全国通用高考数学二轮复习板块四考前回扣回扣4数列.ppt

,剖析题型 提炼方法,实验解读,构建知识网络 强化答题语句,探究高考 明确考向,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,*,*,*,回扣,4,数列,板块四考前回扣,回归教材,易错提醒,内容索引,回扣训练,回归教材,1.,牢记概念与公式,等差数列、等比数列,等差数列,等比数列,通项公式,a,n,a,1,(,n,1),d,a,n,a,1,q,n,1,(,q,0),前,n,项和,2.,活用定理与结论,(1),等差、等比数列,a,n,的常用性质,等差数列,等比数列,性质,若,m,，,n,，,p,，,q,N,*,，且,m,n,p,q,，则,a,m,a,n,a,p,a,q,；,a,n,a,m,(,n,m,),d,；,S,m,，,S,2,m,S,m,，,S,3,m,S,2,m,，,仍成等差数列,若,m,，,n,，,p,，,q,N,*,，且,m,n,p,q,，则,a,m,a,n,a,p,a,q,；,a,n,a,m,q,n,m,；,S,m,，,S,2,m,S,m,，,S,3,m,S,2,m,，,仍成等比数列,(,S,m,0),(2),判断等差数列的常用方法,定义法,a,n,1,a,n,d,(,常数,)(,n,N,*,),a,n,是等差数列,.,通项公式法,a,n,pn,q,(,p,，,q,为常数，,n,N,*,),a,n,是等差数列,.,中项公式法,2,a,n,1,a,n,a,n,2,(,n,N,*,),a,n,是等差数列,.,前,n,项和公式法,S,n,An,2,Bn,(,A,，,B,为常数，,n,N,*,),a,n,是等差数列,.,通项公式法,a,n,cq,n,(,c,，,q,均是不为,0,的常数，,n,N,*,),a,n,是等比数列,.,3.,数列求和的常用方法,(1),等差数列或等比数列的求和，直接利用公式求和,.,(2),形如,a,n,b,n,(,其中,a,n,为等差数列，,b,n,为等比数列,),的数列，利用错位相减法求和,.,(4),通项公式形如,a,n,(,1),n,n,或,a,n,a,(,1),n,(,其中,a,为常数，,n,N,*,),等正负项交叉的数列求和一般用并项法,.,并项时应注意分,n,为奇数、偶数两种情况讨论,.,(5),分组求和法：分组求和法是解决通项公式可以写成,c,n,a,n,b,n,形式的数列求和问题的方法，其中,a,n,与,b,n,是等差,(,比,),数列或一些可以直接求和的数列,.,(6),并项求和法：先将某些项放在一起求和，然后再求,S,n,.,易错提醒,1.,已知数列的前,n,项和求,a,n,，易忽视,n,1,的情形，直接用,S,n,S,n,1,表示,.,事实上，当,n,1,时，,a,1,S,1,；当,n,2,时，,a,n,S,n,S,n,1,.,2.,易混淆几何平均数与等比中项，正数,a,，,b,的等比中项是,.,4.,易忽视等比数列中公比,q,0,导致增解，易忽视等比数列的奇数项或偶数项符号相同造成增解,.,8.,通项中含有,(,1),n,的数列求和时，要把结果写成,n,为奇数和,n,为偶数两种情况的分段形式,.,5.,运用等比数列的前,n,项和公式时，易忘记分类讨论,.,一定分,q,1,和,q,1,两种情况进行讨论,.,6.,利用错位相减法求和时，要注意寻找规律，不要漏掉第一项和最后一项,.,7.,裂项相消法求和时，裂项前后的值要相等，,回扣训练,解析,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,答案,1.,设等差数列,a,n,的前,n,项和为,S,n,，已知,S,13,0,，,S,14,0,，若,a,k,a,k,1,0,，,a,8,0,，,a,7,a,8,0,，,a,6,a,7,0,，,a,7,0,，,a,1,a,13,2,a,7,0,，,S,13,0,的最大自然数,n,的值为,12.,3.,设等差数列,a,n,的前,n,项和为,S,n,，且,a,1,0,，,a,3,a,10,0,，,a,6,a,7,0,的最大自然数,n,的值为,A.6 B.7,C.12 D.13,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,解析,由已知,9,，所以,a,n,1,a,n,2,，,所以数列,a,n,是公差为,2,的等差数列，,a,5,a,7,a,9,(,a,2,3,d,),(,a,4,3,d,),(,a,6,3,d,),(,a,2,a,4,a,6,),9,d,9,9,2,27,，,所以,3.,故选,C.,答案,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,解析,5.,已知正数组成的等比数列,a,n,，若,a,1,a,20,100,，那么,a,7,a,14,的最小值为,A.20 B.25,C.50 D.,不存在,解析,在正数组成的等比数列,a,n,中，,因为,a,1,a,20,100,，由等比数列的性质可得,a,1,a,20,a,7,a,14,100,，,当且仅当,a,7,a,14,10,时取等号，所以,a,7,a,14,的最小值为,20.,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,6.,已知数列,a,n,的前,n,项和为,S,n,，若,S,n,2,a,n,4(,n,N,*,),，则,a,n,等于,A.2,n,1,B.2,n,C.2,n,1,D.2,n,2,解析,a,n,1,S,n,1,S,n,2,a,n,1,4,(2,a,n,4),a,n,1,2,a,n,，再令,n,1,，,S,1,2,a,1,4,a,1,4,，,数列,a,n,是以,4,为首项，,2,为公比的等比数列，,a,n,42,n,1,2,n,1,，故选,A.,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,d,2,a,1,d,，,d,0,，,d,a,1,，,解析,在等差数列,a,n,中，,a,2,，,a,4,，,a,8,成等比数列，,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,8.,已知,S,n,为数列,a,n,的前,n,项和，若,a,n,(4,cos,n,),n,(2,cos,n,)(,n,N,*,),，则,S,20,等于,A.31 B.122,C.324 D.484,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,解析,由题意可知，因为,a,n,(4,cos,n,),n,(2,cos,n,),，,所以,S,20,(,a,1,a,3,a,19,),(,a,2,a,4,a,20,),122,，故选,B.,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,解析,由题意,a,1,，,a,3,，,a,13,成等比数列，可得,(1,2,d,),2,1,12,d,，,解得,d,2,，,故,a,n,2,n,1,，,S,n,n,2,，,当且仅当,n,2,时取得最小值,4.,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,11.,在等差数列,a,n,中，已知,a,3,a,8,10,，则,3,a,5,a,7,_.,20,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,解析,设公差为,d,，,则,a,3,a,8,2,a,1,9,d,10,，,3,a,5,a,7,3(,a,1,4,d,),(,a,1,6,d,),4,a,1,18,d,2,10,20.,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,12.,若等比数列,a,n,的各项均为正数，且,a,10,a,11,a,9,a,12,2e,5,，则,ln,a,1,ln,a,2,ln,a,20,_.,50,答案,解析,解析,数列,a,n,为等比数列，且,a,10,a,11,a,9,a,12,2e,5,，,a,10,a,11,a,9,a,12,2,a,10,a,11,2e,5,，,a,10,a,11,e,5,，,ln,a,1,ln,a,2,ln,a,20,ln,(,a,1,a,2,a,20,),ln,(,a,10,a,11,),10,ln,(e,5,),10,ln,e,50,50.,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,13.,数列,a,n,的前,n,项和为,S,n,，已知,a,1,2,，,S,n,1,(,1),n,S,n,2,n,，则,S,100,_.,198,答案,解析,解析,当,n,为偶数时，,S,n,1,S,n,2,n,，,S,n,2,S,n,1,2,n,2,，,所以,S,n,2,S,n,4,n,2,，,故,S,n,4,S,n,2,4(,n,2),2,，所以,S,n,4,S,n,8,，,由,a,1,2,知，,S,1,2,，又,S,2,S,1,2,，所以,S,2,4,，,因为,S,4,S,2,4,2,2,10,，,所以,S,4,6,，所以,S,8,S,4,8,，,S,12,S,8,8,，,，,S,100,S,96,8,，,所以,S,100,24,8,S,4,192,6,198.,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,14.,若数列,a,n,满足,a,2,a,1,a,3,a,2,a,4,a,3,a,n,1,a,n,，则称数列,a,n,为,“,差递减,”,数列,.,若数列,a,n,是,“,差递减,”,数列，且其通项,a,n,与其前,n,项和,S,n,(,n,N,*,),满足,2,S,n,3,a,n,2,1(,n,N,*,),，则实数,的取值范,围是,_.,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,解析,当,n,1,时，,2,a,1,3,a,1,2,1,，,a,1,1,2,，,当,n,1,时，,2,S,n,1,3,a,n,1,2,1,，,所以,2,a,n,3,a,n,3,a,n,1,，,a,n,3,a,n,1,，,所以,a,n,(1,2,)3,n,1,，,a,n,a,n,1,(1,2,)3,n,1,(1,2,)3,n,2,(2,4,)3,n,2,，,依题意,(2,4,)3,n,2,是一个递减数列，,解答,15.,S,n,为等差数列,a,n,的前,n,项和，且,a,1,1,，,S,7,28.,记,b,n,lg,a,n,，其中,x,表示不超过,x,的最大整数，如,0.9,0,，,lg,99,1.,(1),求,b,1,，,b,11,，,b,101,；,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,解,设,a,n,的公差为,d,，据已知有,7,21,d,28,，,解得,d,1.,所以,a,n,的通项公式为,a,n,n,(,n,N,*,).,b,1,lg,1,0,，,b,11,lg,11,1,，,b,101,lg,101,2.,解答,(2),求数列,b,n,的前,1 000,项和,.,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,所以数列,b,n,的前,1 000,项和为,1,90,2,900,3,1,1 893.,解答,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,由,化简得,(,a,n,a,n,1,)(,a,n,a,n,1,2),0,，,又数列,a,n,各项为正数，,当,n,2,时，,a,n,a,n,1,2,，,故数列,a,n,成等差数列，公差为,2,，,证明,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,