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类型全国通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.ppt

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    全国 通用 高考 数学 一轮 复习 第六 数列 6.2 等差数列 及其
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    ,剖析题型 提炼方法,实验解读,构建知识网络 强化答题语句,探究高考 明确考向,6.2,等差数列及其前,n,项和,第六章数列,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,.,知识梳理,从第,2,项起,每一项与它的前一项的差等于同一个,常数,公差,d,a,n,a,1,(,n,1),d,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,.,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,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,设等差数列,a,n,的公差为,d,,其前,n,项和,S,n,_,或,S,n,_,.,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,.,【,知识拓展,】,题组一思考辨析,1.,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,.(,),基础自测,1,2,3,4,5,6,(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.(,),(5),数列,a,n,为等差数列的充要条件是对任意,n,N,*,,都有,2,a,n,1,a,n,a,n,2,.,(,),(6),已知数列,a,n,的通项公式是,a,n,pn,q,(,其中,p,,,q,为常数,),,则数列,a,n,一定是等差数列,.(,),1,2,3,4,5,6,题组二教材改编,2.P46A,组,T2,设数列,a,n,是等差数列,其前,n,项和为,S,n,,若,a,6,2,且,S,5,30,,则,S,8,等于,A.31 B.32,C.33 D.34,答案,解析,1,2,3,4,5,6,1,2,3,4,5,6,3.P39T5,在等差数列,a,n,中,若,a,3,a,4,a,5,a,6,a,7,450,,则,a,2,a,8,_.,答案,1,2,3,4,5,6,180,解析,由等差数列的性质,得,a,3,a,4,a,5,a,6,a,7,5,a,5,450,,,a,5,90,,,a,2,a,8,2,a,5,180.,解析,题组三易错自纠,4.,一个等差数列的首项为,,从第,10,项起开始比,1,大,则这个等差数列的公差,d,的取值范围是,答案,1,2,3,4,5,6,解析,1,2,3,4,5,6,5.,若等差数列,a,n,满足,a,7,a,8,a,9,0,,,a,7,a,10,0,,,a,2 016,a,2 017,0,,,a,2 016,a,2 017,0,成立的最大正整数,n,是,A.2 016 B.2 017,C.4 032 D.4 033,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,因为,a,1,0,,,a,2 016,a,2 017,0,,,a,2 016,a,2 017,0,,,所以,d,0,,,a,2 017,0,成立的最大正整数,n,是,4 032,,故选,C.,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,7.(2017,安徽省安师大附中、马鞍山二中阶段性测试,),若等差数列,a,n,的前,n,项和为,S,n,,且满足,a,2,S,3,4,,,a,3,S,5,12,,则,a,4,S,7,的值是,_.,24,解析,因为,a,4,和,a,2 016,是,3,x,2,12,x,4,0,的两根,所以,a,4,a,2 016,4.,又,a,4,,,a,1 010,,,a,2 016,成等差数列,所以,2,a,1 010,a,4,a,2 016,,即,a,1 010,2,,,所以,8.,等差数列,a,n,中的,a,4,,,a,2 016,是,3,x,2,12,x,4,0,的两根,则,a,1 010,_.,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,9.(2017,郑州模拟,),张丘建算经卷上第,22,题为:,“,今有女善织,日益功疾,.,初日织五尺,今一月日织九匹三丈,.,”,其意思为今有女子善织布,且从第,2,天起,每天比前一天多织相同量的布,.,若第一天织,5,尺布,现在一个月,(,按,30,天计,),共织,390,尺布,则该女最后一天织,_,尺布,.,21,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,10.,设数列,a,n,的通项公式为,a,n,2,n,10(,n,N,*,),,则,|,a,1,|,|,a,2,|,|,a,15,|,_.,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,130,解析,由,a,n,2,n,10(,n,N,*,),知,a,n,是以,8,为首项,,2,为公差的等差数列,,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.,解答,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,11.(2016,全国,),等差数列,a,n,中,,a,3,a,4,4,,,a,5,a,7,6.,(1),求,a,n,的通项公式;,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解,设数列,a,n,的首项为,a,1,,公差为,d,,,解答,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,(2),设,b,n,a,n,,求数列,b,n,的前,10,项和,其中,x,表示不超过,x,的最大整数,如,0.9,0,,,2.6,2.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,所以数列,b,n,的前,10,项和为,1,3,2,2,3,3,4,2,24.,12.(2018,贵州质检,),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,n,4(,n,N,*,).,(1),求证:数列,a,n,为等差数列;,证明,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解得,a,1,3(,a,1,1,舍去,).,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,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,.,解答,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,(2),求数列,a,n,的通项公式,.,解,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.,13.(2017,郑州一模,),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,的值是,_.,技能提升练,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,2,na,n,(,n,1),a,n,1,(,n,1),a,n,1,,,数列,na,n,是以,a,1,1,为首项,,2,a,2,a,1,5,为公差的等差数列,,20,a,20,1,5,19,96,,,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,15.,设等差数列,a,n,的前,n,项和为,S,n,,若,S,m,1,2,,,S,m,0,,,S,m,1,3,,则,m,_.,拓展冲刺练,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,5,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,数列,a,n,为等差数列,且前,n,项和为,S,n,,,解得,m,5,,经检验符合题意,.,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,121,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,设数列,a,n,的公差为,d,,,化简可得,d,2,a,1,2,,,所以,a,n,1,(,n,1),2,2,n,1,,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,
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