高考数学复习高考专题突破三高考中的数列问题理市赛课公开课一等奖省名师优质课获奖PPT课件.pptx

,高考专题突破三,高考中数列问题,1/47,考点自测,课时作业,题型分类深度剖析,内容索引,2/47,考点自测,3/47,1.(,苏州,月考,),数列,a,n,是公差不为,0,等差数列，且,a,1,，,a,3,，,a,7,为等比数列,b,n,中连续三项，则数列,b,n,公比为,_.,答案,解析,设数列,a,n,公差为,d,(,d,0),，,由,a,1,a,7,，得,(,a,1,2,d,),2,a,1,(,a,1,6,d,),，解得,a,1,2,d,，,2,4/47,2.,已知等差数列,a,n,前,n,项和为,S,n,，,a,5,5,，,S,5,15,，则数列,前,100,项和为,_.,答案,解析,设等差数列,a,n,首项为,a,1,，公差为,d,.,a,n,a,1,(,n,1),d,n,.,a,5,5,，,S,5,15,，,5/47,3.(,南通、淮安模拟,),在等比数列,a,n,中，,a,2,1,，公比,q,1.,若,a,1,，,4,a,3,，,7,a,5,成等差数列，则,a,6,值是,_.,答案,解析,因为,a,n,为等比数列，且,a,2,1,，所以,a,1,，,a,3,q,，,a,5,q,3,，,由,a,1,，,4,a,3,，,7,a,5,成等差数列得,8,q,7,q,3,，,解得,q,2,1(,舍去,),或,q,2,，故,a,6,a,2,q,4,.,6/47,4.(,课标全国,),设,S,n,是数列,a,n,前,n,项和，且,a,1,1,，,a,n,1,S,n,S,n,1,，,则,S,n,_.,答案,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,，,因为,S,n,0,，所以,1,，,所以,1,(,n,1),n,，所以,S,n,.,7/47,5.,已知数列,a,n,前,n,项和为,S,n,，对任意,n,N,*,都有,S,n,，若,1,S,k,9,(,k,N,*,),，则,k,值为,_.,答案,解析,4,由题意，,S,n,，,当,n,2,时，,S,n,1,，,两式相减，得,a,n,，,a,n,是以,1,为首项，以,2,为公比等比数列，,a,n,(,2),n,1,，,由,1,S,k,9,，得,4(,2),k,28,，,又,k,N,*,，,k,4.,a,n,2,a,n,1,，,又,a,1,1,，,8/47,题型分类深度剖析,9/47,题型一等差数列、等比数列综合问题,例,1,(,苏州暑假测试,),已知等差数列,a,n,公差为,2,，其前,n,项和,S,n,pn,2,2,n,，,n,N,*,.,(1),求实数,p,值及数列,a,n,通项公式；,解答,S,n,na,1,na,1,n,(,n,1),n,2,(,a,1,1),n,，,又,S,n,pn,2,2,n,，,n,N,*,，,所以,p,1,，,a,1,1,2,，即,a,1,3,，,所以,a,n,3,2(,n,1),2,n,1.,10/47,(2),在等比数列,b,n,中，,b,3,a,1,，,b,4,a,2,4,，若,b,n,前,n,项和为,T,n,.,求证：,数列,T,n,为等比数列,.,证实,因为,b,3,a,1,3,，,b,4,a,2,4,9,，所以,q,3.,所以,b,n,b,3,q,n,3,3,3,n,3,3,n,2,，所以,b,1,.,所以数列,T,n,是以,为首项，,3,为公比等比数列,.,11/47,等差数列、等比数列综合问题解题策略,(1),分析已知条件和求解目标，为最终处理问题设置中间问题，比如求和需要先求出通项、求通项需要先求出首项和公差,(,公比,),等，确定解题次序,.,(2),注意细节：在等差数列与等比数列综合问题中，假如等比数列公比不能确定，则要看其是否有等于,1,可能，在数列通项问题中第一项和后面项能否用同一个公式表示等，这些细节对解题影响也是巨大,.,思维升华,12/47,跟踪训练,1,在等差数列,a,n,中，,a,10,30,，,a,20,50.,(1),求数列,a,n,通项公式；,解答,设数列,a,n,公差为,d,，则,a,n,a,1,(,n,1),d,，,由,a,10,30,，,a,20,50,，得方程组,解得,所以,a,n,12,(,n,1)2,2,n,10.,13/47,(2),令,b,n,，证实：数列,b,n,为等比数列；,证实,由,(1),，得,b,n,2,a,n,10,2,2,n,10,10,2,2,n,4,n,，,所以,b,n,是首项为,4,，公比为,4,等比数列,.,14/47,(3),求数列,nb,n,前,n,项和,T,n,.,解答,由,nb,n,n,4,n,，得,T,n,1,4,2,4,2,n,4,n,，,4,T,n,1,4,2,(,n,1),4,n,n,4,n,1,，,，得,3,T,n,4,4,2,4,n,n,4,n,1,15/47,题型二数列通项与求和,例,2,已知数列,a,n,前,n,项和为,S,n,，在数列,b,n,中，,b,1,a,1,，,b,n,a,n,a,n,1,(,n,2),，且,a,n,S,n,n,.,(1),设,c,n,a,n,1,，求证：,c,n,是等比数列；,证实,a,n,S,n,n,，,a,n,1,S,n,1,n,1.,，得,a,n,1,a,n,a,n,1,1,，,2,a,n,1,a,n,1,，,2(,a,n,1,1),a,n,1,，,a,n,1,是等比数列,.,首项,c,1,a,1,1,，又,a,1,a,1,1.,又,c,n,a,n,1,，,c,n,是以,为首项，,为公比等比数列,.,16/47,(2),求数列,b,n,通项公式,.,解答,a,n,c,n,1,1,(),n,.,当,n,2,时，,b,n,a,n,a,n,1,又,b,1,a,1,，代入上式也符合，,b,n,(),n,.,17/47,(1),普通求数列通项往往要结构数列，此时要从证结论出发，这是很主要解题信息,.,(2),依据数列特点选择适当求和方法，惯用有错位相减法，分组求和法，裂项求和法等,.,思维升华,18/47,跟踪训练,2,已知,a,n,是等差数列，其前,n,项和为,S,n,，,b,n,是等比数列，且,a,1,b,1,2,，,a,4,b,4,21,，,S,4,b,4,30.,(1),求数列,a,n,和,b,n,通项公式；,解答,设等差数列,a,n,公差为,d,，等比数列,b,n,公比为,q,.,由,a,1,b,1,2,，得,a,4,2,3,d,，,b,4,2,q,3,，,S,4,8,6,d,.,由条件,a,4,b,4,21,，,S,4,b,4,30,，,所以,a,n,n,1,，,b,n,2,n,，,n,N,*,.,19/47,(2),记,c,n,a,n,b,n,，,n,N,*,，求数列,c,n,前,n,项和,.,解答,由题意知,c,n,(,n,1),2,n,.,记,T,n,c,1,c,2,c,3,c,n,.,则,T,n,2,2,3,2,2,4,2,3,n,2,n,1,(,n,1),2,n,，,2,T,n,2,2,2,3,2,3,(,n,1),2,n,1,n,2,n,(,n,1)2,n,1,，,所以,T,n,2,2,(2,2,2,3,2,n,),(,n,1),2,n,1,，,即,T,n,n,2,n,1,，,n,N,*,.,20/47,题型三数列与其它知识交汇,命题点,1,数列与函数交汇,例,3,已知二次函数,f,(,x,),ax,2,bx,图象过点,(,4,n,0),，且,f,(0),2,n,，,n,N,*,，数列,a,n,满足,，且,a,1,4.,(1),求数列,a,n,通项公式；,解答,21/47,f,(,x,),2,ax,b,，由题意知,b,2,n,，,16,n,2,a,4,nb,0,，,a,，,则,f,(,x,),2,nx,，,n,N,*,.,数列,a,n,满足,又,f,(,x,),x,2,n,，,由叠加法可得,2,4,6,2(,n,1),n,2,n,，,化简可得,a,n,(,n,2),，,当,n,1,时，,a,1,4,也符合，,a,n,(,n,N,*,).,22/47,(2),记,b,n,，求数列,b,n,前,n,项和,T,n,.,解答,T,n,b,1,b,2,b,n,23/47,命题点,2,数列与不等式交汇,例,4,设各项均为正数数列,a,n,前,n,项和为,S,n,，且,S,n,满足,(,n,2,n,3),S,n,3(,n,2,n,),0,，,n,N,*,.,(1),求,a,1,值；,令,n,1,代入得,a,1,2(,负值舍去,).,解答,24/47,(2),求数列,a,n,通项公式；,解答,由,(,n,2,n,3),S,n,3(,n,2,n,),0,，,n,N,*,，,得,S,n,(,n,2,n,)(,S,n,3),0.,又已知数列,a,n,各项均为正数，故,S,n,n,2,n,.,当,n,2,时，,a,n,S,n,S,n,1,n,2,n,(,n,1),2,(,n,1),2,n,，,当,n,1,时，,a,1,2,也满足上式，,a,n,2,n,，,n,N,*,.,25/47,证实,k,N,*,，,4,k,2,2,k,(3,k,2,3,k,),k,2,k,k,(,k,1),0,，,4,k,2,2,k,3,k,2,3,k,，,不等式成立,.,26/47,命题点,3,数列应用题,例,5,(,南京模拟,),某企业一下属企业从事某种高科技产品生产,.,该企业第一年年初有资金,2 000,万元，将其投入生产，到当年年底资金增加了,50%.,预计以后每年年增加率与第一年相同,.,企业要求企业从第一年开始，每年年底上缴资金,d,万元，并将剩下资金全部投入下一年生产,.,设第,n,年年底企业上缴资金后剩下资金为,a,n,万元,.,(1),用,d,表示,a,1,，,a,2,，并写出,a,n,1,与,a,n,关系式；,解答,27/47,由题意，得,a,1,2 000(1,50%),d,3 000,d,，,a,2,a,1,(1,50%),d,a,1,d,4 500,，,a,n,1,a,n,(1,50%),d,d,.,28/47,(2),若企业希望经过,m,(,m,3),年使企业剩下资金为,4 000,万元，试确定企业每年上缴资金,d,值,(,用,m,表示,).,解答,29/47,由,(1),，得,a,n,a,n,1,d,(,a,n,2,d,),d,整理，得,a,n,(),n,1,(3 000,d,),2,d,(),n,1,1,(),n,1,(3 000,3,d,),2,d,.,由题意，得,a,m,4 000,，,即,(),m,1,(3 000,3,d,),2,d,4 000.,故该企业每年上缴资金,d,值为,时，经过,m,(,m,3),年企业剩下资金为,4 000,万元,.,30/47,数列与其它知识交汇问题常见类型及解题策略,(1),数列与函数交汇问题,已知函数条件，处理数列问题，这类问题普通利用函数性质、图象研究数列问题；,已知数列条件，处理函数问题，处理这类问题普通要充分利用数列范围、公式、求和方法对式子化简变形,.,另外，解题时要注意数列与函数内在联络，灵活利用函数思想方法求解，在问题求解过程中往往会碰到递推数列，所以掌握递推数列常看法法有利于该类问题处理,.,思维升华,31/47,(2),数列与不等式交汇问题,函数方法：即结构函数，经过函数单调性、极值等得出关于正实数不等式，经过对关于正实数不等式特殊赋值得出数列中不等式；,放缩方法：数列中不等式能够经过对中间过程或者最终结果放缩得到；,比较方法：作差或者作商比较,.,(3),数列应用题,依据题意，确定数列模型；,准确求解模型；,问题作答，不要忽略问题实际意义,.,32/47,跟踪训练,3,设等差数列,a,n,公差为,d,，点,(,a,n,，,b,n,),在函数,f,(,x,),2,x,图象上,(,n,N,*,).,(1),若,a,1,2,，点,(,a,8,，,4,b,7,),在函数,f,(,x,),图象上，求数列,a,n,前,n,项和,S,n,；,解答,由已知，得,b,7,，,b,8,4,b,7,，,有,解得,d,a,8,a,7,2.,所以,S,n,na,1,2,n,n,(,n,1),n,2,3,n,.,33/47,(2),若,a,1,1,，函数,f,(,x,),图象在点,(,a,2,，,b,2,),处切线在,x,轴上截距为,2,，求数列,前,n,项和,T,n,.,解答,f,(,x,),2,x,ln 2,，,f,(,a,2,),2,a,2,ln 2,，,故函数,f,(,x,),2,x,在,(,a,2,，,b,2,),处切线方程为,y,2,a,2,2,a,2,ln 2(,x,a,2,),，,它在,x,轴上截距为,a,2,.,解得,a,2,2.,由题意，得,a,2,2,，,所以,d,a,2,a,1,1.,从而,a,n,n,，,b,n,2,n,.,所以,T,n,.,34/47,课时作业,35/47,1.(,全国甲卷,),等差数列,a,n,中，,a,3,a,4,4,，,a,5,a,7,6.,(1),求,a,n,通项公式；,解答,设数列,a,n,公差为,d,，由题意有,2,a,1,5,d,4,，,a,1,5,d,3.,解得,a,1,1,，,d,.,所以,a,n,通项公式为,a,n,.,1,2,3,4,5,36/47,(2),设,b,n,a,n,，求数列,b,n,前,10,项和，其中,x,表示不超出,x,最大整数，如,0.9,0,，,2.6,2.,解答,由,(1),知，,b,n,.,当,n,1,2,3,时，,1,2,，,b,n,1,；,当,n,4,5,时，,2,3,，,b,n,2,；,所以数列,b,n,前,10,项和为,1,3,2,2,3,3,4,2,24.,当,n,6,7,8,时，,3,4,，,b,n,3,；,当,n,9,10,时，,4,1,，,a,n,前,n,项和为,S,n,，等比数列,b,n,首项为,1,，公比为,q,(,q,0),，前,n,项和为,T,n,.,若存在正整数,m,，使得,T,3,，求,q,.,解答,因为,a,1,1,，所以,a,n,6,n,3,，从而,S,n,3,n,2,.,由,T,3,，得,1,q,q,2,，,整理得,q,2,q,1,0.,因为,1,4(1,),0,，所以,m,2,.,因为,m,N,*,，所以,m,1,或,m,2.,当,m,1,时，,q,(,舍去,),或,q,.,当,m,2,时，,q,0,或,q,1(,均舍去,).,总而言之，,q,.,1,2,3,4,5,44/47,5.,在等比数列,a,n,中，,a,n,0(,n,N,*,),，公比,q,(0,1),，且,a,1,a,5,2,a,3,a,5,a,2,a,8,25,，又,a,3,与,a,5,等比中项为,2.,(1),求数列,a,n,通项公式；,解答,a,1,a,5,2,a,3,a,5,a,2,a,8,25,，,又,a,n,0,，,a,3,a,5,5,，,(,a,3,a,5,),2,25,，,又,a,3,与,a,5,等比中项为,2,，,a,3,a,5,4,，而,q,(0,1),，,a,3,a,5,，,a,3,4,，,a,5,1,，,q,，,a,1,16,，,a,n,16,(),n,1,2,5,n,.,1,2,3,4,5,45/47,(2),设,b,n,log,2,a,n,，求数列,b,n,前,n,项和,S,n,；,解答,b,n,log,2,a,n,5,n,，,b,n,1,b,n,1,，,b,1,log,2,a,1,log,2,16,log,2,2,4,4,，,b,n,是以,b,1,4,为首项，,1,为公差等差数列，,1,2,3,4,5,46/47,(3),是否存在,k,N,*,，使得,0,；当,n,9,时，,0,；,当,n,9,时，,0.,当,n,8,或,n,9,时，,故存在,k,N,*,，使得,对任意,n,N,*,恒成立，,k,最小值为,19.,1,2,3,4,5,47/47,