高考数学复习第六章数列与数学归纳法6.2等差数列及其前n项和市赛课公开课一等奖省名师优质课获奖.pptx

,6.2,等差数列及其前,n,项和,1/58,基础知识自主学习,课时训练,题型分类深度剖析,内容索引,2/58,基础知识自主学习,3/58,1.,等差数列定义,普通地，假如一个数列,，那么这个数列就叫做等差数列，这个常数叫做等差数列,_,，通惯用字母,表示,.,2.,等差数列通项公式,假如等差数列,a,n,首项为,a,1,，公差为,d,，那么它通项公式是,.,3.,等差中项,由三个数,a,，,A,，,b,组成等差数列能够看成最简单等差数列,.,这时，,A,叫做,a,与,b,.,知识梳理,从第,2,项起，每一项与它前一项差等于同一,个常数,公差,d,a,n,a,1,(,n,1),d,等差中项,4/58,4.,等差数列惯用性质,(1),通项公式推广：,a,n,a,m,(,n,，,m,N,*,).,(2),若,a,n,为等差数列，且,k,l,m,n,(,k,，,l,，,m,，,n,N,*,),，则,_,_,.,(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,.,(4),若,a,n,，,b,n,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,等差数列,.,(6),数列,S,m,，,S,2,m,S,m,，,S,3,m,S,2,m,，,组成等差数列,.,(,n,m,),d,a,k,a,l,a,m,a,n,2,d,md,5/58,7.,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,值；若,a,1,0,，则,S,n,存在最,值,.,5.,等差数列前,n,项和公式,设等差数列,a,n,公差为,d,，其前,n,项和,S,n,_,或,S,n,_,.,6.,等差数列前,n,项和公式与函数关系,大,小,6/58,等差数列四种判断方法,(1),定义法：,a,n,1,a,n,d,(,d,是常数,),a,n,是等差数列,.,(2),等差中项法：,2,a,n,1,a,n,a,n,2,(,n,N,*,),a,n,是等差数列,.,(3),通项公式：,a,n,pn,q,(,p,，,q,为常数,),a,n,是等差数列,.,(4),前,n,项和公式：,S,n,An,2,Bn,(,A,，,B,为常数,),a,n,是等差数列,.,知识拓展,7/58,判断以下结论是否正确,(,请在括号中打,“”,或,“”,),(1),若一个数列从第二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),等差数列,a,n,单调性是由公差,d,决定,.(,),(3),等差数列前,n,项和公式是常数项为,0,二次函数,.(,),(4),已知等差数列,a,n,通项公式,a,n,3,2,n,，则它公差为,2.(,),思索辨析,8/58,考点自测,1.,在等差数列,a,n,中，若,a,2,4,，,a,4,2,，则,a,6,等于,A.,1 B.0 C.1 D.6,答案,解析,由等差数列性质，,得,a,6,2,a,4,a,2,2,2,4,0,，故选,B.,9/58,2.(,全国乙卷,),已知等差数列,a,n,前,9,项和为,27,，,a,10,8,，则,a,100,等于,A.100 B.99 C.98 D.97,答案,解析,得,a,5,3,，而,a,10,8,，,a,100,a,10,90,d,98,，故选,C.,10/58,答案,解析,3.(,绍兴一模,),已知数列,a,n,中，,a,3,3,，,a,n,1,a,n,2,，则,a,2,a,4,_,，,a,n,_.,6,2,n,3,由已知得,a,n,1,a,n,2,，,所以,a,n,为公差为,2,等差数列，,由,a,1,2,d,3,，得,a,1,1,，,所以,a,n,1,(,n,1),2,2,n,3,，,a,2,a,4,2,a,3,6.,11/58,答案,解析,4.,若等差数列,a,n,满足,a,7,a,8,a,9,0,，,a,7,a,10,3),，,S,n,100,，则,n,值为,A.8 B.9,C.10 D.11,答案,解析,由,S,n,S,n,3,51,，得,a,n,2,a,n,1,a,n,51,，,所以,a,n,1,17,，又,a,2,3,，,1,2,3,4,5,6,7,8,9,10,11,12,13,42/58,4.(,绍兴柯桥区二模,),各项均不为零等差数列,a,n,中，若,a,n,1,a,n,1,(,n,N,*,，,n,2),，则,S,2 016,等于,A.0 B.2 C.2 015 D.4 032,答案,解析,a,n,各项均不为零，,a,n,2(,n,2),，,又,a,n,为等差数列，,a,n,2,，,S,2 016,4 032.,1,2,3,4,5,6,7,8,9,10,11,12,13,43/58,5.,已知数列,a,n,满足,a,n,1,a,n,，且,a,1,5,，设,a,n,前,n,项和为,S,n,，则使得,S,n,取得最大值序号,n,值为,A.7 B.8,C.7,或,8 D.8,或,9,由题意可知数列,a,n,是首项为,5,，,该数列前,7,项是正数项，第,8,项是,0,，从第,9,项开始是负数项，,所以,S,n,取得最大值时，,n,7,或,n,8,，故选,C.,1,2,3,4,5,6,7,8,9,10,11,12,13,答案,解析,44/58,*6.,设数列,a,n,前,n,项和为,S,n,，若,为常数，则称数列,a,n,为,“,吉祥数列,”.,已知等差数列,b,n,首项为,1,，公差不为,0,，若数列,b,n,为,“,吉祥数列,”,，则数列,b,n,通项公式为,A.,b,n,n,1 B.,b,n,2,n,1,C.,b,n,n,1 D.,b,n,2,n,1,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,45/58,即,2,(,n,1),d,4,k,2,k,(2,n,1),d,，,整理得,(4,k,1),dn,(2,k,1)(2,d,),0.,因为对任意正整数,n,上式均成立，,所以,(4,k,1),d,0,，,(2,k,1)(2,d,),0,，,所以数列,b,n,通项公式为,b,n,2,n,1.,1,2,3,4,5,6,7,8,9,10,11,12,13,46/58,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,47/58,8.,设数列,a,n,通项公式为,a,n,2,n,10(,n,N,*,),，则,|,a,1,|,|,a,2,|,|,a,15,|,_.,答案,解析,130,由,a,n,2,n,10(,n,N,*,),知,a,n,是以,8,为首项，,2,为公差等差数列，,又由,a,n,2,n,10,0,，得,n,5,，,当,n,5,时，,a,n,0,，当,n,5,时，,a,n,0,，,|,a,1,|,|,a,2,|,|,a,15,|,(,a,1,a,2,a,3,a,4,),(,a,5,a,6,a,15,),20,110,130.,1,2,3,4,5,6,7,8,9,10,11,12,13,48/58,答案,解析,a,n,，,b,n,为等差数列，,1,2,3,4,5,6,7,8,9,10,11,12,13,49/58,10.(,浙江新高考预测三,),设数列,a,n,满足：,a,1,1,，,a,2,3,，且,2,na,n,(,n,1),a,n,1,(,n,1),a,n,1,，则,a,20,值是,_.,答案,解析,由,2,na,n,(,n,1),a,n,1,(,n,1),a,n,1,，得,na,n,(,n,1),a,n,1,(,n,1),a,n,1,na,n,，,又因为,1,a,1,1,2,a,2,1,a,1,5,，,所以数列,na,n,是首项为,1,，公差为,5,等差数列，,1,2,3,4,5,6,7,8,9,10,11,12,13,50/58,11.,在等差数列,a,n,中，,a,1,1,，,a,3,3.,(1),求数列,a,n,通项公式；,解答,设等差数列,a,n,公差为,d,，,则,a,n,a,1,(,n,1),d,.,由,a,1,1,，,a,3,3,，可得,1,2,d,3,，解得,d,2.,从而,a,n,1,(,n,1),(,2),3,2,n,.,1,2,3,4,5,6,7,8,9,10,11,12,13,51/58,(2),若数列,a,n,前,k,项和,S,k,35,，求,k,值,.,解答,由,(1),可知,a,n,3,2,n,，,由,S,k,35,，可得,2,k,k,2,35,，,即,k,2,2,k,35,0,，解得,k,7,或,k,5.,又,k,N,*,，故,k,7.,1,2,3,4,5,6,7,8,9,10,11,12,13,52/58,证实,当,n,2,时，由,a,n,2,S,n,S,n,1,0,，,1,2,3,4,5,6,7,8,9,10,11,12,13,53/58,(2),求数列,a,n,通项公式,.,解答,1,2,3,4,5,6,7,8,9,10,11,12,13,54/58,*13.,已知数列,a,n,各项均为正数，前,n,项和为,S,n,，且满足,2,S,n,n,4,(,n,N,*,).,(1),求证：数列,a,n,为等差数列；,证实,1,2,3,4,5,6,7,8,9,10,11,12,13,55/58,解得,a,1,3(,a,1,1,舍去,).,所以,a,n,1,a,n,1,或,a,n,1,a,n,1,.,1,2,3,4,5,6,7,8,9,10,11,12,13,56/58,若,a,n,1,a,n,1,，则,a,n,a,n,1,1.,而,a,1,3,，,所以,a,2,2,，这与数列,a,n,各项均为正数相矛盾，,所以,a,n,1,a,n,1,，即,a,n,a,n,1,1,，,所以数列,a,n,是首项为,3,，公差为,1,等差数列,.,1,2,3,4,5,6,7,8,9,10,11,12,13,57/58,(2),求数列,a,n,通项公式,.,解答,1,2,3,4,5,6,7,8,9,10,11,12,13,由,(1),知,a,1,3,，,d,1,，,所以数列,a,n,通项公式,a,n,3,(,n,1),1,n,2,，,即,a,n,n,2.,58/58,