高中数学第二章数列2.2等差数列2.2.1等差数列的概念及通项公式省公开课一等奖新名师优质课获奖PP.pptx

-,*,-,2,.2.1,等差数列概念及通项公式,1/28,2/28,1,2,3,1,.,等差数列,普通地,假如一个数列从,第,2,项,起,每一项与它,前一项,差等于,同一个常数,那么这个数列就叫做等差数列,这个常数叫做等差数列,公差,公差通惯用字母,d,表示,.,3/28,1,2,3,4/28,1,2,3,名师点拨,对等差数列定义了解,:,(1),等差数列定义中关键词是,:“,从第,2,项起,”,与,“,同一个常数,”,.,假如一个数列,不是从第,2,项起,而是从第,3,项或第,4,项起,每一项与前一项差是同一个常数,那么此数列不是等差数列,.,假如一个数列,从第,2,项起,每一项与前一项差,尽管是常数,但这个数列也不一定是等差数列,.,这是因为这些常数可能不相同,必须是同一个常数,才是等差数列,.,等差数列中最少有三项,.,(2),定义中,“,每一项与它前一项差,”,含义有两个,:,其一是强调作差次序,即后面项减前面项,;,其二是强调这两项必须相邻,.,5/28,1,2,3,2,.,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,等差中项,.,这三个数满足关系式,2,A=a+b,.,练一练,2,若,a,b,是方程,x,2,-,2,x-,3,=,0,两根,则,a,b,等差中项为,(,),解析,:,由已知得,a+b=,2,所以,a,b,等差中项为,答案,:,C,6/28,1,2,3,3,.,等差数列通项公式,以,a,1,为首项,d,为公差等差数列,a,n,通项公式为,a,n,=a,1,+,(,n-,1),d,.,练一练,3,等差数列,a,n,公差,d=-,1,a,1,=,2,则,a,3,等于,(,),A,.,1,B,.,0,C,.,2,D,.,3,解析,:,a,3,=a,1,+,(3,-,1),d=,2,+,2,(,-,1),=,0,.,答案,:,B,7/28,探究一,探究二,探究三,探究四,探究一等差数列通项公式及应用,1,.,等差数列通项公式可由首项与公差确定,所以要求等差数列通项公式,只需求出首项与公差,.,2,.,等差数列,a,n,通项公式,a,n,=a,1,+,(,n-,1),d,中共含有四个参数,即,a,1,d,n,a,n,假如知道了其中任意三个数,就能够由通项公式求出第四个数,这一求未知量过程,我们通常称之为,“,知三求一,”,.,3,.,通项公式可变形为,a,n,=dn+,(,a,1,-d,),可把,a,n,看作自变量为,n,一次函数,.,8/28,探究一,探究二,探究三,探究四,经典例题,1,(1),在等差数列,a,n,中,已知,a,5,=,10,a,12,=,31,求通项公式,a,n,;,(2),已知数列,a,n,为等差数列,求,a,15,值,.,思绪分析,:,设出首项与公差,列出方程组,并求出首项与公差,再写出通项公式,.,9/28,探究一,探究二,探究三,探究四,10/28,探究一,探究二,探究三,探究四,变式训练,1,已知等差数列,2,5,8,11,则,23,是这个数列,(,),A,.,第,5,项,B,.,第,6,项,C,.,第,7,项,D,.,第,8,项,解析,:,由已知得等差数列首项,a,1,=,2,公差,d=,3,令,2,+,(,n-,1),3,=,23,解得,n=,8,.,答案,:,D,11/28,探究一,探究二,探究三,探究四,探究二等差数列判定与证实,判断一个数列是否是等差数列是常见题型,常见判断方法以下,:,(1),若,a,n,-a,n-,1,=d,(,常数,)(,n,2,且,n,N,*,),a,n,为等差数列,;,(2),若,2,a,n,=a,n-,1,+a,n+,1,(,n,2,且,n,N,*,),a,n,为等差数列,;,(3),若,a,n,=kn+b,(,k,b,为常数,n,N,*,),a,n,为等差数列,.,12/28,探究一,探究二,探究三,探究四,经典例题,2,(1),求证,:,数列,b,n,是等差数列,;,(2),求数列,a,n,通项公式,.,思绪分析,:,先用,a,n,表示,b,n+,1,b,n,再验证,b,n+,1,-b,n,为常数,最终可求出数列,a,n,通项公式,.,13/28,探究一,探究二,探究三,探究四,14/28,探究一,探究二,探究三,探究四,方法总结,用定义判断数列是等差数列是最惯用方法,这类题目往往紧接着求通项,所以,前面判断某个数列是等差数列能够看作是一个提醒,.,15/28,探究一,探究二,探究三,探究四,变式训练,2,已知数列,a,n,通项,a,n,=,4,n-,1,n,N,*,.,(1),求,log,2,a,2,log,2,a,3,;,(2),求证,:,数列,log,2,a,n,是等差数列,.,(1),解,:,log,2,a,2,=,log,2,4,=,2,log,2,a,3,=,log,2,16,=,4,.,(2),证实,:,log,2,a,n,=,log,2,4,n-,1,=,(,n-,1)log,2,4,=,2,n-,2,log,2,a,n+,1,=,2(,n+,1),-,2,=,2,n.,log,2,a,n+,1,-,log,2,a,n,=,2,n-,(2,n-,2),=,2,.,又,log,2,a,1,=,log,2,4,0,=,0,故数列,log,2,a,n,是首项为,0,公差为,2,等差数列,.,16/28,探究一,探究二,探究三,探究四,探究三等差中项应用,1,.a,A,b,成等差数列,A-a=b-A,A=.,2,.,利用等差中项概念可判断一个数列是否为等差数列,如若,a,n,a,n+,1,a,n+,2,满足,2,a,n+,1,=a,n,+a,n+,2,则数列,a,n,为等差数列,这是因为,2,a,n+,1,=a,n,+a,n+,2,等价于,a,n+,1,-a,n,=a,n+,2,-a,n+,1,由定义知数列,a,n,为等差数列,.,17/28,探究一,探究二,探究三,探究四,18/28,探究一,探究二,探究三,探究四,19/28,探究一,探究二,探究三,探究四,20/28,探究四易错辨析,易错点,对等差数列定义了解不透致错,经典例题,4,若数列,a,n,通项公式为,a,n,=,10,+,lg 2,n,(,n,N,*,),求证,:,数列,a,n,为等差数列,.,错解,:,因为,a,n,=,10,+,lg,2,n,=,10,+n,lg,2,所以,a,1,=,10,+,lg,2,a,2,=,10,+,2lg,2,a,3,=,10,+,3lg,2,所以,a,2,-a,1,=,lg,2,a,3,-a,2,=,lg,2,则,a,2,-a,1,=a,3,-a,2,故数列,a,n,为等差数列,.,错因分析,:,a,3,-a,2,=a,2,-a,1,=,常数,不能满足等差数列定义中,“,从第,2,项起,每一项与它前一项差等于同一个常数,”,要求,.,探究一,探究二,探究三,探究四,21/28,探究一,探究二,探究三,探究四,正解,:,因为,a,n,=,10,+,lg,2,n,=,10,+n,lg,2,所以,a,n+,1,=,10,+,(,n+,1)lg,2,.,所以,a,n+,1,-a,n,=,10,+,(,n+,1)lg,2,-,(10,+n,lg,2),=,lg,2(,n,N,*,),.,所以数列,a,n,为等差数列,.,22/28,1 2 3 4 5,1,.,数列,a,n,通项公式,a,n,=,2,n+,5,则此数列,(,),A,.,是公差为,2,等差数列,B,.,是公差为,5,等差数列,C,.,是首项为,5,等差数列,D,.,是公差为,n,等差数列,解析,:,a,n+,1,-a,n,=,2(,n+,1),+,5,-,(2,n+,5),=,2,数列,a,n,是公差为,2,等差数列,.,答案,:,A,23/28,1 2 3 4 5,2,.,在数列,a,n,中,a,1,=,2,a,n+,1,=a,n,+,2,则,a,20,=,(,),A,.,38B,.,40C,.-,36D,.-,38,解析,:,a,n+,1,=a,n,+,2,a,n+,1,-a,n,=,2,数列,a,n,是公差为,2,等差数列,.,a,1,=,2,a,20,=,2,+,(20,-,1),2,=,40,.,答案,:,B,24/28,1 2 3 4 5,3,.,已知等差数列,a,n,前,3,项依次是,x-,1,x+,1,2,x+,3,则其通项公式为,.,解析,:,x-,1,x+,1,2,x+,3,是等差数列,a,n,前,3,项,2(,x+,1),=x-,1,+,2,x+,3,解得,x=,0,.,a,1,=x-,1,=-,1,a,2,=,1,a,3,=,3,.,d=,2,a,n,=-,1,+,2(,n-,1),=,2,n-,3,.,答案,:,a,n,=,2,n-,3,25/28,1 2 3 4 5,4,.,已知等差数列第,3,项是,7,第,11,项是,-,1,则它第,7,项是,.,解析,:,设首项为,a,1,公差为,d,由,a,3,=,7,a,11,=-,1,得,a,1,+,2,d=,7,a,1,+,10,d=-,1,所以,a,1,=,9,d=-,1,则,a,7,=,3,.,答案,:,3,26/28,1 2 3 4 5,5,.,已知数列,a,n,满足,a,1,=,2,(1),数列,是否为等差数列,?,请说明理由,;,(2),求,a,n,.,27/28,1 2 3 4 5,28/28,