高三数学复习第六章数列第二节等差数列及其前n项和文.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,*,*,文数,课标版,第二节等差数列及其前,n,项和,1/30,1.等差数列定义,假如一个数列从,第二项,起,每一项与前一项差等于,同一个常数,那么这个数列就叫做等差数列,这个常数叫做等差数列,公差,通惯用字母,d,表示,定义表示式为,a,n,+1,-,a,n,=,d,(,n,N,*,).,教材研读,2.等差数列通项公式,等差数列,a,n,通项公式是,a,n,=,a,1,+(,n,-1),d,.,2/30,3.等差中项,假如,A,=,那么,A,叫做,a,与,b,等差中项.,4.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,是等差数列,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,(,p,q,是常数)仍是等差,数列.,(5)若,a,n,是等差数列,则,a,k,a,k,+,m,a,k,+2,m,(,k,m,N,*,)组成公差为,md,等差数列.,3/30,5.等差数列前,n,项和公式,等差数列,a,n,前,n,项和,S,n,=,或,S,n,=,na,1,+,.,6.等差数列前,n,项和公式与函数关系,S,n,=,n,2,+,n,.,数列,a,n,是等差数列,S,n,=,An,2,+,Bn,(,A,、,B,为常数).,7.等差数列前,n,项和最值,在等差数列,a,n,中,若,a,1,0,d,0,则,S,n,存在最,大,值;若,a,1,0,则,S,n,存在最,小,值.,4/30,判断以下结论正误(正确打“”,错误打“,”),(1)若一个数列从第2项起每一项与它前一项差都是常数,则这个数,列是等差数列.,(,),(2)数列,a,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,.,(),(3)等差数列,a,n,单调性是由公差,d,决定.,(),(4)已知数列,a,n,通项公式是,a,n,=,pn,+,q,(其中,p,q,为常数),则数列,a,n,一定,是等差数列.,(),(5)若数列,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,组成一,个新数列,则此新数列一定是等差数列.,(),5/30,1.在等差数列,a,n,中,若,a,2,=4,a,4,=2,则,a,6,=,(),A.-1B.0C.1D.6,答案,B设数列,a,n,公差为,d,由,a,4,=,a,2,+2,d,a,2,=4,a,4,=2,得2=4+2,d,d,=-1,a,6,=,a,4,+2,d,=0.故选B.,2.(东北师大附中摸底考试)等差数列,a,n,中,a,1,+,a,5,=10,a,4,=7,则数列,a,n,公差为,(),A.1B.2C.3D.4,答案,B设公差为,d,.,a,1,+,a,5,=2,a,3,=10,a,3,=5,又,a,4,=7,d,=2.故选B.,6/30,3.(广东惠州调研)等差数列,a,n,前,n,项和为,S,n,且,S,3,=6,a,1,=4,则公差,d,等于,(),A.1B.,C.-2D.3,答案,C,S,3,=6=,(,a,1,+,a,3,),且,a,3,=,a,1,+2,d,a,1,=4,d,=-2,故选C.,4.已知等差数列,a,n,满足,a,1,+,a,2,+,a,3,+,+,a,101,=0,则有,(),A.,a,1,+,a,101,0B.,a,2,+,a,100,0,S,5,=,S,12,”,怎样,求解?,解析,解法一:由,S,5,=,S,12,得5,a,1,+10,d,=12,a,1,+66,d,即,d,=-,a,1,0,n,N,*,所以当,n,=8或,n,=9时,S,n,取得最大值.,解法二:同解法一得,d,=-,a,1,0.,设此数列前,k,项和最大,则,即,解得,即8,k,9,27/30,又,k,N,*,所以,k,=8或9,所以当,n,=8或,n,=9时,S,n,取得最大值.,解法三:同解法一得,d,=-,a,1,0,S,13,0”,怎样求解?,解析,因为,a,3,=,a,1,+2,d,=12,所以,a,1,=12-2,d,因为,所以,解得-,d,-3.,故公差,d,取值范围为,.,解法一:由,d,0可知,a,n,为递减数列,所以,在1,n,12中,必存在一个自然数,n,使得,a,n,0,a,n,+1,0,此时对应,S,n,就是,S,1,S,2,S,12,中最大值.,因为,所以,a,7,0,所以,S,6,最大.,29/30,解法二:由,d,0可知,a,n,是递减数列,令,可得,由-,d,-3,可得,所以5.5,n,7,故,n,=6,由此可知,S,6,最大.,30/30,