高中数学第一章数列1.2.2.2an与Sn的关系及裂项求和法省公开课一等奖新名师优质课获奖PPT课件.pptx

,-,*,-,1.1,数列的概念,-,*,-,-,*,-,1.1,数列的概念,自主预习,合作学习,当堂检测,首页,-,*,-,1.1,数列的概念,自主预习,合作学习,当堂检测,首页,-,*,-,1.1,数列的概念,自主预习,合作学习,当堂检测,首页,-,*,-,1.1,数列的概念,合作学习,自主预习,当堂检测,首页,-,*,-,1.1,数列的概念,当堂检测,自主预习,合作学习,首页,第,2,课时,a,n,与,S,n,关系及裂项求和法,1/32,2/32,1,.a,n,与,S,n,关系,因为,S,n,=a,1,+a,2,+a,3,+,+a,n,当,n,2,且,n,N,+,时,S,n-,1,=a,1,+a,2,+,+a,n-,1,所以,S,n,=S,n-,1,+a,n,即,a,n,=S,n,-S,n-,1,;,而当,n=,1,时,a,1,=S,1,即,S,1,为数列,a,n,首项,.,所以,假如已知数列,a,n,前,n,项和,S,n,公式,那么这个数列是确定,而且,【,做一做,1,】,已知数列,a,n,前,n,项和,S,n,=n,2,+,2,n,则,a,n,通项公式为,.,答案,:,a,n,=,2,n+,1,3/32,名师点拨,利用,S,n,求,a,n,方法,已知数列,a,n,前,n,项和求通项公式,a,n,普通要使用公式,a,n,=S,n,-S,n-,1,(,n,2),但必须注意它成立条件是,n,2,除此之外还要注意以下几点,:,(1),求,a,1,时不能使用,a,n,=S,n,-S,n-,1,因为,S,0,在数列前,n,项和中无意义,而应该是,a,1,=S,1,;,(2),由,a,n,=S,n,-S,n-,1,求得,a,n,代入,n=,1,时,若恰好,a,1,=S,1,则,a,n,=S,n,-S,n-,1,就是其通项公式,;,(3),由,a,n,=S,n,-S,n-,1,求得,a,n,代入,n=,1,时,若,a,1,S,1,则数列通项公式就用分段形式来表示,即,4/32,2,.,裂项求和法,裂项法求和是数列求和一个惯用方法,它基本思想是设法将数列每一项拆成两项,(,裂成两项,),并使它们在相加时除了首尾各有一项或少数几项外,其余各项都能前后相抵消,进而可求出数列前,n,项和,.,【,做一做,2,】,5/32,6/32,思索辨析,判断以下说法是否正确,正确在后面括号内打,“,”,错误打,“,”,.,(1),已知数列,a,n,前,n,项和为,S,n,=,2,n,+,2,则,a,n,通项公式为,a,n,=,2,n+,1,.,(,),(2),已知数列,a,n,通项公式为,a,n,=,18,-,3,n,S,n,是,a,n,前,n,项和,T,n,是,|a,n,|,前,n,项和,则一定有,T,n,S,n,.,(,),(3),数列,-,1,2,-,3,4,-,5,6,-,7,8,(,-,1),n,n,前,n,项和为,.,(,),答案,:,(1),(2),(3),7/32,探究一,探究二,探究三,探究四,思维辨析,【例,1,】,(1),已知数列,a,n,前,n,项和为,S,n,=,2,n,2,-,8,n+,10,求通项公式,a,n,并判断数列是否为等差数列,;,(2),已知数列,a,n,前,n,项和公式,求其通项公式,.,分析,:,依据,a,n,与,S,n,关系求,a,n,要注意分类讨论,.,解,:,(1),当,n,2,时,S,n-,1,=,2(,n-,1),2,-,8(,n-,1),+,10,=,2,n,2,-,12,n+,20,a,n,=S,n,-S,n-,1,=,2,n,2,-,8,n+,10,-,2,n,2,+,12,n-,20,=,4,n-,10,.,当,n=,1,时,a,1,=S,1,=,2,-,8,+,10,=,4,当,n,2,时,a,n,-a,n-,1,=,4,n-,10,-,4(,n-,1),+,10,=,4,数列,a,n,从第,2,项起组成等差数列,但,a,n,不是等差数列,.,8/32,探究一,探究二,探究三,探究四,思维辨析,9/32,探究一,探究二,探究三,探究四,思维辨析,反思感悟,1,.,已知,S,n,求,a,n,时,应分为以下三步,:,(1),当,n,2,时,由,a,n,=S,n,-S,n-,1,求出,a,n,;,(2),当,n=,1,时,由,a,1,=S,1,求出,a,1,并判断,a,1,值是否适合,(1),中求得,a,n,;,(3),写出,a,n,表示式,.,2,.,在由,S,n,求,a,n,时,若忽略对,n=,1,时情况讨论,将可能造成错误,.,10/32,探究一,探究二,探究三,探究四,思维辨析,变式训练,1,已知数列,a,n,前,n,项和为,S,n,(,S,n,0),11/32,探究一,探究二,探究三,探究四,思维辨析,12/32,探究一,探究二,探究三,探究四,思维辨析,【例,2,】,已知正项数列,a,n,前,n,项和为,S,n,且,8,S,n,=,(,a,n,+,2),2,.,(1),求证,:,a,n,为等差数列,;,(2),求,a,n,通项公式,.,分析,:,(1),依据,a,n,=S,n,-S,n-,1,消去,S,n,得到,a,n,与,a,n-,1,关系后进行判断,;(2),由,a,1,=S,1,代入求出,a,1,值,结合,(1),求得通项公式,.,13/32,探究一,探究二,探究三,探究四,思维辨析,(1),证实,:,因为,8,S,n,=,(,a,n,+,2),2,所以当,n,2,时,8,S,n-,1,=,(,a,n-,1,+,2),2,所以,(,a,n,+a,n-,1,)(,a,n,-a,n-,1,-,4),=,0,.,又,a,n,为正项数列,所以,a,n,+a,n-,1,0,从而,a,n,-a,n-,1,-,4,=,0,即,a,n,-a,n-,1,=,4,故,a,n,是公差为,4,等差数列,.,(2),解,:,当,n=,1,时,得,8,S,1,=,(,a,1,+,2),2,即,8,a,1,=,(,a,1,+,2),2,解得,a,1,=,2,所以,a,n,通项公式,a,n,=,2,+,(,n-,1),4,即,a,n,=,4,n-,2,.,14/32,探究一,探究二,探究三,探究四,思维辨析,反思感悟,在给出数列,a,n,与,S,n,关系式时,可依据,a,n,=S,n,-S,n-,1,(,n,2),将关系式中,S,n,(,或,a,n,),消去,从而求得,a,n,与,a,n-,1,(,或,S,n,与,S,n-,1,),关系,然后借助等差数列或其它特殊数列中方法求解,.,15/32,探究一,探究二,探究三,探究四,思维辨析,变式训练,2,已知在各项均为正数数列,a,n,中,a,1,=,1,S,n,是数列,a,n,前,n,项和,对任意,n,N,+,有,2,S,n,=+pa,n,-p,(,p,R,),.,(1),求常数,p,值,;,(2),求数列,a,n,通项公式,.,16/32,探究一,探究二,探究三,探究四,思维辨析,17/32,探究一,探究二,探究三,探究四,思维辨析,【例,3,】,已知等差数列,a,n,满足,a,3,=,7,a,5,+a,7,=,26,a,n,前,n,项和为,S,n,.,(1),求,a,n,及,S,n,;,(2),令,(,n,N,+,),求数列,b,n,前,n,项和,T,n,.,分析,:,(1),设出公差,依据已知条件结构方程组可求出首项和公差,进而求出,a,n,及,S,n,;(2),先,由,(1),求出,b,n,通项公式,再依据通项特点选择求和方法,.,18/32,探究一,探究二,探究三,探究四,思维辨析,19/32,探究一,探究二,探究三,探究四,思维辨析,反思感悟,1,.,通常情况下,当数列通项公式是分式形式,且分子是一个常数,分母是两个相邻正整数之积时,可考虑用裂项法求和,.,2,.,用裂项法求和时,首先要将通项公式进行变形,化为两项相减形式,然后将数列各项用改写后通项公式形式表示,最终将正、负项抵消即得前,n,项和,.,20/32,探究一,探究二,探究三,探究四,思维辨析,变式训练,3,21/32,探究一,探究二,探究三,探究四,思维辨析,【例,4,】,在数列,a,n,中,a,1,=,8,a,4,=,2,且满足,a,n+,2,=,2,a,n+,1,-a,n,(,n,N,+,),.,(1),求数列,a,n,通项公式,;,(2),设,S,n,=|a,1,|+|a,2,|+,+|a,n,|,求,S,n,.,分析,:,(1),依据等差数列定义可知,a,n,是等差数列,.,(2),先找出数列,a,n,中,非负,项,再分类讨论,.,22/32,探究一,探究二,探究三,探究四,思维辨析,解,:,(1),由题意知,a,n+,2,-a,n+,1,=a,n+,1,-a,n,所以,a,n,为等差数列,.,设公差为,d,由,a,1,=,8,a,4,=,2,得,2,=,8,+,3,d,解得,d=-,2,所以,a,n,=,8,-,2(,n-,1),=,10,-,2,n.,(2),由,(1),知,a,n,=,10,-,2,n,令,10,-,2,n,0,得,n,5,即数列,a,n,前,5,项为非负数,后面为负数,所以当,n,5,时,23/32,探究一,探究二,探究三,探究四,思维辨析,反思感悟,对数列,|a,n,|,求和问题,首先要明确数列类型,然后要清楚,a,1,+a,2,+a,3,+,+a,n,与,|a,1,|+|a,2,|+,+|a,n,|,区分与联络,找出数列,a,n,中出现正负转换时临界是处理问题关键,.,24/32,探究一,探究二,探究三,探究四,思维辨析,因错用裂项求和法而犯错,25/32,探究一,探究二,探究三,探究四,思维辨析,纠错心得,1,.,在用裂项法求和时,要注意最终剩下项不一定就是最前面一项和最终面一项,.,2,.,对于错解,1,显然通项公式变形是错误,.,抵消项时也出现了错误,;,错解,2,对通项公式变形即使正确,但抵消项时出现了错误,.,26/32,探究一,探究二,探究三,探究四,思维辨析,变式训练,已知数列,a,n,是递增等差数列,a,1,a,2,是方程,x,2,-,3,x+,2,=,0,两根,.,(1),求数列,a,n,通项公式,;,(2),求数列,前,n,项和,S,n,.,解,:,(1),方程,x,2,-,3,x+,2,=,0,两根为,1,2,由题意得,a,1,=,1,a,2,=,2,.,设数列,a,n,公差为,d,则,d=a,2,-a,1,=,1,所以数列,a,n,通项公式为,a,n,=n.,27/32,1,2,3,4,5,1,.,设数列,a,n,前,n,项和,S,n,=n,2,则,a,8,值为,(,),A.15B.16,C.49D.64,解析,:,a,8,=S,8,-S,7,=,8,2,-,7,2,=,15,.,答案,:,A,28/32,1,2,3,4,5,2,.,已知数列,a,n,前,n,项和,S,n,=,2,n,-,1,则其通项公式为,(,),A.,a,n,=,2,n,B.,a,n,=,2,n-,1,C.,a,n,=,2,n+,1,D.,a,n,=,2,n-,1,-,1,解析,:,当,n,2,时,a,n,=S,n,-S,n-,1,=,2,n,-,1,-,(2,n-,1,-,1),=,2,n-,1,当,n=,1,时,a,1,=S,1,=,2,1,-,1,=,1,适合上式,故,a,n,=,2,n-,1,(,n,N,+,),.,答案,:,B,29/32,1,2,3,4,5,30/32,1,2,3,4,5,4,.,已知数列,a,n,通项公式为,a,n,=,26,-,6,n,则数列,|a,n,|,前,10,项和为,.,解析,:,令,26,-,6,n,0,得,n,4,即数列,a,n,前,4,项为非负数,后面为负数,所以当,n,4,时,答案,:,158,31/32,1,2,3,4,5,5,.,已知,在,等差数列,a,n,中,公差,d,0,又,a,2,a,3,=,15,a,1,+a,4,=,8,.,(1),求数列,a,n,通项公式,;,(2),记数列,数列,b,n,前,n,项和记为,S,n,求,S,n,.,解,:,(1),由等差数列性质得,a,2,+a,3,=a,1,+a,4,=,8,.,又,a,2,a,3,=,15,a,2,a,3,是方程,x,2,-,8,x+,15,=,0,两根,结合,d,0,解得,a,2,=,3,a,3,=,5,d=,2,a,n,=a,2,+,(,n-,2),d=,2,n-,1,.,32/32,