2019届江苏专用高考数学大一轮复习第六章数列6.2等差数列及其前n项和讲义文苏教版.ppt

,6.2,等差数列及其前,n,项和,基础知识自主学习,课时作业,题型分类深度剖析,内容索引,基础知识自主学习,1.,等差数列的定义,一般地，如果一个数列,_,，那么这个数列就叫做等差数列，这个常数叫做等差数列的,，通常用字母,表示,.,2.,等差数列的通项公式,如果等差数列,a,n,的首项为,a,1,，公差为,d,，那么它的通项公式是,_,.,3.,等差中项,由三个数,a,，,A,，,b,组成的等差数列可以看成最简单的,等差数列,.,这时，,A,叫做,a,与,b,的,.,知识梳理,从第二项起，每一项减去它的前一项所得的差都,等于同一个常数,公差,d,a,n,a,1,(,n,1),d,等差中项,4.,等差数列的常用性质,(1),通项公式的推广：,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,的等差数列,.,(6),数列,S,m,，,S,2,m,S,m,，,S,3,m,S,2,m,，,构成等差数列,.,(,n,m,),d,a,k,a,l,a,m,a,n,2,d,md,5.,等差数列的前,n,项和公式,设等差数列,a,n,的公差为,d,，其前,n,项和,S,n,或,S,n,.,6.,等差数列的前,n,项和公式与函数的关系,数列,a,n,是等差数列,S,n,An,2,Bn,(,A,，,B,为常数,).,7.,等差数列的前,n,项和的最值,在等差数列,a,n,中，,a,1,0,，,d,0,，则,S,n,存在最,值；若,a,1,0,，则,S,n,存在最,值,.,大,小,知识拓展,等差数列的四种判断方法,(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,.,思考辨析,判断下列结论是否正确,(,请在括号中打,“,”,或,“”,),(1),若一个数列从第二项起每一项与它的前一项的差都是常数，则这个数列是等差数列,.(,),(2),等差数列,a,n,的单调性是由公差,d,决定的,.(,),(3),等差数列的前,n,项和公式是常数项为,0,的二次函数,.(,),(4),已知等差数列,a,n,的通项公式,a,n,3,2,n,，则它的公差为,2.(,),考点自测,1.(,教材改编,),设,S,n,为等差数列,a,n,的前,n,项和，若,a,3,3,，,S,9,S,6,27,，,则该数列的首项,a,1,_.,答案,解析,2.(,教材改编,),已知五个数成等差数列，它们的和为,5,，平方和为,，,则这五个数的积为,_.,设第三个数为,a,，公差为,d,，则这五个数分别为,a,2,d,，,a,d,，,a,，,a,d,，,a,2,d,，,由已知条件得,答案,解析,3.(2016,全国乙卷,),已知等差数列,a,n,前,9,项的和为,27,，,a,10,8,，则,a,100,_.,得,a,5,3,，而,a,10,8,，因此公差,d,1,，,a,100,a,10,90,d,98.,答案,解析,98,4.,设数列,a,n,是等差数列，若,a,3,a,4,a,5,12,，则,a,1,a,2,a,7,_.,a,3,a,4,a,5,3,a,4,12,，,a,4,4,，,a,1,a,2,a,7,7,a,4,28.,答案,解析,28,5.,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,的值为,_.,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,，,10,1,2,3,4,5,6,7,8,9,10,11,12,13,14,7.(2015,安徽,),已知数列,a,n,中，,a,1,1,，,a,n,a,n,1,(,n,2),，则数列,a,n,的前,9,项和等于,_.,答案,解析,由题意知数列,a,n,是以,1,为首项，以,为公差的等差数列，,27,1,2,3,4,5,6,7,8,9,10,11,12,13,14,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,1,2,3,4,5,6,7,8,9,10,11,12,13,14,9.,设等差数列,a,n,，,b,n,的前,n,项和分别为,S,n,，,T,n,，若对任意自然数,n,都,有,的值为,_.,答案,解析,a,n,，,b,n,为等差数列，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,10.,设等差数列,a,n,的前,n,项和为,S,n,，若,a,1,3,，,a,k,1,，,S,k,12,，则正整数,k,.,13,答案,解析,解得,k,13.,11.(2016,苏州暑假测试,),已知数列,a,n,满足,a,1,1,，,a,2,，且,a,n,(,a,n,1,a,n,1,),2,a,n,1,a,n,1,(,n,2),，则,a,2 016,_.,答案,解析,又,a,1,1,，,a,2,，所以数列,是以,1,为首项，,1,为公差的等差数列,.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,12.,已知等差数列,a,n,前三项的和为,3,，前三项的积为,8.,(1),求等差数列,a,n,的通项公式；,解答,1,2,3,4,5,6,7,8,9,10,11,12,13,14,设等差数列,a,n,的公差为,d,，则,a,2,a,1,d,，,a,3,a,1,2,d,.,所以由等差数列通项公式可得,a,n,2,3(,n,1),3,n,5,或,a,n,4,3(,n,1),3,n,7.,故,a,n,3,n,5,或,a,n,3,n,7.,(2),若,a,2,，,a,3,，,a,1,成等比数列，求数列,|,a,n,|,的前,n,项和,.,解答,1,2,3,4,5,6,7,8,9,10,11,12,13,14,当,a,n,3,n,5,时，,a,2,，,a,3,，,a,1,分别为,1,，,4,2,，不成等比数列；,当,a,n,3,n,7,时，,a,2,，,a,3,，,a,1,分别为,1,2,，,4,，成等比数列，满足条件,.,当,n,1,时，,S,1,|,a,1,|,4,；当,n,2,时，,S,2,|,a,1,|,|,a,2,|,5,；,当,n,3,时，,S,n,S,2,|,a,3,|,|,a,4,|,|,a,n,|,5,(3,3,7),(3,4,7),(3,n,7),记数列,|,a,n,|,的前,n,项和为,S,n,.,当,n,2,时，满足此式，当,n,1,时，不满足此式,.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,*13.,已知数列,a,n,的各项均为正数，前,n,项和为,S,n,，且满足,2,S,n,a,n,4(,n,N,*,).,(1),求证：数列,a,n,为等差数列；,证明,1,2,3,4,5,6,7,8,9,10,11,12,13,14,当,n,1,时，有,2,a,1,1,4,，,即,2,a,1,3,0,，,当,n,2,时，有,2,S,n,1,n,5,，,又,2,S,n,n,4,，,因此,a,n,1,a,n,1,或,a,n,1,a,n,1,.,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,的各项均为正数相矛盾，,所以,a,n,1,a,n,1,，即,a,n,a,n,1,1,，,因此数列,a,n,是首项为,3,，公差为,1,的等差数列,.,解得,a,1,3(,a,1,1,舍去,).,1,2,3,4,5,6,7,8,9,10,11,12,13,14,(2),求数列,a,n,的通项公式,.,解答,由,(1),知,a,1,3,，,d,1,，,所以数列,a,n,的通项公式,a,n,3,(,n,1),1,n,2,，,即,a,n,n,2.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,14.(2016,苏北四市摸底,),已知数列,a,n,满足,2,a,n,1,a,n,a,n,2,k,(,n,N,*,，,k,R,),，且,a,1,2,，,a,3,a,5,4.,(1),若,k,0,，求数列,a,n,的前,n,项和,S,n,；,解答,当,k,0,时，,2,a,n,1,a,n,a,n,2,，即,a,n,2,a,n,1,a,n,1,a,n,，,所以数列,a,n,是等差数列,.,设数列,a,n,的公差为,d,，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,(2),若,a,4,1,，求数列,a,n,的通项公式,.,解答,1,2,3,4,5,6,7,8,9,10,11,12,13,14,由题意得,2,a,4,a,3,a,5,k,，即,2,4,k,，所以,k,2.,当,n,1,时，,2,a,2,a,1,a,3,2,，,当,n,2,时，,2,a,3,a,2,a,4,2,，,所以,a,4,2,a,3,a,2,2,3,a,2,2,a,1,6,1,，所以,a,2,3,由,2,a,n,1,a,n,a,n,2,2,，得,(,a,n,2,a,n,1,),(,a,n,1,a,n,),2,，,所以数列,a,n,1,a,n,是以,a,2,a,1,1,为首项，,2,为公差的等差数列，,所以,a,n,1,a,n,2,n,3.,当,n,2,时，有,a,n,a,n,1,2(,n,1),3,，,于是,a,n,1,a,n,2,2(,n,2),3,，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,a,n,2,a,n,3,2(,n,3),3,，,a,3,a,2,2,2,3,，,a,2,a,1,2,1,3,，,叠加得，,a,n,a,1,21,2,(,n,1),3(,n,1)(,n,2),，,所以,a,n,2,3(,n,1),2,n,2,4,n,1(,n,2).,又当,n,1,时，,a,1,2,也适合上式,.,所以数列,a,n,的通项公式为,a,n,n,2,4,n,1,，,n,N,*,.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,