高考数学复习第六章数列第三节等比数列及其前n项和市赛课公开课一等奖省名师优质课获奖PPT课件.pptx

,总纲目录教材研读考点突破,栏目索引,总纲目录教材研读考点突破,栏目索引,*,*,总纲目录教材研读考点突破,栏目索引,总纲目录,总纲目录教材研读考点突破,栏目索引,教材研读,总纲目录教材研读考点突破,栏目索引,考点突破,*,*,总纲目录教材研读考点突破,栏目索引,*,*,总纲目录教材研读考点突破,栏目索引,*,*,第三节等比数列及其前,n,项和,1/26,总纲目录,教材研读,1.,等比数列定义,考点突破,2.,等比数列通项公式,3.,等比中项,考点二等比数列性质及其应用,考点一等比数列基本运算,4.,等比数列惯用性质,5.,等比数列前,n,项和公式,6.,等比数列前,n,项和性质,考点三等比数列判定与证实,2/26,1.等比数列定义,假如一个数列从,第二项,起,每一项与前一项比等于,同一个常数,那么这个数列就叫做等比数列,这个常数叫做等比数列,公比,通惯用字母,q,表示,定义表示式为,=,q,(,n,N,*,).,教材研读,2.等比数列通项公式,等比数列,a,n,通项公式为,a,n,=,a,1,q,n,-1,.,3/26,3.等比中项,若,G,2,=,ab,(,ab,0),那么,G,叫做,a,与,b,等比中项.,4.等比数列惯用性质,(1)通项公式推广:,a,n,=,a,m,q,n,-,m,(,n,m,N,*,).,(2)若,a,n,为等比数列,且,k,+,l,=,m,+,n,(,k,l,m,n,N,*,),则,a,k,a,l,=,a,m,a,n,.,(3)若,a,n,b,n,(项数相同)是等比数列,则,a,n,(,0),a,n,b,n,仍是等比数列.,4/26,5.等比数列前,n,项和公式,等比数列,a,n,公比为,q,(,q,0),其前,n,项和为,S,n,当,q,=1时,S,n,=,na,1,;,当,q,1时,S,n,=,=,.,6.等比数列前,n,项和性质,公比不为-1等比数列,a,n,前,n,项和为,S,n,则,S,n,S,2,n,-,S,n,S,3,n,-,S,2,n,仍成等比数,列,其公比为,q,n,.,5/26,与等比数列相关结论,(1),a,n,=,a,m,q,n,-,m,a,n,+,m,=,a,n,q,m,=,a,m,q,n,(,m,n,N,*,).,(2),a,1,a,2,a,3,a,m,a,m,+1,a,m,+2,a,2,m,a,2,m,+1,a,2,m,+2,a,3,m,成等比数列(,m,N,*,).,(3)若等比数列项数为2,n,(,n,N,*,),公比为,q,奇数项之和为,S,奇,偶数项之,和为,S,偶,则,=,q,.,(4)三个数成等比数列,通常设为,x,xq,;四个数成等比数列,通常设为,xq,xq,3,.,6/26,1.已知,a,n,是等比数列,a,2,=2,a,5,=,则公比,q,=(),A.-,B.-2C.2D.,答案,D由通项公式及已知得,a,1,q,=2,a,1,q,4,=,由,得,q,3,=,解得,q,=,.故选D.,D,7/26,2.已知等比数列,a,n,前三项依次为,a,-1,a,+1,a,+4,则,a,n,=,(),A.4,B.4,C.4,D.4,答案,B由题意得(,a,+1),2,=(,a,-1)(,a,+4),解得,a,=5,故,a,1,=4,a,2,=6,所以,q,=,则,a,n,=4,.,B,8/26,3.在等比数列,a,n,中,已知,a,7,a,12,=5,则,a,8,a,9,a,10,a,11,=,(),A.10B.25C.50D.75,答案,B,a,7,a,12,=5,a,8,a,9,a,10,a,11,=(,a,8,a,11,)(,a,9,a,10,)=(,a,7,a,12,),2,=25.,4.(北京丰台一模)已知等比数列,a,n,中,a,1,=1,且,=8,那么,S,5,值是,(),A.15B.31C.63D.64,答案,B,=,=,=,q,3,=8,q,=2.,又,a,1,=1,S,5,=,=31.,B,B,9/26,5.(北京海淀一模)已知等比数列,a,n,中,a,2,a,4,=,a,5,a,4,=8,则公比,q,=,其前4项和,S,4,=,.,答案,2;15,解析,设等比数列,a,n,公比为,q,.,a,2,a,4,=,a,5,a,4,=8,8,a,2,=,a,2,q,3,q,=2.,a,1,=1,S,4,=,=15.,10/26,6.(北京朝阳期中)各项均为正数等比数列,a,n,前,n,项和为,S,n,若,a,3,=2,S,4,=5,S,2,则,a,1,值为,S,4,值为,.,答案,;,解析,当等比数列公比等于1时,由,a,3,=2,得,S,4,=4,a,3,=4,2=8,5,S,2,=5,2,a,3,=5,2,2=20,与题意不符.,设各项均为正数等比数列公比为,q,(,q,0且,q,1),由,a,3,=2,S,4,=5,S,2,得,整理得,解得,或,(舍).,则,S,4,=,=,.,11/26,典例1,(北京,15,13分)已知等差数列,a,n,和等比数列,b,n,满足,a,1,=,b,1,=1,a,2,+,a,4,=10,b,2,b,4,=,a,5,.,(1)求,a,n,通项公式;,(2)求和:,b,1,+,b,3,+,b,5,+,+,b,2,n,-1,.,考点一等比数列基本运算,考点突破,12/26,解析,本题考查等差数列及等比数列通项公式,数列求和.考查运算,求解能力.,(1)设等差数列,a,n,公差为,d,.,因为,a,2,+,a,4,=10,所以2,a,1,+4,d,=10.,解得,d,=2.,所以,a,n,=2,n,-1.,(2)设等比数列,b,n,公比为,q,.,因为,b,2,b,4,=,a,5,所以,b,1,qb,1,q,3,=9.,解得,q,2,=3.,所以,b,2,n,-1,=,b,1,q,2,n,-2,=3,n,-1,.,从而,b,1,+,b,3,+,b,5,+,+,b,2,n,-1,=1+3+3,2,+,+3,n,-1,=,.,13/26,方法指导,处理等比数列相关问题惯用思想方法,(1)方程思想:等比数列中有五个量,a,1,n,q,a,n,S,n,普通能够“知三求二”.,(2)分类讨论思想:等比数列前,n,项和公式包括对公比,q,分类讨论,当,q,=1时,a,n,前,n,项和,S,n,=,na,1,;当,q,1时,a,n,前,n,项和,S,n,=,=,.,14/26,1-1,(北京西城期末)已知数列,a,n,是等比数列,而且,a,1,a,2,+1,a,3,是公,差为-3等差数列.,(1)求数列,a,n,通项公式;,(2)设,b,n,=,a,2,n,记,S,n,为数列,b,n,前,n,项和,证实:,S,n,.,15/26,解析,(1)设等比数列,a,n,公比为,q,因为,a,1,a,2,+1,a,3,是公差为-3等差数列,所以,即,解得,a,1,=8,q,=,.,所以,a,n,=,a,1,q,n,-1,=8,=2,4-,n,.,(2)证实:因为,=,=,所以数列,b,n,是以,b,1,=,a,2,=4为首项,为公比等比数列.,所以,S,n,=,=,.,16/26,典例2,(1)(北京海淀期中)若等比数列,a,n,满足,a,1,+,a,3,=5,且公比,q,=2,则,a,3,+,a,5,=,(),A.10B.13C.20D.25,(2)若等比数列,a,n,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,=,.,(3)设等比数列,a,n,前,n,项和为,S,n,若,S,6,S,3,=12,则,S,9,S,3,=,.,考点二等比数列性质及其应用,17/26,答案,(1)C(2)50(3)34,解析,(1),a,3,+,a,5,=,a,1,q,2,+,a,3,q,2,=(,a,1,+,a,3,),q,2,=5,2,2,=20.,(2)因为等比数列,a,n,中,a,10,a,11,=,a,9,a,12,所以由,a,10,a,11,+,a,9,a,12,=2e,5,可得,a,10,a,11,=e,5,.,所以ln,a,1,+ln,a,2,+,+ln,a,20,=ln(,a,1,a,2,a,20,)=ln(,a,10,a,11,),10,=10ln(,a,10,a,11,)=10ln e,5,=50.,(3)由题意可知,q,-1,故由等比数列性质知,S,3,S,6,-,S,3,S,9,-,S,6,仍成等比数列,于是(,S,6,-,S,3,),2,=,S,3,(,S,9,-,S,6,),将,S,6,=,S,3,代入可得,=,.,18/26,易错警示,(1)在处理等比数列相关问题时,要注意挖掘隐含条件,利用性质,尤其,是“若,m,+,n,=,p,+,q,(,m,、,n,、,p,、,q,N,*,),则,a,m,a,n,=,a,p,a,q,”,能够降低运算量,提,高解题速度.(2)在应用对应性质解题时,要注意性质成立前提,有时需,要进行适当变形.另外,解题时注意对设而不求思想利用.,19/26,2-1,已知,x,y,z,R,若-1,x,y,z,-3成等比数列,则,xyz,值为,(),A.-3B.,3,C.-3,D.,3,答案,C由题意知,y,2,=3,y,=,又,y,与-1,-3符号相同,y,=-,又,y,2,=,xz,所以,xyz,=,y,3,=-3,.,C,20/26,2-2,记等比数列,a,n,前,n,项积为,T,n,(,n,N,*,),已知,a,m,-1,a,m,+1,-2,a,m,=0,且,T,2,m,-1,=,128,则,m,值为,(),A.4B.7C.10D.12,答案,A因为,a,n,是等比数列,所以,a,m,-1,a,m,+1,=,故由,a,m,-1,a,m,+1,-2,a,m,=0,可知,a,m,=2(,a,m,=0舍去).,由等比数列性质可知前(2,m,-1)项积,T,2,m,-1,=,2,2,m,-1,=128,故,m,=4.,A,21/26,典例3,(北京西城二模)已知数列,a,n,前,n,项和,S,n,满足4,a,n,-3,S,n,=2,其,中,n,N,*,.,(1)求证:数列,a,n,为等比数列;,(2)设,b,n,=,a,n,-4,n,求数列,b,n,前,n,项和,T,n,.,考点三等比数列判定与证实,解析,(1)证实:4,a,n,-3,S,n,=2,当,n,=1时,4,a,1,-3,S,1,=2,所以,a,1,=2;,22/26,当,n,2时,4,a,n,-1,-3,S,n,-1,=2,由-,得4,a,n,-4,a,n,-1,-3(,S,n,-,S,n,-1,)=0,所以,a,n,=4,a,n,-1,由,a,1,=2,得,a,n,0,所以,=4,其中,n,2.,故,a,n,是首项为2,公比为4等比数列.,(2)由(1)得,a,n,=2,4,n,-1,.,所以,b,n,=,a,n,-4,n,=4,n,-1,-4,n,.,则,b,n,前,n,项和,T,n,=(4,0,-4)+(4,1,-8)+,+(4,n,-1,-4,n,),=(4,0,+4,1,+,+4,n,-1,)-(4+8+,+4,n,),=,-,=,-2,n,2,-2,n,.,23/26,方法技巧,证实数列,a,n,(各项不为零)是等比数列惯用方法:一是定义法,证实,=,q,(,n,2,q,为非零常数);二是等比中项法,证实,=,a,n,-1,a,n,+1,(,n,2).若判,定一个数列不是等比数列,则能够举反例,也能够用反证法.,24/26,3-1,(北京朝阳一模)已知数列,a,n,满足,a,1,=1,a,n,+1,=,a,n,设,b,n,=,n,N,*,.,(1)证实:,b,n,是等比数列;,(2)求数列log,2,b,n,前,n,项和,T,n,.,25/26,解析,(1)证实:由,a,n,+1,=,a,n,得,=2,.,因为,b,n,=,所以,b,n,+1,=2,b,n,即,=2.,又因为,b,1,=,=1,所以数列,b,n,是以1为首项,2为公比等比数列.,(2)由(1)可知,b,n,=12,n,-1,=2,n,-1,所以log,2,b,n,=log,2,2,n,-1,=,n,-1.,则数列log,2,b,n,前,n,项和,T,n,=0+1+2+3+,+(,n,-1)=,.,26/26,