高三数学二轮复习第一篇专题突破专题四数列第1讲等差数列、等比数列理名师公开课.ppt

*,*,考情分析总纲目录,考点聚焦,随堂检测,栏目索引,高考导航,考情分析总纲目录,考点聚焦,随堂检测,栏目索引,高考导航,*,*,考情分析总纲目录,考点聚焦,随堂检测,栏目索引,高考导航,考情分析,考情分析总纲目录,考点聚焦,随堂检测,栏目索引,高考导航,总纲目录,*,*,考情分析总纲目录,考点聚焦,随堂检测,栏目索引,高考导航,考点聚焦,考情分析总纲目录,考点聚焦,随堂检测,栏目索引,高考导航,随堂检测,*,*,考情分析总纲目录,考点聚焦,随堂检测,栏目索引,高考导航,典题精练,考情分析总纲目录,考点聚焦,随堂检测,栏目索引,高考导航,题型特点,考情分析总纲目录,考点聚焦,随堂检测,栏目索引,高考导航,题组训练,考情分析总纲目录,考点聚焦,随堂检测,栏目索引,高考导航,真题回访,第,1,讲 等差数列、等比数列,考情分析,总纲目录,考点一 等差、等比数列的基本运算,考点二 等差、等比数列的判定与证明,考点三 等差、等比数列的性质,考点一等差、等比数列的基本运算,(1)通项公式:,等差数列:,a,n,=,a,1,+(,n,-1),d,;,等比数列:,a,n,=,a,1,q,n,-1,(,q,0).,(2)求和公式:,等差数列:,S,n,=,=,na,1,+,d,;,等比数列:当,q,=1时,S,n,=,na,1,;当,q,1时,S,n,=,=,.,典型例题,(1)(2017课标全国,4,5分)记,S,n,为等差数列,a,n,的前,n,项和.若,a,4,+,a,5,=,24,S,6,=48,则,a,n,的公差为,(),A.1B.2C.4D.8,(2)(2017课标全国,17,12分)已知等差数列,a,n,的前,n,项和为,S,n,等比数,列,b,n,的前,n,项和为,T,n,a,1,=-1,b,1,=1,a,2,+,b,2,=2.,(i)若,a,3,+,b,3,=5,求,b,n,的通项公式;,(ii)若,T,3,=21,求,S,3,.,解析,(1)设等差数列,a,n,的公差为,d,S,6,=,=48,则,a,1,+,a,6,=16=,a,2,+,a,5,又,a,4,+,a,5,=24,所以,a,4,-,a,2,=2,d,=24-16=8,得,d,=4,故选C.,(2)设,a,n,的公差为,d,b,n,的公比为,q,则,a,n,=-1+(,n,-1),d,b,n,=,q,n,-1,.由,a,2,+,b,2,=2得,d,+,q,=3.,(i)由,a,3,+,b,3,=5得2,d,+,q,2,=6.,联立和解得,(舍去)或,因此,b,n,的通项公式为,b,n,=2,n,-1,.,(ii)由,b,1,=1,T,3,=21得,q,2,+,q,-20=0.,解得,q,=-5或,q,=4.,当,q,=-5时,由得,d,=8,则,S,3,=21.,当,q,=4时,由得,d,=-1,则,S,3,=-6.,答案,(1)C,2.(2017昆明教学质量检测)已知数列,a,n,的前,n,项和为,S,n,且2,S,n,a,n,成等差,数列,则,S,17,=,(),A.0B.2C.-2D.34,答案,B由2,S,n,a,n,成等差数列,得2,S,n,=,a,n,+2,2,S,n,+1,=,a,n,+1,+2,-,整理得,=-1,又2,a,1,=,a,1,+2,所以,a,1,=2,所以数列,a,n,是首项为2,公比为-1的等比数列,所以,S,17,=,=2,故选B.,3.(2017湖北七市(州)联考)已知等比数列,a,n,的前,n,项和,S,n,=2,n,+1,+,a,数列,b,n,满足,b,n,=2-log,2,.,(1)求常数,a,的值;,(2)求数列,b,n,的前,n,项和,T,n,.,解析,(1)当,n,=1时,a,1,=,S,1,=2,2,+,a,=4+,a,当,n,2时,a,n,=,S,n,-,S,n,-1,=2,n,+1,+,a,-(2,n,+,a,)=2,n,a,n,为等比数列,=,a,1,a,3,即(2,2,),2,=(4+,a,)2,3,解得,a,=-2.,(2)由(1)知,a,n,=2,n,则,b,n,=2-log,2,2,3,n,=2-3,n,b,n,+1,-,b,n,=-3对一切,n,N,*,都成立,b,n,是以-1为首项,-3为公差的等差数列,即,b,1,=-1,d,=-3,T,n,=,nb,1,+,d,=,.,考点二等差、等比数列的判定与证明,1.证明数列,a,n,是等差数列的两种基本方法,(1)利用定义证明,a,n,+1,-,a,n,(,n,N,*,)为一常数;,(2)利用等差中项,即证明2,a,n,=,a,n,-1,+,a,n,+1,(,n,2).,2.证明数列,a,n,是等比数列的两种基本方法,(1)利用定义证明,(,n,N,*,)为一常数;,(2)利用等比中项,即证明,=,a,n,-1,a,n,+1,(,n,2).,典型例题,(2017贵州适应性考试)已知数列,a,n,满足,a,1,=1,且,na,n,+1,-(,n,+1),a,n,=2,n,2,+2,n,.,(1)求,a,2,a,3,;,(2)证明数列,是等差数列,并求,a,n,的通项公式.,解析,(1)由已知得,a,2,-2,a,1,=4,则,a,2,=2,a,1,+4,又,a,1,=1,所以,a,2,=6.,由2,a,3,-3,a,2,=12得2,a,3,=12+3,a,2,所以,a,3,=15.,(2)由已知,na,n,+1,-(,n,+1),a,n,=2,n,(,n,+1),得,=2,即,-,=2,所以数列,是首项为1,公差为2的等差数列.,则,=1+2(,n,-1)=2,n,-1,所以,a,n,=2,n,2,-,n,.,方法归纳,(1)判定一个数列是等差(比)数列,可以利用通项公式或前,n,项和公式,但,不能将其作为证明方法;,(2),=,q,和,=,a,n,-1,a,n,+1,(,n,2)都是数列,a,n,为等比数列的必要不充分条,件,判定时还要看各项是否为零.,跟踪集训,1.已知,S,n,是等比数列,a,n,的前,n,项和,且,S,3,S,9,S,6,成等差数列,下列结论正确,的是,(),A.,a,1,a,7,a,4,成等差数列B.,a,1,a,7,a,4,成等比数列,C.,a,1,2,a,7,a,4,成等差数列D.,a,1,2,a,7,a,4,成等比数列,答案,A显然,q,=1时不合题意,依题意得,S,3,+,S,6,=2,S,9,即,(1-,q,3,)+,(1-,q,6,)=,(1-,q,9,),1+,q,3,=2,q,6,a,1,+,a,1,q,3,=2,a,1,q,6,a,1,+,a,4,=2,a,7,a,1,a,7,a,4,成等差数列.,2.(2017课标全国,17,12分)记,S,n,为等比数列,a,n,的前,n,项和.已知,S,2,=2,S,3,=-6.,(1)求,a,n,的通项公式;,(2)求,S,n,并判断,S,n,+1,S,n,S,n,+2,是否成等差数列.,解析,(1)设,a,n,的公比为,q,由题设可得,解得,q,=-2,a,1,=-2.故,a,n,的通项公式为,a,n,=(-2),n,.,(2)由(1)可得,S,n,=,=-,+(-1),n,.,由于,S,n,+2,+,S,n,+1,=-,+(-1),n,=2,=2,S,n,故,S,n,+1,S,n,S,n,+2,成等差数列.,考点三等差、等比数列的性质,等差数列,等比数列,性,质,(1)若,m,n,p,q,N,*,且,m,+,n,=,p,+,q,则,a,m,+,a,n,=,a,p,+,a,q,;,(2),a,n,=,a,m,+(,n,-,m,),d,;,(3),S,m,S,2,m,-,S,m,S,3,m,-,S,2,m,仍成等差数列,(1)若,m,n,p,q,N,*,且,m,+,n,=,p,+,q,则,a,m,a,n,=,a,p,a,q,;,(2),a,n,=,a,m,q,n,-,m,;,(3),S,m,S,2,m,-,S,m,S,3,m,-,S,2,m,仍成等比数列(,q,-1),典型例题,(1)(2017云南11校跨区调研)已知数列,a,n,是等比数列,S,n,为其前,n,项,和,若,a,1,+,a,2,+,a,3,=4,a,4,+,a,5,+,a,6,=8,则,S,12,=,(),A.40B.60C.32D.50,(2)设,S,n,为等差数列,a,n,的前,n,项和,(,n,+1),S,n,nS,n,+1,(,n,N,*,).若,-1,则,(),A.,S,n,的最大值是,S,8,B.,S,n,的最小值是,S,8,C.,S,n,的最大值是,S,7,D.,S,n,的最小值是,S,7,解析,(1)由等比数列的性质可知,数列,S,3,S,6,-,S,3,S,9,-,S,6,S,12,-,S,9,是等比数列,即,数列4,8,S,9,-,S,6,S,12,-,S,9,是等比数列,则,S,9,-,S,6,=16,S,12,-,S,9,=32,因此,S,12,=4+8+16+32,=60,故选B.,(2)由(,n,+1),S,n,nS,n,+1,得,(,n,+1),n,整理得,a,n,a,n,+1,所以等差数列,a,n,是递增数列,又,0,a,7,0,所以数列,a,n,的前7项为负值,即,S,n,的最小值是,S,7,故选D.,答案,(1)B(2)D,方法归纳,应用数列性质解题的方法,(1)解决此类问题的关键是抓住项与项之间的关系及项的序号之间的关,系,从这些特点入手选择恰当的性质进行求解.,(2)应牢固掌握等差、等比数列的性质,特别是等差数列中,“若,m,+,n,=,p,+,q,则,a,m,+,a,n,=,a,p,+,a,q,(,m,n,p,q,N,*,)”这一性质与求和公式,S,n,=,的综合,应用.,跟踪集训,1.(2017太原模拟试题)已知,S,n,是等差数列,a,n,的前,n,项和,2(,a,1,+,a,3,+,a,5,)+3(,a,8,+,a,10,)=36,则,S,11,=,(),A.66B.55C.44D.33,答案,D因为,a,1,+,a,5,=2,a,3,a,8,+,a,10,=2,a,9,所以2(,a,1,+,a,3,+,a,5,)+3(,a,8,+,a,10,)=6,a,3,+6,a,9,=36,所以,a,3,+,a,9,=6,所以,S,11,=,=,=33,故选D.,2.一个项数为偶数的等比数列,a,n,全部各项之和为偶数项之和的4倍,前3项之积为64,则,a,1,=,(),A.11B.12C.13D.14,答案,B设数列,a,n,的公比为,q,全部奇数项、偶数项之和分别记为,S,奇,、,S,偶,由题意知,S,奇,+,S,偶,=4,S,偶,即,S,奇,=3,S,偶,.因为数列,a,n,的项数为偶数,所以,q,=,=,.,又,a,1,(,a,1,q,)(,a,1,q,2,)=64,所以,q,3,=64,故,a,1,=12.,3.在数列,a,n,中,a,1,=5,(,a,n,+1,-2)(,a,n,-2)=3(,n,N,*,),则该数列的前2 016项的和,是,.,答案,8 064,解析,因为(,a,n,+1,-2)(,a,n,-2)=3,所以(,a,n,+2,-2)(,a,n,+1,-2)=3,因此,a,n,+2,-2=,a,n,-2,即,a,n,+2,=,a,n,所以数列,a,n,是以2为周期的数列.又,a,1,=5,因此(,a,2,-2)(,a,1,-2)=3,故,a,2,=3,a,1,+,a,2,=8.注意到2 016=2,1 008,因此该数列的前2 016项的和等于1 008(,a,1,+,a,2,)=8 064.,1.(2017贵州适应性考试)已知数列,a,n,满足,a,n,=,a,n,+1,若,a,3,+,a,4,=2,则,a,4,+,a,5,=,(),A.,B.1C.4D.8,随堂检测,答案,C解法一:因为,a,n,=,a,n,+1,a,3,+,a,4,=2,所以,a,n,0,可得,=2,所以,a,n,为等比数列,由,a,n,=,a,m,q,n,-,m,得,a,3,+,a,3,2,4-3,=2,解得,a,3,=,所以,a,n,=,a,3,q,n,-3,=,2,n,-3,由,此可得,a,4,=,a,3,2=,a,5,=,a,3,2,2,=,所以,a,4,+,a,5,=,+,=,=4.,解法二:已知,a,n,=,a,n,+1,可得,a,n,+1,=2,a,n,所以,a,4,+,a,5,=2,a,3,+2,a,4,=2(,a,3,+,a,4,)=2,2=4.,2.已知等比数列,a,n,的公比为,q,前,n,项和为,S,n,若点(,n,S,n,)在函数,y,=2,x,+1,+,m,的图象上,则,m,=,(),A.-2B.2C.-3D.3,答案,A易知,q,1,S,n,=,=,-,q,n,=,-,q,n,+1,又点(,n,S,n,)在函数,y,=2,x,+1,+,m,的图象上,所以,S,n,=2,n,+1,+,m,所以,q,=2,解得,m,=-2.,3.(2017北京,10,5分)若等差数列,a,n,和等比数列,b,n,满足,a,1,=,b,1,=-1,a,4,=,b,4,=,8,则,=,.,答案,1,解析,设等差数列,a,n,的公差为,d,等比数列,b,n,的公比为,q,.,a,1,=,b,1,=-1,a,4,=,b,4,=8,a,2,=2,b,2,=2.,=,=1.,4.(2017广西三市第一次联考)已知数列,a,n,的前,n,项和为,S,n,且,S,n,=2,n,-1(,n,N,*,).,(1)求数列,a,n,的通项公式;,(2)设,b,n,=log,4,a,n,+1,求,b,n,的前,n,项和,T,n,.,解析,(1)当,n,2时,a,n,=,S,n,-,S,n,-1,=2,n,-1,.,当,n,=1时,a,1,=2-1=1,满足,a,n,=2,n,-1,数列,a,n,的通项公式为,a,n,=2,n,-1,(,n,N,*,).,(2)由(1)得,b,n,=log,4,a,n,+1=,则,b,n,+1,-,b,n,=,-,=,数列,b,n,是首项为1,公差,d,=,的等差数列,T,n,=,nb,1,+,d,=,.,