高三数学复习第六章数列第二节等差数列及其前n项和理.pptx

,教材研读,考点突破,栏目索引,教材研读,考点突破,栏目索引,教材研读,考点突破,栏目索引,*,*,教材研读,考点突破,栏目索引,*,*,教材研读,考点突破,栏目索引,教材研读,考点突破,栏目索引,*,*,理数,课标版,第二节等差数列及其前,n,项和,1/25,1.等差数列相关概念,(1)定义:假如一个数列从,第2项,起,每一项与它前一项,差,都等于同一个常数,那么这个数列就叫做等差数列.符号表示为,a,n,+1,-,a,n,=,d,(,n,N,*,d,为常数).,教材研读,(2)等差中项:数列,a,A,b,成等差数列充要条件是,A,=,其中,A,叫做,a,b,等差中项,.,2/25,2.等差数列相关公式,(1)通项公式:,a,n,=,a,1,+(,n,-1),d,.,(2)前,n,项和公式:,S,n,=,na,1,+,d,=,.,3.等差数列惯用性质,(1)通项公式推广:,a,n,=,a,m,+,(,n,-,m,),d,(,n,m,N,*,).,(2)若,a,n,为等差数列,且,k,+,l,=,m,+,n,(,k,l,m,n,N,*,),则,a,k,+,a,l,=,a,m,+,a,n,.,(3)若,a,n,是等差数列,公差为,d,则,a,2,n,也是等差数列,公差为,2,d,.,(4)若,a,n,b,n,(项数相同)是等差数列,则,pa,n,+,qb,n,也是等差数列.,(5)若,a,n,是等差数列,公差为,d,则,a,k,a,k,+,m,a,k,+2,m,(,k,m,N,*,)是公差为,md,等差数列.,3/25,判断以下结论正误.(正确打“”,错误打“,”),(1)数列,a,n,(,n,N,*,)为等差数列充要条件是对任意,n,N,*,都有2,a,n,+1,=,a,n,+,a,n,+2,.,(),(2)等差数列,a,n,单调性是由公差,d,决定.,(),(3)等差数列前,n,项和公式可看作常数项为0二次函数.(),(4)在等差数列,a,n,中,若,a,m,+,a,n,=,a,p,+,a,q,则一定有,m,+,n,=,p,+,q,.,(,),(5)数列,a,n,b,n,(项数相同)都是等差数列,则数列,a,n,+,b,n,也一定是等差,数列.,(),(6)等差数列,a,n,首项为,a,1,公差为,d,取出数列中全部奇数项,使其按,原次序组成一个新数列,则此新数列一定是等差数列,.,(),4/25,1.若等差数列,a,n,前5项之和,S,5,=25,且,a,2,=3,则,a,7,=,(),A.12B.13C.14D.15,答案,B由,S,5,=,25=,a,4,=7,所以7=3+2,d,d,=2,所,以,a,7,=,a,4,+3,d,=7+3,2=13,故选B,.,5/25,2.(课标全国,3,5分)已知等差数列,a,n,前9项和为27,a,10,=8,则,a,100,=,(),A.100B.99C.98D.97,答案,C设,a,n,公差为,d,由等差数列前,n,项和公式及通项公式,得,解得,a,n,=,a,1,+(,n,-1),d,=,n,-2,a,100,=100-2=98.故选,C.,6/25,3.(广东,10,5分)在等差数列,a,n,中,若,a,3,+,a,4,+,a,5,+,a,6,+,a,7,=25,则,a,2,+,a,8,=,.,答案,10,解析,利用等差数列性质可得,a,3,+,a,7,=,a,4,+,a,6,=2,a,5,从而,a,3,+,a,4,+,a,5,+,a,6,+,a,7,=5,a,5,=25,故,a,5,=5,所以,a,2,+,a,8,=2,a,5,=10.,4.(北京,12,5分)已知,a,n,为等差数列,S,n,为其前,n,项和.若,a,1,=6,a,3,+,a,5,=,0,则,S,6,=,.,答案,6,解析,设等差数列,a,n,公差为,d,a,1,=6,a,3,+,a,5,=0,6+2,d,+6+4,d,=0,d,=-2,S,6,=6,6+,(-2)=6.,7/25,考点一等差数列基本运算,考点突破,典例1,(1)(课标,7,5分)已知,a,n,是公差为1等差数列,S,n,为,a,n,前,n,项和.若,S,8,=4,S,4,则,a,10,=,(),A.,B.,C.10D.12,(2)等差数列,a,n,前,n,项和为,S,n,已知,a,5,=8,S,3,=6,则,S,10,-,S,7,值是,(),A.24B.48C.60D.72,8/25,答案,(1)B(2)B,解析,(1)由,S,8,=4,S,4,得8,a,1,+,1=4,解得,a,1,=,a,10,=,a,1,+9,d,=,故选B.,(2)设等差数列,a,n,公差为,d,由题意可得,解得,则,S,10,-,S,7,=,a,8,+,a,9,+,a,10,=3,a,1,+24,d,=48,.,9/25,方法技巧,处理等差数列运算问题思想方法,(1)方程思想:等差数列基本量为首项,a,1,和公差,d,通常利用已知条件及,通项公式或前,n,项和公式列方程(组)求解.,(2)整体思想:当所给条件只有一个时,可将已知和所求都用,a,1,d,表示,寻,求二者间联络,整体代换即可求解.,(3)利用性质:利用等差数列性质能够化繁为简、优化解题过程.,10/25,1-1,设,S,n,为等差数列,a,n,前,n,项和,a,12,=-8,S,9,=-9,则,S,16,=,.,答案,-72,解析,设等差数列,a,n,公差为,d,由已知,得,解得,S,16,=16,3+,(-1)=-72.,11/25,1-2,已知等差数列,a,n,中,a,1,=1,a,3,=-3.,(1)求数列,a,n,通项公式;,(2)若数列,a,n,前,k,项和,S,k,=-35,求,k,值.,解析,(1)设等差数列,a,n,公差为,d,则,a,n,=,a,1,+(,n,-1),d,.,因为,a,1,=1,a,3,=-3,所以1+2,d,=-3,解得,d,=-2,则,a,n,=1+(,n,-1)(-2)=3-2,n,(,n,N,*,),.,(2)由(1)知,a,n,=3-2,n,则,S,n,=,=2,n,-,n,2,.,由,S,k,=-35,得2,k,-,k,2,=-35,即,k,2,-2,k,-35=0,解得,k,=7或,k,=-5,又,k,N,*,所以,k,=7.,12/25,考点二等差数列判断与证实,典例2,(天津,18改编)已知,a,n,是各项均为正数等差数列,公差,为,d,.对任意,n,N,*,b,n,是,a,n,和,a,n,+1,等比中项.,(1)设,c,n,=,-,n,N,*,求证:数列,c,n,是等差数列;,(2)设,a,1,=,d,T,n,=,(-1),k,n,N,*,求,T,n,.,解析,(1)证实:由题意得,=,a,n,a,n,+1,有,c,n,=,-,=,a,n,+1,a,n,+2,-,a,n,a,n,+1,=2,da,n,+1,所以,c,n,+1,-,c,n,=2,d,(,a,n,+2,-,a,n,+1,)=2,d,2,所以,c,n,是等差数列.,(2),T,n,=(-,+,)+(-,+,)+,+(-,+,),=2,d,(,a,2,+,a,4,+,+,a,2,n,),=2,d,=2,d,2,n,(,n,+1).,13/25,方法技巧,1.等差数列识别依据,(1)若数列,a,n,是等差数列,则数列,a,n,+,b,仍为等差数列,公差为,d,.,(2)若,b,n,a,n,(项数相同)都是等差数列,则,a,n,b,n,仍为等差数列.,(3),a,n,=,pn,+,q,(,p,q,为常数),a,n,是等差数列.,(4)数列,a,n,前,n,项和,S,n,=,An,2,+,Bn,(,A,B,为常数),a,n,是等差数列.,2.证实等差数列两种基本方法,(1)定义法:证实,a,n,-,a,n,-1,=,d,(,n,2,d,为常数).,(2)等差中项法:证实2,a,n,=,a,n,-1,+,a,n,+1,(,n,2),.,14/25,2-1,(东营模拟)已知数列,a,n,前,n,项和为,S,n,且满足,a,n,+2,=2,a,n,+1,-,a,n,a,5,=4-,a,3,则,S,7,=,(),A.7B.12C.14D.21,答案,C由,a,n,+2,=2,a,n,+1,-,a,n,得,a,n,+2,+,a,n,=2,a,n,+1,即数列,a,n,为等差数列,由,a,5,=4-,a,3,得,a,5,+,a,3,=4,则,S,7,=,=,=14.,15/25,2-2,已知公差大于零等差数列,a,n,前,n,项和为,S,n,且满足,a,3,a,4,=117,a,2,+,a,5,=22.,(1)求数列,a,n,通项公式;,(2)若数列,b,n,满足,b,n,=,是否存在非零实数,c,使得,b,n,为等差数列?若,存在,求出,c,值;若不存在,请说明理由.,解析,(1)设等差数列,a,n,公差为,d,则,d,0,由等差数列性质,得,a,3,+,a,4,=,a,2,+,a,5,=22,所以,a,3,a,4,是关于,x,方程,x,2,-22,x,+117=0,解,所以,a,3,=9,a,4,=13,易得,a,1,=1,d,=4,故,a,n,=1+(,n,-1),4=4,n,-3.,16/25,(2)解法一:存在.,由(1)知,S,n,=,=2,n,2,-,n,所以,b,n,=,=,(,c,0).,所以,b,1,=,b,2,=,b,3,=,.,令2,b,2,=,b,1,+,b,3,解得,c,=-,.,当,c,=-,时,b,n,=,=2,n,此时,b,n,-,b,n,-1,=2(,n,2).,故存在,c,=-,使数列,b,n,为等差数列.,17/25,解法二:存在.,b,n,=,=,=,c,0,要满足题意,需,c,=-,得到,b,n,=2,n,.,此时,b,n,+1,-,b,n,=2(,n,+1)-2,n,=2(,n,N,*,),即数列,b,n,是公差为2等差数列.,存在,c,=-,使数列,b,n,为等差数列.,18/25,考点三等差数列性质及最值,典例3,(1)在等差数列,a,n,中,a,1,=29,S,10,=,S,20,则数列,a,n,前,n,项和中最大,为,(),A.,S,15,B.,S,16,C.,S,15,和,S,16,D.,S,17,(2)设等差数列,a,n,前,n,项和为,S,n,已知前6项和为36,最终6项和为18,0,S,n,=324(,n,6),则,n,=,.,(3)等差数列,a,n,前,m,项和为30,前3,m,项和为90,则它前2,m,项和为,.,答案,(1)A(2)18(3)60,解析,(1),S,10,=,S,20,10,a,1,+,d,=20,a,1,+,d,又,a,1,=29,d,=-2,19/25,S,n,=29,n,+,(-2)=-,n,2,+30,n,=-(,n,-15),2,+225.,当,n,=15时,S,n,取得最大值.,(2)由题意知,a,1,+,a,2,+,+,a,6,=36,a,n,+,a,n,-1,+,a,n,-2,+,+,a,n,-5,=180,+得(,a,1,+,a,n,)+(,a,2,+,a,n,-1,)+,+(,a,6,+,a,n,-5,)=6(,a,1,+,a,n,)=216,a,1,+,a,n,=36,又,S,n,=,=324,18,n,=324,n,=18.,(3)由,S,m,S,2,m,-,S,m,S,3,m,-,S,2,m,成等差数列,可得2(,S,2,m,-,S,m,)=,S,m,+,S,3,m,-,S,2,m,即,S,2,m,=,=,=60.,20/25,方法技巧,1.等差数列和性质,(1),S,2,n,=,n,(,a,1,+,a,2,n,)=,=,n,(,a,n,+,a,n,+1,).,(2),S,2,n,-1,=(2,n,-1),a,n,.,(3)当项数为偶数2,n,时,S,偶,-,S,奇,=,nd,;项数为奇数2,n,-1时,S,奇,-,S,偶,=,a,n,S,奇,S,偶,=,n,(,n,-1).,21/25,2.求等差数列前,n,项和,S,n,最值两种方法,(1)函数法:等差数列前,n,项和,S,n,=,An,2,+,Bn,经过配方,借助求二次函数最值,方法求解.,(2)邻项变号法:,a,1,0,d,0时,满足,项数,m,使得,S,n,取得最大值,S,m,;,当,a,1,0时,满足,项数,m,使得,S,n,取得最小值,S,m,.,22/25,3-1,设,S,n,是等差数列,a,n,前,n,项和,若,=,则,=,(),A.1B.-1C.2D.,答案,A,=,=,=,=1.,变式3-2,若将本例(1)中条件“,a,1,=29,S,10,=,S,20,”改为“,a,1,0,S,5,=,S,12,”,如,何求解?,解析,解法一:由,S,5,=,S,12,得5,a,1,+10,d,=12,a,1,+66,d,则,d,=-,a,1,0,n,N,*,所以当,n,=8或,n,=9时,S,n,有最大值.,解法二:同解法一得,d,=-,a,1,0.,设此数列前,k,项和最大,则,即,24/25,解得,即8,k,9,又,k,N,*,所以,k,=8或9,所以当,n,=8或,n,=9时,S,n,有最大值.,解法三:同解法一得,d,=-,a,1,0,因为,S,n,=,na,1,+,d,=,n,2,+,n,设,f,(,x,)=,x,2,+,x,则函数,y,=,f,(,x,)图象为开口向下抛物线,由,S,5,=,S,12,知,抛物线对称轴为,x,=,=,易知当1,n,8时,S,n,单调递增;,当,n,9时,S,n,单调递减,且,S,8,=,S,9,所以当,n,=8或,n,=9时,S,n,最大,.,25/25,