高考数学复习第七章数列推理与证明40等比数列.pptx

,单击此处编辑母版文本样式,课前热身,课堂导学,课堂评价,第七章数列、推理与证明,高考总复习 一轮复习导学案 数学文科,单击此处编辑母版文本样式,第七章数列、推理与证明,1/45,第40课等比数列,2/45,课 前 热 身,3/45,1.(,必修5P49习题1改编),已知数列,a,n,为正项等比数列，,a,2,9，,a,4,4，那么数列,a,n,通项公式,a,n,_.,激活思维,4/45,2.(,必修5P49习题1改编,)假如1，,a,，,b,，,c,，9成等比数列，那么,b,_，,a,c,_.,【解析】,由等比数列性质可得,ac,(1)(9)9，,b,b,9，且,b,与奇数项符号相同，故,b,3.,3,9,5/45,3.(,必修5P58练习6改编,)若对于实数,x,，有,a,n,x,n,，则数列,a,n,前,n,项和,S,n,_.,6/45,4.(,必修5P61习题3改编,)若等比数列通项公式为,a,n,43,1,n,，则数列,a,n,是_数列(填“递增”或“递减”),递减,7/45,1.等比数列定义及通项,假如一个数列从第二项起，每一项与它前一项_都等于_，那么这个数列就叫作等比数列这个常数叫作等比数列_,推广：,a,n,a,m,q,n,m,.,知识梳理,比,同一个常数,公比,8/45,9/45,3.等比数列性质,设数列,a,n,是等比数列，公比为,q,.,(1)若,m,n,p,q,(,m,，,n,，,p,，,q,N,*,)，则_；,a,m,a,n,a,p,a,q,10/45,课 堂 导 学,11/45,等比数列基本量运算,例 1,12/45,【,精关键点评,】,此题主要考查等比数列通项公式求等比数列通项就是要求基本量,a,1,和,q,，要注意,q,1情况,13/45,1,14/45,2.,(苏州期中),已知等比数列,a,n,公比大于1，若,a,5,a,1,15，,a,4,a,2,6，则,a,3,_.,4,15/45,3.,(苏北四市摸底),在等比数列,a,n,中，若,a,1,1，,a,3,a,5,4(,a,4,1)，则,a,7,_.,【解析】,方法一：,设等比数列,a,n,公比为,q,，因为,a,1,1，,a,3,a,5,4(,a,4,1)，所以,q,2,q,4,4(,q,3,1)，即,q,6,4,q,3,40，,q,3,2，所以,a,7,q,6,4.,4,16/45,4.,(全国卷),在数列,a,n,中，已知,a,1,2，,a,n,1,2,a,n,，,S,n,为,a,n,前,n,项和若,S,n,126，则,n,_.,6,17/45,(1),(广州期末),在正项等比数列,a,n,中，若,a,1,和,a,19,为方程,x,2,10,x,160两根，则,a,8,a,10,a,12,_.,等比数列性质及应用,例 2,64,18/45,(2)(南通一调)已知等比数列,an,前,n,项和为,Sn,，若,S,23，,S,415，则,S,6值为_,63,19/45,方法三：,由,S,2,，,S,4,S,2,，,S,6,S,4,成等比数列可得(,S,4,S,2,),2,S,2,(,S,6,S,4,)，所以,S,6,63.,20/45,【思维引导】,(1)考虑一元二次方程根与系数关系；(2)依据性质,S,2,，,S,4,S,2,，,S,6,S,4,成等比数列列式计算,S,6,.,21/45,(1),(梁丰中学),已知等比数列,a,n,前,n,项和为,S,n,，若,a,2,a,8,2,a,3,a,6,，,S,5,62，则,a,1,_.,【解析】,设等比数列,a,n,公比为,q,，由,a,2,a,8,2,a,3,a,6,，得,a,2,a,5,a,4,，因为,a,5,0，所以,a,5,2,a,4,，所以,q,2.又因为,S,5,62，所以,a,1,(124816)62，即31,a,1,62，解得,a,1,2.,变式,2,22/45,23/45,已知数列,a,n,前,n,项和,S,n,满足,S,n,2,a,n,(1),n,.,(1)求数列,a,n,前3项,a,1,，,a,2,，,a,3,值；,等比数列判定和证实,例 3,24/45,25/45,【思维引导】,第(1)小问中只需要分别令,n,1,2,3解方程即可；第(2)小问先依据,S,n,2,a,n,(1),n,得到,a,n,与,a,n,1,关系，再依据定义得出结论，最终写出通项公式,26/45,27/45,变式,28/45,29/45,等比数列判定和证实,例 4,【解答】,(1)设等比数列,a,n,公比为,q,，因为,a,4,S,4,，,a,5,S,5,，,a,6,S,6,成等差数列，,所以,a,5,S,5,a,4,S,4,a,6,S,6,a,5,S,5,，,即2,a,6,3,a,5,a,4,0，所以2,q,2,3,q,10.,30/45,31/45,(2)对,n,N,*,，在,a,n,与,a,n,1,之间插入3,n,个数，使这3,n,2个数成等差数列，记插入这3,n,个数和为,b,n,，求数列,b,n,前,n,项和,T,n,.,32/45,【,精关键点评,】,本题主要考查等比数列前,n,项和公式利用，同时考查结构新数列求通项、求和方法,33/45,(四川卷),设数列,a,n,前,n,项和,S,n,满足,S,n,2,a,n,a,3,，且,a,1,，,a,2,1，,a,3,成等差数列,(1)求数列,a,n,通项公式；,【解答】,(1)因为,S,n,2,a,n,a,3,，所以,a,n,S,n,S,n,1,2,a,n,2,a,n,1,(,n,2，,n,N,*,)，即,a,n,2,a,n,1,(,n,2，,n,N,*,)，,从而,a,2,2,a,1,，,a,3,2,a,2,4,a,1,.,又因为,a,1,，,a,2,1，,a,3,成等差数列，,即,a,1,a,3,2(,a,2,1)，,所以,a,1,4,a,1,2(2,a,1,1),，解得,a,1,2,，,所以数列,a,n,是首项为2、公比为2等比数列，所以,a,n,2,n,.,变式,34/45,35/45,(南通二调),设,a,n,是公差为,d,等差数列，,b,n,是公比为,q,(,q,1)等比数列，记,c,n,a,n,b,n,.,(1)求证：数列,c,n,1,c,n,d,为等比数列；,【解答】,(1)由题意得,c,n,1,c,n,d,(,a,n,1,b,n,1,)(,a,n,b,n,),d,(,a,n,1,a,n,),d,(,b,n,1,b,n,),b,n,(,q,1)0，,又因为,c,2,c,1,d,b,1,(,q,1)0，,所以,c,n,1,c,n,d,是首项为,b,1,(,q,1)，公比为,q,等比数列,备用例题,36/45,(2)已知数列,c,n,前4项分别为4,10,19,34，求数列,a,n,和,b,n,通项公式,【解答】,方法一：,c,n,1,c,n,d,前3项为6,d,9,d,15,d,，则(9,d,),2,(6,d,)(15,d,)，,解得,d,3，从而,q,2.,解得,a,1,1，,b,1,3，,所以,a,n,3,n,2，,b,n,32,n,1,.,37/45,消去,b,1,，得,q,2，,从而解得,a,1,1，,b,1,3，,d,3，,所以,a,n,3,n,2，,b,n,32,n,1,.,38/45,【,精关键点评,】,等差数列与等比数列综合利用主要还是考查基本量运算，同时考查学生综合分析问题能力,39/45,课 堂 评 价,40/45,1.,(泰州二模),在等比数列,a,n,中，已知,a,3,4，,a,7,2,a,5,320，那么,a,7,_.,【解析】,设公比为,q,，则有,a,3,q,4,2,a,3,q,2,320，即,q,4,2,q,2,80，解得,q,2,4(负值舍去)，所以,a,7,a,3,q,4,64.,2.在正项等比数列,a,n,中，若,a,3,a,11,16，则log,2,a,2,log,2,a,12,_.,【解析】,因为等比数列,a,n,中，,a,3,a,11,16，所以,a,2,a,12,a,3,a,11,16，所以log,2,a,2,log,2,a,12,log,2,(,a,2,a,12,)log,2,164.,64,4,41/45,3.在等比数列,a,n,中，若,S,5,4，,S,10,12，则,S,15,_.,【解析】,由等比数列性质知,S,5,，,S,10,S,5,，,S,15,S,10,成等比数列，,S,5,4，,S,10,S,5,8，所以,S,15,S,10,16，则,S,15,28.,4.,(浙江卷),已知数列,a,n,前,n,项和为,S,n,.若,S,2,4，,a,n,1,2,S,n,1，则,S,5,_.,28,121,42/45,43/45,5.,(安徽卷),已知数列,a,n,是递增等比数列，且,a,1,a,4,9，,a,2,a,3,8.,(1)求数列,a,n,通项公式；,【解答】,(1)由题设知,a,1,a,4,a,2,a,3,8，,联立,a,1,a,4,9，,由,a,4,a,1,q,3,，得公比,q,2，,所以,a,n,a,1,q,n,1,2,n,1,.,44/45,45/45,