2019届江苏专用高考数学大一轮复习第六章数列6.3等比数列及其前n项和讲义理苏教版.ppt

,6.3,等比数列及其前,n,项和,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.,等比数列的定义,一般地，如果一个数列,_,，那么这个数列叫做等比数列，这个常数叫做等比数列的,，,通常用字母,表示,(,q,0).,2.,等比数列的通项公式,设等比数列,a,n,的首项为,a,1,，公比为,q,，则它的通项,a,n,.,3.,等比中项,如果在,a,与,b,中间插入一个数,G,，使,a,，,G,，,b,成等比数列，那么,G,叫做,a,与,b,的,.,知识梳理,从第二项起，每一项与它的前一项的比都等于同,一个常数,公比,q,a,1,q,n,1,等比中项,4.,等比数列的常用性质,(1),通项公式的推广：,a,n,a,m,(,n,，,m,N,*,).,(2),若,a,n,为等比数列，且,k,l,m,n,(,k,，,l,，,m,，,n,N,*,),，则,.,(3),若,a,n,，,b,n,(,项数相同,),是等比数列，则,a,n,(,0),，,仍是等比数列,.,5.,等比数列的前,n,项和公式,等比数列,a,n,的公比为,q,(,q,0),，其前,n,项和为,S,n,，,当,q,1,时，,S,n,na,1,；,当,q,1,时，,S,n,q,n,m,a,k,a,l,a,m,a,n,6.,等比数列前,n,项和的性质,公比不为,1,的等比数列,a,n,的前,n,项和为,S,n,，则,S,n,，,S,2,n,S,n,，,S,3,n,S,2,n,仍成等比数列，其公比为,.,q,n,等比数列,a,n,的单调性,(4),当,q,0),，由正项等比数列,a,n,满足,a,2 015,2,a,2 013,a,2 014,，,可得,a,2 013,q,2,2,a,2 013,a,2 013,q,，,q,2,q,2,0,，,q,0,，,q,2.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,10.(2016,苏锡常镇一调,),设数列,a,n,是首项为,1,，公差不为零的等差数列，,S,n,为其前,n,项和，若,S,1,，,S,2,，,S,4,成等比数列，则数列,a,n,的公差为,_.,设公差为,d,，其中,d,0,，则,S,1,，,S,2,，,S,4,分别为,1,2,d,4,6,d,.,由,S,1,，,S,2,，,S,4,成等比数列，得,(2,d,),2,4,6,d,，,即,d,2,2,d,.,因为,d,0,，所以,d,2.,答案,解析,2,1,2,3,4,5,6,7,8,9,10,11,12,13,14,*11.(2016,苏北四市期末,),已知各项均为正数的数列,a,n,的首项,a,1,1,，,S,n,是数列,a,n,的前,n,项和，且满足,a,n,S,n,1,a,n,1,S,n,a,n,a,n,1,a,n,a,n,1,(,0,，,n,N,*,).,(1),若,a,1,，,a,2,，,a,3,成等比数列，求实数,的值；,解答,令,n,1,，得,a,2,.,令,n,2,，得,a,2,S,3,a,3,S,2,a,2,a,3,a,2,a,3,，,因为,0,，所以,1.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,(2),若,，求,S,n,.,解答,1,2,3,4,5,6,7,8,9,10,11,12,13,14,当,时，,a,n,S,n,1,a,n,1,S,n,a,n,a,n,1,a,n,a,n,1,，,所以数列,是以,2,为首项，,为公差的等差数列，,当,n,2,时，,S,n,1,1,(,1),a,n,1,，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,1,2,3,4,5,6,7,8,9,10,11,12,13,14,12.,已知,a,n,是首项为,1,，公差为,2,的等差数列，,S,n,表示,a,n,的前,n,项和,.,(1),求,a,n,及,S,n,；,因为,a,n,是首项,a,1,1,，公差,d,2,的等差数列，,所以,a,n,a,1,(,n,1),d,2,n,1.,故,S,n,1,3,(2,n,1),解答,1,2,3,4,5,6,7,8,9,10,11,12,13,14,(2),设,b,n,是首项为,2,的等比数列，公比,q,满足,q,2,(,a,4,1),q,S,4,0,，求,b,n,的通项公式及其前,n,项和,T,n,.,由,(1),得,a,4,7,，,S,4,16.,因为,q,2,(,a,4,1),q,S,4,0,，即,q,2,8,q,16,0,，,所以,(,q,4),2,0,，从而,q,4.,又因为,b,1,2,，,b,n,是公比,q,4,的等比数列，,所以,b,n,b,1,q,n,1,24,n,1,2,2,n,1,.,解答,1,2,3,4,5,6,7,8,9,10,11,12,13,14,13.(2016,全国丙卷,),已知各项都为正数的数列,a,n,满足,a,1,1,，,(2,a,n,1,1),a,n,2,a,n,1,0.,(1),求,a,2,，,a,3,；,解答,(2),求,a,n,的通项公式,.,解答,由,(2,a,n,1,1),a,n,2,a,n,1,0,，得,2,a,n,1,(,a,n,1),a,n,(,a,n,1).,因为,a,n,的各项都为正数，所以,故,a,n,是首项为,1,，公比为,的等比数列，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,14.,已知数列,a,n,中，,a,1,1,，,a,n,a,n,1,，记,T,2,n,为,a,n,的前,2,n,项的和，,b,n,a,2,n,a,2,n,1,，,n,N,*,.,(1),判断数列,b,n,是否为等比数列，并求出,b,n,；,解答,a,n,a,n,1,，,a,n,1,a,n,2,，,b,n,a,2,n,a,2,n,1,，,a,1,1,，,a,1,a,2,，,a,2,b,1,a,1,a,2,.,b,n,是首项为,，公比为,的等比数列,.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,(2),求,T,2,n,.,解答,由,(1),可知，,a,n,2,，,a,1,，,a,3,，,a,5,，,是以,a,1,1,为首项，以,为公比的等比数列；,a,2,，,a,4,，,a,6,，,是以,a,2,为首项，以,为公比的等比数列，,T,2,n,(,a,1,a,3,a,2,n,1,),(,a,2,a,4,a,2,n,),1,2,3,4,5,6,7,8,9,10,11,12,13,14,