高中数学第一章数列本章整合省公开课一等奖新名师优质课获奖课件.pptx

,-,*,-,1.1,数列的概念,-,*,-,-,*,-,1.1,数列的概念,自主预习,合作学习,当堂检测,首页,-,*,-,1.1,数列的概念,自主预习,合作学习,当堂检测,首页,-,*,-,1.1,数列的概念,自主预习,合作学习,当堂检测,首页,-,*,-,1.1,数列的概念,合作学习,自主预习,当堂检测,首页,-,*,-,1.1,数列的概念,当堂检测,自主预习,合作学习,首页,-,*,-,1.1,数列的概念,知识网络,核心归纳,高考体验,-,*,-,1.1,数列的概念,核心归纳,知识网络,高考体验,-,*,-,1.1,数列的概念,知识网络,核心归纳,高考体验,本章整合,第一章 数列,1/55,2/55,专题一,专题二,专题一,数列通项公式求法,数列通项公式是给出数列主要方式,其本质就是函数解析式围绕数列通项公式,不但能够判断数列类型,研究数列项改变趋势与规律,而且有利于求数列前,n,项和,.,求数列通项公式是数列关键问题之一,.,下面介绍几个惯用求法,.,3/55,专题一,专题二,1,.,观察法,已知数列前若干项,求该数列通项公式时,普通对所给项观察分析,寻找规律,从而依据规律写出此数列一个通项公式,.,【,例,1,】,写出以下两个数列一个通项公式,:,解,:,(1),这是一个分数数列,分子是偶数,而分母是,1,3,3,5,5,7,7,9,9,11,为两个连续奇数乘积,.,故所求数列通项公式为,(2),各项分别加上,2,变为,10,100,1,000,10,000,a,n,=,10,n,-,2,.,4/55,专题一,专题二,变式训练,1,写出以下数列一个通项公式,.,5/55,专题一,专题二,解,:,(1),数列可记为,2,1,+,1,2,2,+,1,2,3,+,1,2,4,+,1,2,5,+,1,所以数列一个通项公式为,a,n,=,2,n,+,1,.,6/55,专题一,专题二,7/55,专题一,专题二,2,.,依据,a,n,与,S,n,关系求通项公式法,(1),若已知,S,n,表示式,则可直接利用,求得,a,n,注意对,n=,1,与,n,2,讨论,.,(2),若已知,S,n,与,a,n,关系式,则可依据,a,n,=S,n,-S,n-,1,消去,S,n,(,或,a,n,),得到,a,n,与,a,n-,1,(,或,S,n,与,S,n-,1,),关系式,然后用其它方法求解,.,8/55,专题一,专题二,9/55,专题一,专题二,变式训练,2,已知数列,a,n,前,n,项和,S,n,=,3,-,2,a,n,(,n,N,+,),求通项公式,a,n,.,解,:,因为,S,n,=,3,-,2,a,n,所以当,n,2,时,S,n-,1,=,3,-,2,a,n-,1,.,两式相减得,a,n,=-,2,a,n,+,2,a,n-,1,10/55,专题一,专题二,3,.,累加法,对于形如,a,n+,1,-a,n,=f,(,n,),型递推公式求通项公式,.,(1),当,f,(,n,),=d,为常数时,a,n,为等差数列,则,a,n,=a,1,+,(,n-,1),d,;,(2),当,f,(,n,),为关于,n,函数时,用叠加法,.,方法以下,:,由,a,n+,1,-a,n,=f,(,n,),得,当,n,2,时,a,n,-a,n-,1,=f,(,n-,1),a,n-,1,-a,n-,2,=f,(,n-,2),a,3,-a,2,=f,(2),a,2,-a,1,=f,(1),.,以上,(,n-,1),个等式叠加,得,a,n,-a,1,=f,(,n-,1),+f,(,n-,2),+,+f,(2),+f,(1),a,n,=f,(,n-,1),+f,(,n-,2),+,+f,(2),+f,(1),+a,1,.,11/55,专题一,专题二,为了书写方便,也能够用横式来写,:,当,n,2,时,a,n,-a,n-,1,=f,(,n-,1),a,n,=,(,a,n,-a,n-,1,),+,(,a,n-,1,-a,n-,2,),+,+,(,a,2,-a,1,),+a,1,=f,(,n-,1),+f,(,n-,2),+,+f,(2),+f,(1),+a,1,.,12/55,专题一,专题二,【,例,3,】,已知数列,a,n,中,a,1,=,1,且,a,n+,1,-a,n,=,3,n,-n,求数列,a,n,通项公式,.,解,:,由,a,n+,1,-a,n,=,3,n,-n,得,a,n,-a,n-,1,=,3,n-,1,-,(,n-,1),a,n-,1,-a,n-,2,=,3,n-,2,-,(,n-,2),a,3,-a,2,=,3,2,-,2,a,2,-a,1,=,3,-,1,.,当,n,2,时,以上,(,n-,1),个等式两端分别相加,得,(,a,n,-a,n-,1,),+,(,a,n-,1,-a,n-,2,),+,+,(,a,2,-a,1,),=,3,n-,1,+,3,n-,2,+,+,3,-,(,n-,1),+,(,n-,2),+,+,1,13/55,专题一,专题二,14/55,专题一,专题二,变式训练,3,15/55,专题一,专题二,16/55,专题一,专题二,17/55,专题一,专题二,变式训练,4,18/55,专题一,专题二,5,.,结构辅助数列法,(1),取倒数结构法,19/55,专题一,专题二,变式训练,5,20/55,专题一,专题二,(2),加常数结构法,对于满足,a,n,=pa,n-,1,+q,型数列,a,n,可利用待定系数法将其变形为,a,n,+,=p,(,a,n-,1,+,),再设,a,n,+,=b,n,则,b,n,即为以,b,1,=a,1,+,为首项,p,为公比等比数列,求出,b,n,通项公式后即得,a,n,.,21/55,专题一,专题二,(3),其它结构法,例,7,已知数列,a,n,满足,a,n+,1,=,2,a,n,+,32,n,a,1,=,2,求,a,n,通项公式,.,22/55,专题一,专题二,变式训练,6,23/55,专题一,专题二,专题二,数列求和惯用方法,数列求和是数列部分主要内容,也是高考主要考点之一,.,对于数列求和问题,普通是先观察数列特点和规律,假如通项公式能够求出,那么,可先求出通项公式再决定使用哪种求和方法,.,下面介绍几个惯用求和方法,.,24/55,专题一,专题二,1,.,公式法,公式法是数列求和最惯用方法之一,可直接利用等差数列、等比数列求和公式,也可利用常见求前,n,项和公式,如,:,解,:,数列,a,n,前,n,项和为,S,n,=,2,n,-,1,a,n,为等比数列,.,a,1,=S,1,=,2,1,-,1,=,1,a,2,=S,2,-S,1,=,3,-,1,=,2,数列,a,n,公比,q=,2,.,25/55,专题一,专题二,变式训练,7,求数列,n,(,n+,1),前,n,项和,S,n,.,解,:,设,a,n,=n,(,n+,1),=n,2,+n,则,S,n,=a,1,+a,2,+,+a,n,=,(1,2,+,1),+,(2,2,+,2),+,+,(,n,2,+n,),=,(1,2,+,2,2,+,+n,2,),+,(1,+,2,+,+n,),26/55,专题一,专题二,2,.,裂项相消法,对于裂项后显著有能够相消项一类数列,在求和时惯用,“,裂项法,”,对于分式求和多利用此法,.,解题时可用待定系数法对通项公式进行拆项,相消时应注意消去项规律,即消去哪些项,保留哪些项,.,例,9,已知数列,a,n,通项公式为,求其前,n,项和,T,n,.,27/55,专题一,专题二,变式训练,8,等差数列,a,n,各项均为正数,a,1,=,3,前,n,项和为,S,n,b,n,为等比数列,b,1,=,1,且,b,2,S,2,=,64,b,3,S,3,=,960,.,(1),求,a,n,与,b,n,;,解,:,(1),设,a,n,公差为,d,b,n,公比为,q,则,d,为正数,a,n,=,3,+,(,n-,1),d,b,n,=q,n-,1,.,28/55,专题一,专题二,29/55,专题一,专题二,3,.,错位相减法,若在数列,a,n,b,n,中,a,n,是等差数列,b,n,是等比数列,则可采取错位相减法求和,.,例,10,求和,:,30/55,专题一,专题二,变式训练,9,已知首项都是,1,两个数列,a,n,b,n,(,b,n,0,n,N,+,),满足,a,n,b,n+,1,-a,n+,1,b,n,+,2,b,n+,1,b,n,=,0,.,解,:,(1),因为,a,n,b,n+,1,-a,n+,1,b,n,+,2,b,n+,1,b,n,=,0,b,n,0(,n,N,+,),所以数列,c,n,是以,1,为首项,2,为公差等差数列,故,c,n,=,2,n-,1,.,(2),由,b,n,=,3,n-,1,知,a,n,=c,n,b,n,=,(2,n-,1)3,n-,1,于是数列,a,n,前,n,项和,S,n,=,13,0,+,33,1,+,53,2,+,+,(2,n-,1)3,n-,1,3,S,n,=,13,1,+,33,2,+,+,(2,n-,3)3,n-,1,+,(2,n-,1)3,n,相减得,-,2,S,n,=,1,+,2(3,1,+,3,2,+,+,3,n-,1,),-,(2,n-,1)3,n,=-,2,-,(2,n-,2)3,n,所以,S,n,=,(,n-,1)3,n,+,1,.,31/55,专题一,专题二,4,.,并项转化法,对于形如,(,-,1),n,a,n,(,其中,a,n,为等差数列,),数列,通常将数列中相邻两项合并,再进行求解,注意对项数,n,分奇数和偶数进行讨论,.,例,11,已知,S,n,=-,1,+,3,-,5,+,7,-,+,(,-,1),n,(2,n-,1),求,S,n,.,解,:,当,n,为奇数时,S,n,=,(,-,1,+,3),+,(,-,5,+,7),+,(,-,9,+,11),+,+,(,-,2,n+,1),32/55,专题一,专题二,变式训练,10,数列,1,1,+,2,1,+,2,+,2,2,1,+,2,+,2,2,+,2,3,1,+,2,+,2,2,+,+,2,n-,1,前,n,项和为,.,解析,:,此数列第,n,项为,1,+,2,+,2,2,+,+,2,n-,1,=,2,n,-,1,答案,:,2,n+,1,-n-,2,33/55,专题一,专题二,5,.,分组求和法,分组求和法是处理通项公式能够写成,c,n,=a,n,+b,n,形式数列求和问题方法,其中,a,n,与,b,n,均是等差数列或等比数列等一些能够直接求和数列,.,34/55,专题一,专题二,例,12,在等差数列,a,n,中,已知公差,d=,2,a,2,是,a,1,与,a,4,等比中项,.,(1),求数列,a,n,通项公式,;,解,:,(1),由题意知,(,a,1,+d,),2,=a,1,(,a,1,+,3,d,),即,(,a,1,+,2),2,=a,1,(,a,1,+,6),解得,a,1,=,2,所以数列,a,n,通项公式为,a,n,=,2,n.,35/55,专题一,专题二,变式训练,11,等比数列,a,n,各项均为正数,且,2,a,1,+,3,a,2,=,1,(1),求数列,a,n,通项公式,;,(2),设,b,n,=,log,3,a,1,+,log,3,a,2,+,+,log,3,a,n,求数列,b,n,前,n,项和,.,解,:,(1),设等比数列,a,n,公比为,q.,36/55,考点,1,考点,2,考点,3,考点,4,考点,1,等差数列,1,.,(,全国乙高考,),已知等差数列,a,n,前,9,项和为,27,a,10,=,8,则,a,100,=,(,),A.100B.99C.98D.97,解析,:,(,方法一,),设等差数列,a,n,公差为,d,答案,:,C,37/55,2,.,(,北京高考,),已知,a,n,为等差数列,S,n,为其前,n,项和,.,若,a,1,=,6,a,3,+a,5,=,0,则,S,6,=,.,解析,:,a,n,是等差数列,a,3,+a,5,=,2,a,4,=,0,.,a,4,=,0,.,a,4,-a,1,=,3,d=-,6,.,d=-,2,.,S,6,=,6,a,1,+,15,d=,6,6,+,15,(,-,2),=,6,.,答案,:,6,3,.,(,江苏高考,),已知,a,n,是等差数列,S,n,是其前,n,项和,.,若,S,5,=,10,则,a,9,值是,.,解析,:,由,S,5,=,10,得,a,3,=,2,所以,2,-,2,d+,(2,-d,),2,=-,3,d=,3,a,9,=,2,+,3,6,=,20,.,答案,:,20,考点,1,考点,2,考点,3,考点,4,38/55,考点,1,考点,2,考点,3,考点,4,考点,2,等比数列,4,.,(,四川高考,),某企业为激励创新,计划逐年加大研发资金投入,.,若该企业,年整年投入研发资金,130,万元,在此基础上,每年投入研发资金比上一年增加,12%,则该企业整年投入研发资金开始超出,200,万元年份是,(,),(,参考数据,:lg 1,.,120,.,05,lg 1,.,30,.,11,lg 20,.,30),A.,年,B.,年,C.,年,D.,年,解析,:,设从,年后第,n,年该企业整年投入研发资金开始超出,200,万元,由已知得,130,(1,+,12%),n,200,答案,:,B,39/55,考点,1,考点,2,考点,3,考点,4,5,.,(,浙江高考,),设数列,a,n,前,n,项和为,S,n,若,S,2,=,4,a,n+,1,=,2,S,n,+,1,n,N,+,则,a,1,=,S,5,=,.,解析,:,由题意,可得,a,1,+a,2,=,4,a,2,=,2,a,1,+,1,所以,a,1,=,1,a,2,=,3,.,再由,a,n+,1,=,2,S,n,+,1,a,n,=,2,S,n-,1,+,1(,n,2),得,a,n+,1,-a,n,=,2,a,n,即,a,n+,1,=,3,a,n,(,n,2),.,又因为,a,2,=,3,a,1,所以数列,a,n,是以,1,为首项,3,为公比等比数列,.,答案,:,1,121,40/55,考点,1,考点,2,考点,3,考点,4,6,.,(,全国丙高考,),已知数列,a,n,前,n,项和,S,n,=,1,+,a,n,其中,0,.,(1),证实,:,a,n,是等比数列,并求其通项公式,;,41/55,考点,1,考点,2,考点,3,考点,4,考点,3,数列综合应用,7,.,(,课标全国,高考,),设,S,n,是等差数列,a,n,前,n,项和,若,a,1,+a,3,+a,5,=,3,则,S,5,=,(,),A.5B.7C.9D.11,解析,:,由,a,1,+a,3,+a,5,=,3,得,3,a,3,=,3,解得,a,3,=,1,.,故,答案,:,A,8,.,(,课标全国,高考,),设,S,n,是数列,a,n,前,n,项和,且,a,1,=-,1,a,n+,1,=S,n,S,n+,1,则,S,n,=,.,42/55,考点,1,考点,2,考点,3,考点,4,9,.,(,全国,1,高考,),设,S,n,为等比数列,a,n,前,n,项和,已知,S,2,=,2,S,3,=-,6,.,(1),求,a,n,通项公式,;,(2),求,S,n,并判断,S,n+,1,S,n,S,n+,2,是否成等差数列,.,43/55,考点,1,考点,2,考点,3,考点,4,10,.,(,全国,2,高考,),已知等差数列,a,n,前,n,项和为,S,n,等比数列,b,n,前,n,项和为,T,n,a,1,=-,1,b,1,=,1,a,2,+b,2,=,2,.,(1),若,a,3,+b,3,=,5,求,b,n,通项公式,;(2),若,T,3,=,21,求,S,3,.,解,:,设,a,n,公差为,d,b,n,公比为,q,则,a,n,=-,1,+,(,n-,1),d,b,n,=q,n-,1,.,由,a,2,+b,2,=,2,得,d+q=,3,.,(1),由,a,3,+b,3,=,5,得,2,d+q,2,=,6,.,所以,b,n,通项公式为,b,n,=,2,n-,1,.,(2),由,b,1,=,1,T,3,=,21,得,q,2,+q-,20,=,0,解得,q=-,5,或,q=,4,.,当,q=-,5,时,由,得,d=,8,则,S,3,=,21,.,当,q=,4,时,由,得,d=-,1,则,S,3,=-,6,.,44/55,考点,1,考点,2,考点,3,考点,4,11,.,(,全国,3,高考,),设数列,a,n,满足,a,1,+,3,a,2,+,+,(2,n-,1),a,n,=,2,n.,(1),求,a,n,通项公式,;,45/55,考点,1,考点,2,考点,3,考点,4,46/55,考点,1,考点,2,考点,3,考点,4,12,.,(,全国甲高考,),S,n,为等差数列,a,n,前,n,项和,且,a,1,=,1,S,7,=,28,.,记,b,n,=,lg,a,n,其中,x,表示不超出,x,最大整数,如,0,.,9,=,0,lg99,=,1,.,(1),求,b,1,b,11,b,101,;,(2),求数列,b,n,前,1 000,项和,.,解,:,(1),设,a,n,公差为,d,据已知有,7,+,21,d=,28,解得,d=,1,.,所以,a,n,通项公式为,a,n,=n.,b,1,=,lg,1,=,0,b,11,=,lg,11,=,1,b,101,=,lg,101,=,2,.,47/55,考点,1,考点,2,考点,3,考点,4,13,.,(,全国乙高考,),已知,a,n,是公差为,3,等差数列,数列,b,n,满足,b,1,=,1,a,n,b,n+,1,+b,n+,1,=nb,n,.,(1),求,a,n,通项公式,;,(2),求,b,n,前,n,项和,.,48/55,考点,1,考点,2,考点,3,考点,4,14,.,(,浙江高考,),设数列,a,n,前,n,项和为,S,n,.,已知,S,2,=,4,a,n+,1,=,2,S,n,+,1,n,N,+,.,(1),求通项公式,a,n,;(2),求数列,|a,n,-n-,2,|,前,n,项和,.,49/55,考点,1,考点,2,考点,3,考点,4,50/55,考点,1,考点,2,考点,3,考点,4,51/55,考点,3,考点,4,考点,1,考点,2,考点,4,数列探究题型,16,.,(,课标全国,高考,),已知数列,a,n,前,n,项和为,S,n,a,1,=,1,a,n,0,a,n,a,n+,1,=,S,n,-,1,其中,为常数,.,(1),证实,:,a,n+,2,-a,n,=,;,(2),是否存在,使得,a,n,为等差数列,?,并说明理由,.,(1),证实,:,由题设,a,n,a,n+,1,=,S,n,-,1,a,n+,1,a,n+,2,=,S,n+,1,-,1,两式相减,得,a,n+,1,(,a,n+,2,-a,n,),=,a,n+,1,.,因为,a,n+,1,0,所以,a,n+,2,-a,n,=,.,(2),解,:,由题设,a,1,=,1,a,1,a,2,=,S,1,-,1,可得,a,2,=,-,1,.,由,(1),知,a,3,=,+,1,.,令,2,a,2,=a,1,+a,3,解得,=,4,.,故,a,n+,2,-a,n,=,4,.,由此可得,a,2,n-,1,是首项为,1,公差为,4,等差数列,a,2,n-,1,=,4,n-,3;,a,2,n,是首项为,3,公差为,4,等差数列,a,2,n,=,4,n-,1,.,所以,a,n,=,2,n-,1,a,n+,1,-a,n,=,2,.,所以存在,=,4,使得数列,a,n,为等差数列,.,52/55,考点,3,考点,4,考点,1,考点,2,17,.,(,江苏高考,),记,U=,1,2,100,.,对数列,a,n,(,n,N,+,),和,U,子集,T,若,T=,定义,S,T,=,0;,若,T=,t,1,t,2,t,k,定义,比如,:,T=,1,3,66,时,S,T,=a,1,+a,3,+a,66,.,现设,a,n,(,n,N,+,),是公比为,3,等比数列,且当,T=,2,4,时,S,T,=,30,.,(1),求数列,a,n,通项公式,;,(2),对任意正整数,k,(1,k,100),若,T,1,2,k,求证,:,S,T,0,n,N,+,所以,S,T,a,1,+a,2,+,+a,k,=,1,+,3,+,+,3,k-,1,所以,S,T,a,k+,1,.,53/55,考点,3,考点,4,考点,1,考点,2,(3),解,:,下面分三种情况证实,.,若,D,是,C,子集,则,S,C,+S,C,D,=S,C,+S,D,S,D,+S,D,=,2,S,D,.,若,C,是,D,子集,则,S,C,+S,C,D,=S,C,+S,C,=,2,S,C,2,S,D,.,若,D,不是,C,子集,且,C,不是,D,子集,.,令,E=C,U,D,F=D,U,C,则,E,F,E,F=,.,于是,S,C,=S,E,+S,C,D,S,D,=S,F,+S,C,D,进而由,S,C,S,D,得,S,E,S,F,.,设,k,为,E,中最大数,l,为,F,中最大数,则,k,1,l,1,k,l.,由,(2),知,S,E,a,k+,1,.,于是,3,l-,1,=a,l,S,F,S,E,a,k+,1,=,3,k,所以,l-,1,k,即,l,k.,又,k,l,故,l,k-,1,.,54/55,考点,3,考点,4,考点,1,考点,2,从而,S,F,a,1,+a,2,+,+a,l,=,1,+,3,+,+,3,l-,1,故,S,E,2,S,F,+,1,所以,S,C,-S,C,D,2(,S,D,-S,C,D,),+,1,即,S,C,+S,C,D,2,S,D,+,1,.,综上,得,S,C,+S,C,D,2,S,D,.,55/55,