高中数学第二章数列2.2等差数列2.2.2等差数列的性质及应用省公开课一等奖新名师优质课获奖PPT课.pptx

2,.2.2,等差数列性质及应用,1/27,2/27,等差数列主要性质,(1),a,n,=a,m,+,(,n-m,),d,(,m,n,N,*,),.,(2),若,m+n=p+q,(,m,n,p,q,N,*,),则,a,m,+a,n,=a,p,+a,q,.,尤其地,若,m+n=,2,p,则,a,m,+a,n,=,2,a,p,.,(3),a,n,公差为,d,则,d,0,a,n,为递增数列,;,d,0,d=,1,故所求四个数为,-,2,0,2,4,.,14/27,探究一,探究二,探究三,探究四,15/27,探究一,探究二,探究三,探究四,探究三等差数列综合问题,经典例题,3,在数列,a,n,中,a,1,=,2,a,n+,1,=a,n,+,2,n,+,1,.,(1),求证,:,数列,a,n,-,2,n,为等差数列,;,(2),设数列,b,n,满足,b,n,=,2log,2,(,a,n,+,1,-n,),求,b,n,通项公式,.,思绪分析,:,先用定义证实数列,a,n,-,2,n,是等差数列,并求出其通项,进而求出,a,n,再代入,b,n,得,b,n,通项公式,.,16/27,探究一,探究二,探究三,探究四,(1),证实,:,a,n+,1,=a,n,+,2,n,+,1,a,n+,1,-,2,n+,1,-,(,a,n,-,2,n,),=a,n,+,2,n,+,1,-,2,n+,1,a,n,+,2,n,=,1,.,数列,a,n,-,2,n,是以,a,1,-,2,=,0,为首项,以,1,为公差等差数列,.,(2),解,:,由,(1),知,a,n,-,2,n,=n-,1,a,n,=,2,n,+n-,1,.,b,n,=,2log,2,(,a,n,+,1,-n,),=,2log,2,(2,n,+n-,1,+,1,-n,),=,2,n.,方法总结,已知数列递推公式求数列通项时,要对递推公式进行合理变形,结构出等差数列求通项,需掌握常见几个变形形式,考查推理能力与分析问题能力,.,17/27,探究一,探究二,探究三,探究四,变式训练,3,已知等差数列,a,n,首项,a,1,=,3,公差,d=-,5,依次取出项序号被,4,除余,3,项组成数列,b,n,.,(1),求,b,1,和,b,2,.,(2),求数列,b,n,通项公式,.,(3),数列,b,n,中第,110,项是数列,a,n,中第几项,?,18/27,探究一,探究二,探究三,探究四,解,:,(1),由题意,知等差数列,a,n,通项公式为,a,n,=,3,-,5(,n-,1),=,8,-,5,n,设数列,b,n,第,n,项是数列,a,n,第,m,项,则需满足,m=,4,n-,1,n,N,*,所以,b,1,=a,3,=,8,-,5,3,=-,7,b,2,=a,7,=,8,-,5,7,=-,27,.,(2),由,(1),知b,n+,1,-b,n,=a,4(,n+,1),-,1,-a,4,n-,1,=,4,d=-,20,所以新数列,b,n,也为等差数列,且首项为,b,1,=-,7,公差为,d=-,20,所以,b,n,=b,1,+,(,n-,1),d=-,7,+,(,n-,1),(,-,20),=,13,-,20,n.,(3),因为,m=,4,n-,1,n,N,*,所以当,n=,110,时,m=,4,110,-,1,=,439,所以数列,b,n,中第,110,项是数列,a,n,中第,439,项,.,19/27,探究一,探究二,探究三,探究四,探究四易错辨析,易错点,错用等差数列性质致错,经典例题,4,设数列,a,n,是等差数列,a,p,=q,a,q,=p,(,p,q,),试求,a,p+q,.,错解,:,数列,a,n,是等差数列,a,p+q,=a,p,+a,q,=p+q.,错因分析,:,性质,a,m,+a,n,=a,p,+a,q,中必须是两项相加等于两项相加,如,a,7,+a,8,=a,6,+a,9,并不是下标和相等即可,如普通情况下,a,15,=a,7,+,8,a,7,+a,8,.,20/27,探究一,探究二,探究三,探究四,正解,:,设数列,a,n,公差为,d,a,p,=a,q,+,(,p-q,),d,从而,a,p+q,=a,p,+qd=q+q,(,-,1),=,0,a,p+q,=,0,.,21/27,探究一,探究二,探究三,探究四,22/27,1 2 3 4 5,1,.,在等差数列,a,n,中,已知,a,4,+a,8,=,16,则,a,2,+a,10,=,(,),A.12B.16C.20D.24,解析,:,由等差数列性质知,a,2,+a,10,=a,4,+a,8,=,16,故选,B,.,答案,:,B,23/27,1 2 3 4 5,2,.,已知数列,a,n,b,n,为等差数列,且公差分别为,d,1,=,2,d,2,=,1,则数列,2,a,n,-,3,b,n,公差为,(,),A,.,7B,.,5C,.,3D,.,1,解析,:,2,a,n+,1,-,3,b,n+,1,-,(2,a,n,-,3,b,n,),=,2(,a,n+,1,-a,n,),-,3(,b,n+,1,-b,n,),=,2,d,1,-,3,d,2,=,4,-,3,=,1,.,答案,:,D,24/27,1 2 3 4 5,3,.,已知数列,a,n,是等差数列,若,a,1,-a,5,+a,9,-a,13,+a,17,=,117,则,a,3,+a,15,=,.,解析,:,a,3,+a,15,=a,1,+a,17,=a,5,+a,13,所以,a,9,=,117,.,所以,a,3,+a,15,=a,9,+a,9,=,234,.,答案,:,234,25/27,1 2 3 4 5,4,.,在等差数列,a,n,中,已知,a,3,+a,8,=,10,则,3,a,5,+a,7,=,.,解析,:,设等差数列公差为,d,则,a,3,+a,8,=,2,a,1,+,9,d=,10,3,a,5,+a,7,=,4,a,1,+,18,d=,2(2,a,1,+,9,d,),=,20,.,答案,:,20,26/27,1 2 3 4 5,5,.,已知三个数成等差数列,其和为,15,首末两项积为,9,求这三个数,.,解,:,由题意,可设这三个数分别为,a-d,a,a+d,所以,当,d=,4,时,这三个数为,1,5,9;,当,d=-,4,时,这三个数为,9,5,1,.,所以这三个数为,1,5,9,或,9,5,1,.,27/27,