新高考数学复习考点知识讲义ppt课件46---数列中的综合问题

强化,训练,6,数列中的综合问题,第,六,章,数列,新高考数学复习考点知识讲义课件,强化训练6数列中的综合问题第六章数列新高考数学复习考点,1.,(2020,东三省四市模拟,),等比数列,a,n,中，,a,5,，,a,7,是函数,f,(,x,),x,2,4,x,3,的两个零点，则,a,3,a,9,等于,A.,3,B.3 C,.,4,D.4,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,基础保分练,解析,a,5,，,a,7,是函数,f,(,x,),x,2,4,x,3,的两个零点,，,a,5,，,a,7,是方程,x,2,4,x,3,0,的两个根,，,a,5,a,7,3,，由等比数列的性质可得,a,3,a,9,a,5,a,7,3.,1.(2020东三省四市模拟)等比数列an中，a5，a,2.,已知等差数列,a,n,的前,n,项和为,S,n,，公差为,2,，且,a,7,是,a,3,与,a,9,的等比中项，则,S,10,的值,为,A.,110,B,.,90,C.90 D.110,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,2.已知等差数列an的前n项和为Sn，公差为2，且a7,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,a,7,是,a,3,与,a,9,的等比中项，,又数列,a,n,的公差为,2,，,(,a,1,12),2,(,a,1,4)(,a,1,16),，解得,a,1,20,，,a,n,20,(,n,1),(,2),22,2,n,，,12345678910111213141516解析a7是,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,设等差数列的首项为,a,1,，公差为,d,，,则,a,3,a,1,2,d,，,a,7,a,1,6,d,.,因为,a,1,，,a,3,，,a,7,成等比数列，,所以,(,a,1,2,d,),2,a,1,(,a,1,6,d,),，,12345678910111213141516解析设等差,4.,某病毒研究所为了更好地研究,“,新冠,”,病毒，计划改建十个实验室，每个实验室的改建费用分为装修费和设备费，每个实验室的装修费都一样，,设备费从第一到第十实验室依次构成等比数列，已知第五实验室比第二实验室的改建费用高,42,万元，第七实验室比第四实验室的改建费用高,168,万元,，,并要求每个实验室改建费用不能超过,1 700,万元,.,则该研究所改建这十个实验室投入的总费用最多,需要,A.3 233,万元,B.4,706,万元,C.4 709,万元,D.4,808,万元,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,4.某病毒研究所为了更好地研究“新冠”病毒，计划改建十个实验,解析,设每个实验室的装修费用为,x,万元，设备费为,a,n,万元,(,n,1,2,3,，,，,10),，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,依题意,x,1 536,1 700,，即,x,164.,解析设每个实验室的装修费用为x万元，设备费为an万元(n,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,5.,(2021,重庆模拟,),某食品加工厂,2019,年获利,20,万元，经调整食品结构，开发新产品，计划从,2020,年开始每年比上一年获利增加,20%,，则从,(,),年开始这家加工厂年获利超过,60,万元，已知,lg 2,0.301 0,，,lg 3,0.477,1,A.2024,年,B.2025,年,C.2026,年,D.2027,年,123456789101112131415165.(2021,解析,由题意，设从,2019,年开始，第,n,年的获利为,a,n,(,n,N,*,),万元，,则数列,a,n,为等比数列，其中,2019,年的获利为首项，即,a,1,20.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析由题意，设从2019年开始，第n年的获利为an(nN,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,n,8,，,从,2026,年开始这家加工厂年获利超过,60,万元,.,12345678910111213141516n8，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,6.(,多选,),已知数列,a,n,是公差不为,0,的等差数列，前,n,项和为,S,n,，满足,a,1,5,a,3,S,8,，下列选项正确的有,A.,a,10,0,B.,S,10,最小,C.,S,7,S,12,D.,S,20,0,123456789101112131415166.(多选)已,解析,根据题意，数列,a,n,是等差数列，若,a,1,5,a,3,S,8,，,即,a,1,5,a,1,10,d,8,a,1,28,d,，变形可得,a,1,9,d,，,又由,a,n,a,1,(,n,1),d,(,n,10),d,，,则有,a,10,0,，故,A,一定正确；,不能确定,a,1,和,d,的符号，不能确定,S,10,最小，故,B,不正确；,则有,S,7,S,12,，故,C,一定正确；,d,0,，,S,20,0,，则,D,不正确,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析根据题意，数列an是等差数列，若a15a3S8,所以,a,n,是周期为,4,的数列，,因为,2 021,505,4,1,，,所以,a,2 021,a,1,2.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,2,所以an是周期为4的数列，123456789101112,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,8.,(,2021,江苏海头中学月考,),已知数列,a,n,的前,n,项和为,S,n,，,a,1,1,，,a,n,1,2,S,n,1,S,n,0,，则,S,n,_.,123456789101112131415168.(2021,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,因为,a,n,1,S,n,1,S,n,，,则,a,n,1,2,S,n,1,S,n,0,，可化简为,S,n,1,S,n,2,S,n,1,S,n,0,，,等式两边同时除以,S,n,1,S,n,，,12345678910111213141516解析因为an,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,4,当,n,2,时，,a,n,1,a,n,0,，即,a,n,1,a,n,，,当,n,1,时，,a,2,a,1,0,，,数列,a,n,中，从,a,2,开始是递增的，,又,a,2,0,，,b,0,，,n,N,*,).,(1),当,a,2,，,b,3,时，求,u,n,；,1234567891011121314151612.(202,解,当,a,2,，,b,3,时，,u,n,2,n,2,n,1,3,2,n,2,3,2,23,n,1,3,n,(,n,N,*,),，,两边除以,2,n,，得,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,所以,u,n,3,n,1,2,n,1,.,解当a2，b3时，un2n2n132n2,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,(2),若,a,b,，求数列,u,n,的前,n,项和,S,n,.,12345678910111213141516(2)若ab,解,若,a,b,，则,u,n,(,n,1),a,n,，,所以,S,n,2,a,3,a,2,4,a,3,(,n,1),a,n,，,当,a,0,，,a,1,时，在,的两边同乘以,a,，,得,aS,n,2,a,2,3,a,3,4,a,4,(,n,1),a,n,1,，,与,式作差，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解若ab，则un(n1)an，当a0，a1时，在,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,12345678910111213141516,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,技能提升练,12345678910111213141516技能提升练,且,a,n,0,，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解得,a,1,1,或,a,1,0(,舍去,).,即,(,a,n,a,n,1,)(,a,n,a,n,1,1),0,，,且an0，12345678910111213141516解,k,T,n,恒成立，,k,1,，即,k,的最小值为,1.,a,n,0,，,a,n,a,n,1,1,，,a,n,是以,1,为首项，,1,为公差的等差数列，,a,n,n,，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,kTn恒成立，an0，anan11，1234,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,14.,(2020,长治质检,),各项均为正数且公比,q,1,的等比数列,a,n,的前,n,项,和,8,1234567891011121314151614.(202,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,由题意,a,1,a,5,a,2,a,4,4,，,又,a,2,a,4,5,，公比,q,1,，,a,n,2,n,2,，,12345678910111213141516解析由题意a,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,令,t,2,n,1,1,2,2,2,2,3,，,，,当且仅当,t,2,n,1,2,，即,n,2,时取等号,.,12345678910111213141516令t2n1,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,拓展冲刺练,15.,(,2021,江苏丰县中学模拟,),如图所示，正方形,ABCD,的边长为,5 cm,，取正方形,ABCD,各边的中点,E,，,F,，,G,，,H,，作第,2,个正方形,EFGH,，然后再取正方形,EFGH,各边的中点,I,，,J,，,K,，,L,，作第,3,个,正,方形,IJKL,，依此方法一直继续下去,.,如果这个,作图,过程,可以一直继续下去，那么所有这些正方形,的,面积,之和将趋近于,_,cm,2,.,50,12345678910111213141516拓展冲刺练15,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,记第,1,个正方形的面积为,S,1,，第,2,个正方形的面积为,S,2,，,，第,n,个正方形的面积为,S,n,，,12345678910111213141516解析记第1个,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,如果这个作图过程可以一直继续下去，那么所有这些正方形的面积之和将趋近于,50 cm,2,.,12345678910111213141516如果这个作图,故数列,a,n,为等比数列，首项为,1,，公比为,x,，,则,a,n,x,n,1,.,16.,已知数列,a,n,的首项为,1,，,S,n,为数列,a,n,的前,n,项和，,S,n,1,xS,n,1,，其中,x,0,，,n,N,*,.,(1),求,a,n,的通项公式；,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解,S,n,1,xS,n,1,，,S,n,xS,n,1,1(,n,2),，,得,a,n,1,xa,n,，,故数列an为等比数