案,解析,A.2 017 B.2 019 C.1 009 D.1 008,由倒序相加，得,2,a,n,2(,n,1),，,a,n,n,1,，,所以,a,2 018,2 018,1,2 017,，故选,A.,7,8,9,10,11,12,答案,解析,11.,设,S,n,是数列,a,n,的前,n,项和，且,a,1,1,，,a,n,1,S,n,S,n,1,，则,S,n,_.,解析,由题意，得,S,1,a,1,1,，,又由,a,n,1,S,n,S,n,1,，得,S,n,1,S,n,S,n,S,n,1,，,7,8,9,10,11,12,答案,解析,7,8,9,10,11,12,12.,数列,a,n,的前,n,项和记为,S,n,，,a,1,1,，,a,n,1,2,S,n,1(,n,N,*,),，则数列,a,n,的通项公式是,_.,a,n,3,n,1,解析,由,a,n,1,2,S,n,1,，可得,a,n,2,S,n,1,1(,n,2,),，,两式相减得,a,n,1,a,n,2,a,n,，即,a,n,1,3,a,n,(,n,2,).,又,a,2,2,S,1,1,3,，,a,2,3,a,1,，故,a,n,是首项为,1,，公比为,3,的等比数列，,a,n,3,n,1,.,答案,解析,考点三等差、等比数列的综合应用,方法技巧,巧用性质，整体考虑，减少换算量,.,13,14,15,16,17,18,答案,解析,14.(2017,石家庄一模,),已知函数,f,(,x,),的图象关于,x,1,对称，且,f,(,x,),在,(,1,，,),上单调，若数列,a,n,是公差不为,0,的等差数列，且,f,(,a,50,),f,(,a,51,),，则,a,n,的前,100,项的和为,A.,200 B.,100,C.,50,D.0,解析,可得,a,50,a,51,2,，又,a,n,是等差数列，,所以,a,1,a,100,a,50,a,51,2,，,答案,解析,13,14,15,16,17,18,15.(2017,全国,),我国古代数学名著算法统宗中有如下问题：,“,远望巍巍塔七层，红光点点倍加增，共灯三百八十一，请问尖头几盏灯？,”,意思是：一座,7,层塔共挂了,381,盏灯，且相邻两层中的下一层灯数是上一层灯数的,2,倍，则塔的顶层共有灯,A.1,盏,B.3,盏,C.5,盏,D.9,盏,解析,设塔的顶层的灯数为,a,1,，七层塔的总灯数为,S,7,，公比为,q,，,则由题意知,S,7,381,，,q,2,，,解得,a,1,3.,故选,B.,13,14,15,16,17,18,答案,解析,16.,若数列,a,n,对任意的正整数,n,和,m,等式,a,n,a,n,2,m,都成立，则称数列,a,n,为,m,阶梯等比数列,.,若,a,n,是,3,阶梯等比数列且,a,1,1,，,a,4,2,，则,a,10,_.,解析,由题意可知，当,a,n,是,3,阶梯等比数列时，,答案,解析,8,13,14,15,16,17,18,13,14,15,16,17,18,(,a,n,1,3,a,n,)(,a,n,1,2,a,n,),0,，,a,n,0,，,a,n,1,3,a,n,.,又,a,1,2,，,a,n,是首项为,2,，公比为,3,的等比数列，,答案,解析,3,n,1,18.(2017,湘潭市雨湖区模拟,),已知数列,a,n,是各项均为正整数的等差数列，公差,d,N,*,，且,a,n,中任意两项之和也是该数列中的一项，若,a,1,6,m,，其中,m,为给定的正整数，则,d,的所有可能取值的和为,_.,解析,公差,d,是,a,1,6,m,的约数，,d,2,i,3,j,(,i,，,j,0,，,1,，,2,，,，,m,),，,13,14,15,16,17,18,答案,解析,明辨是非,易错易混专项练,1.,在数列,a,n,中，,a,1,1,，,a,2,2,，当整数,n,1,时，,S,n,1,S,n,1,2(,S,n,S,1,),都成立，则,S,15,等于,A.210,B.211,C.224,D.225,解析,当,n,1,时，,S,n,1,S,n,S,n,S,n,1,2,，,a,n,1,a,n,2,，,a,n,1,a,n,2.,数列,a,n,从第二项开始组成公差为,2,的等差数列，,1,2,3,4,答案,解析,A.16,B.20,C.33,D.120,1,2,3,4,解析,a,2,2,a,1,2,，,a,3,a,2,1,3,，,a,4,2,a,3,6,，,a,5,a,4,1,7,，,a,6,2,a,5,14,，,所以前,6,项和,S,6,1,2,3,6,7,14,33,，故选,C.,答案,解析,3.,已知数列,a,n,的前,n,项和为,S,n,，且,S,n,2,n,2,3,n,k,，则,a,n,_.,解析,当,n,1,时，,a,1,S,1,2,1,2,3,1,k,k,1,，,当,n,2,时，,a,n,S,n,S,n,1,2,n,2,3,n,k,2(,n,1),2,3(,n,1),k,4,n,5.,显然，当,k,0,时，,a,1,1,，适合,a,n,4,n,5,，,所以数列,a,n,的通项公式为,a,n,4,n,5.,当,k,0,时，,a,1,k,1,1,，显然不适合,a,n,4,n,5.,1,2,3,4,答案,解析,解析,由题意，得,a,2,a,1,2,，,a,3,a,2,4,，,，,a,n,a,n,1,2(,n,1),，,累加整理可得,a,n,n,2,n,33,，,1,2,3,4,答案,解析,演练模拟,高考押题冲刺练,1.,已知在等比数列,a,n,中，,a,2,a,8,4,a,5,，在等差数列,b,n,中，,b,4,b,6,a,5,，则数列,b,n,的前,9,项和,S,n,等于,A.9,B.18,C.36,D.72,解析,a,2,a,8,4,a,5,，即,4,a,5,，,a,5,4.,又,a,5,b,4,b,6,2,b,5,4,，,b,5,2.,S,9,9,b,5,18,，故选,B.,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,2.(2017,自贡模拟,),设数列,a,n,的前,n,项和为,S,n,，若,S,n,1,，,S,n,，,S,n,2,成等差数列，且,a,2,2,，则,a,7,等于,A.16,B.32,C.64,D.128,1,2,3,4,5,6,7,8,9,10,11,12,解析,数列,a,n,的前,n,项和为,S,n,，若,S,n,1,，,S,n,，,S,n,2,成等差数列，且,a,2,2,，,由题意得,S,n,2,S,n,1,2,S,n,，得,a,n,2,a,n,1,a,n,1,0,，,即,a,n,2,2,a,n,1,，,a,n,从第二项起是公比为,2,的等比数列，,a,7,a,2,q,5,64.,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,3.,已知数列,2 016,，,2 017,，,1,，,2 016,，,，从第二项起，每一项都等于它的前后两项之和，则该数列的前,2 017,项和等于,A.2,016,B.2,017,C.1,D.0,解析,根据数列的规律可知，,该数列为,2 016,，,2 017,，,1,，,2 016,，,2 017,，,1,，,2 016,，,2 017,，,，,可知该数列是周期为,6,的数列，一个周期的和为,0,，,所以,S,2,017,S,1,2 016.,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,5.,设各项均为正数的数列,a,n,的前,n,项和为,S,n,，且,S,n,满足,(,3,n,2,n,4),S,n,2(,3,n,2,n,),0,，,n,N,*,，则数列,a,n,的通项公式是,A.,a,n,3,n,2,B.,a,n,4,n,3,C.,a,n,2,n,1,D.,a,n,2,n,1,解析,由,(,3,n,2,n,4),S,n,2(,3,n,2,n,),0,，,n,N,*,，,因式分解可得,2,S,n,(,3,n,2,n,)(,S,n,2),0,，,因为数列,a,n,的各项均为正数，所以,2,S,n,3,n,2,n,.,当,n,1,时，,2,a,1,3,1,，解得,a,1,1.,当,n,2,时，,2,a,n,2,S,n,2,S,n,1,3,n,2,n,3(,n,1),2,(,n,1),6,n,4,，即,a,n,3,n,2.,当,n,1,时，上式成立,.,所以,a,n,3,n,2(,n,N,*,).,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,A.143,B.156,C.168,D.195,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,7.,已知,S,n,是等差数列,a,n,的前,n,项和，且,S,6,S,7,S,5,，给出下列五个命题：,d,0,；,使,S,n,0,的最大,n,值为,12,；,数列,S,n,中的最大项为,S,11,；,|,a,6,|,a,7,|,，其中正确命题的个数是,A.5,B.4,C.3,D.1,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,解析,S,6,S,7,S,5,，,a,7,0,，,a,6,a,7,0,，,因此,|,a,6,|,a,7,|,；,d,a,7,a,6,0,；,而,S,13,13,a,7,0,的最大,n,值为,12.,由于,a,7,0,，数列,S,n,中的最大项为,S,6,，,错，,正确，故选,B.,1,2,3,4,5,6,7,8,9,10,11,12,8.(2017,永州二模,),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,A.(,，,2)B.(,，,3)C.(,，,4)D.(,，,5),解析,S,n,3,n,(,n,),6,，,S,n,1,3,n,1,(,n,1),6,，,n,1,，,由,，得,a,n,3,n,1,(,2,2,n,1)(,n,1,，,n,N,*,).,数列,a,n,为单调递减数列，,a,n,a,n,1,，,3,n,1,(,2,2,n,1),3,n,(,2,2,n,3),，化为,n,2(,n,1),，,3.,又,a,1,a,2,，,2.,综上，,2.,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,9.,设数列,a,n,的前,n,项和为,S,n,，满足,S,n,2,a,n,2,，则,_.,解析,由,S,n,2,a,n,2,，得,S,n,1,2,a,n,1,2(,n,2,),，,所以,a,n,2,a,n,2,a,n,1,，即,a,n,2,a,n,1,(,n,2,),，,所以数列,a,n,为等比数列，公比为,2,，,则,2,2,4.,答案,解析,4,10.,公差不为,0,的等差数列,a,n,的部分项,构成等比数列，,且,k,1,1,，,k,2,2,，,k,3,6,，则,k,4,_.,解析,根据题意可知，等差数列的,a,1,，,a,2,，,a,6,项成等比数列，,设等差数列的公差为,d,，,则有,(,a,1,d,),2,a,1,(,a,1,5,d,),，,解得,d,3,a,1,，,故,a,2,4,a,1,，,a,6,16,a,1,a,1,(,n,1)(,3,a,1,),64,a,1,，,解得,n,22,，即,k,4,22.,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,22,11.(2017,上海青浦区一模,),设数列,a,n,的通项公式为,a,n,n,2,bn,，若数列,a,n,是单调递增数列，则实数,b,的取值范围是,_.,解析,数列,a,n,是单调递增数列，,n,N,*,，,a,n,1,a,n,，,(,n,1),2,b,(,n,1),n,2,bn,，,化为,b,(,2,n,1).,数列,(,2,n,1),是单调递减数列，,当,n,1,时，,(,2,n,1),取得最大值,3,，,b,3.,答案,解析,(,3,，,),1,2,3,4,5,6,7,8,9,10,11,12,1,从而数列,a,n,的前,n,项和为,显然当,n,1,时，,S,n,取得最小值,1.,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,本课结束,