高中数学第二章数列2.2.2等差数列的前n项和资料.pptx

,第二章,2.2,等差数列,2.2.2,等差数列前,n,项和（二）,1/31,1.,深入熟练掌握等差数列通项公式和前,n,项和公式,.,2.,会解等差数列前,n,项和最值问题,.,3.,了解,a,n,与,S,n,关系，能依据,S,n,求,a,n,.,学习目标,2/31,题型探究,问题导学,内容索引,当堂训练,3/31,问题导学,4/31,思索,知识点一数列中,a,n,与,S,n,关系,已知数列,a,n,前,n,项和,S,n,n,2,，怎样求,a,1,，,a,n,?,答案,a,1,S,1,1,；,当,n,2,时，,a,n,S,n,S,n,1,n,2,(,n,1),2,2,n,1,，,又,n,1,时也适合上式，,所以,a,n,2,n,1,，,n,N,.,5/31,梳理,对任意数列,a,n,，,S,n,与,a,n,关系能够表示为,a,n,(,n,1,),，,(,n,2,，,n,N,),.,S,1,S,n,S,n,1,6/31,知识点二等差数列前,n,项和最值,由二次函数性质能够得出：当,a,1,0,，,d,0,时，,S,n,先减后增，有最小值；当,a,1,0,，,d,0,，,d,0,，则数列前面若干项为正项,(,或,0),，所以将这些项相加即得,S,n,最大值,(2),若,a,1,0,，则数列前面若干项为负项,(,或,0),，所以将这些项相加即得,S,n,最小值,(3),若,a,1,0,，,d,0,，则,S,n,是递增数列，,S,1,是,S,n,最小值,(4),若,a,1,0,，,d,1,，,n,N,),，,11/31,引申探究,例,1,中前,n,项和改为,S,n,n,2,n,1,，求通项公式,解答,12/31,已知前,n,项和,S,n,求通项,a,n,，先由,n,1,时，,a,1,S,1,求得,a,1,，再由,n,2,时，,a,n,S,n,S,n,1,求得,a,n,，最终验证,a,1,是否符合,a,n,，若符合则统一用一个解析式表示,反思与感悟,13/31,跟踪训练,1,已知数列,a,n,前,n,项和,S,n,3,n,，求,a,n,.,当,n,1,时，,a,1,S,1,3,；,当,n,2,时，,a,n,S,n,S,n,1,3,n,3,n,1,23,n,1,.,当,n,1,时，代入,a,n,23,n,1,得,a,1,2,3.,解答,14/31,类型二等差数列前,n,项和最值,例,2,已知等差数列,5,，,前,n,项和为,S,n,，求使得,S,n,最大序号,n,值,解答,15/31,所以当,n,取与,最靠近整数即,7,或,8,时，,S,n,取得最大值,故前,n,项和是从第,9,项开始减小，而第,8,项为,0,，,所以前,7,项或前,8,项和最大,16/31,在等差数列中，求,S,n,最大,(,小,),值，其思绪是找出某一项，使这项及它前面项皆取正,(,负,),值或零，而它后面各项皆取负,(,正,),值，则从第,1,项起到该项各项和为最大,(,小,),因为,S,n,为关于,n,二次函数，也可借助二次函数图象或性质求解,反思与感悟,17/31,跟踪训练,2,在等差数列,a,n,中，,a,n,2,n,14,，试用两种方法求该数列前,n,项和,S,n,最小值,解答,方法一,a,n,2,n,14,，,a,1,12,，,d,2.,a,1,a,2,a,6,a,7,0,a,8,a,9,0,，此时,T,n,S,n,n,2,10,n,；,当,n,5,时，,a,n,0,，此时,T,n,2,S,5,S,n,n,2,10,n,50.,解答,23/31,当堂训练,24/31,当,n,1,时，,a,1,S,1,2,，,当,n,2,时，,a,n,S,n,S,n,1,2,n,，,又因为,a,1,2,符合,a,n,2,n,，,所以,a,n,2,n,.,1.,已知数列,a,n,前,n,项和,S,n,n,2,n,，则,a,n,等于,A.4,n,2 B.,n,2,C.2,n,1 D.2,n,答案,解析,1,2,3,4,25/31,等差数列前,n,项和,S,n,形式为,S,n,an,2,bn,，,故,1.,2.,已知数列,a,n,为等差数列，它前,n,项和为,S,n,，若,S,n,(,n,1),2,，则,值是,A.,2 B.,1 C.0 D.1,答案,解析,1,2,3,4,26/31,S,3,S,8,，,S,8,S,3,a,4,a,5,a,6,a,7,a,8,5,a,6,0,，,a,6,0.,a,1,0,，,a,1,a,2,a,3,a,4,a,5,a,6,0,，,a,7,0.,故当,n,5,或,6,时，,S,n,最大,.,1,2,3,3.,首项为正数等差数列，前,n,项和为,S,n,，且,S,3,S,8,，当,n,_,时，,S,n,取到最大值,.,5,或,6,答案,解析,4,27/31,1,2,3,当,n,1,时，,a,1,S,1,3,2,5.,当,n,2,时，,S,n,1,3,2,n,1,，,又,S,n,3,2,n,，,a,n,S,n,S,n,1,2,n,2,n,1,2,n,1,.,又当,n,1,时，,a,1,5,2,1,1,1,，,解答,4,4.,已知数列,a,n,前,n,项和,S,n,3,2,n,，求,a,n,.,28/31,规律与方法,1.,因为,a,n,S,n,S,n,1,只有,n,2,时才有意义，所以由,S,n,求通项公式,a,n,f,(,n,),时，要分,n,1,和,n,2,两种情况分别计算，然后验证两种情况可否用统一解析式表示，若不能，则用分段函数形式表示,.,2.,求等差数列前,n,项和最值方法：,(1),二次函数法：用求二次函数最值方法来求其前,n,项和最值，但要注意,n,N,，结合二次函数图象对称性来确定,n,值，愈加直观,.,29/31,3.,求等差数列,a,n,前,n,项绝对值之和，关键是找到数列,a,n,正负项分界点,.,30/31,本课结束,31/31,