高考数学复习专题四数列推理与证明第2讲数列的求和问题理市赛课公开课一等奖省名师优质课获奖PPT课件.pptx

第,2,讲数列求和问题,专题四数列、推理与证实,1/53,热点分类突破,真题押题精练,2/53,热点分类突破,3/53,热点一分组转化求和,有些数列，既不是等差数列，也不是等比数列，若将数列通项拆开或变形，可转化为几个等差、等比数列或常见数列，即先分别求和，然后再合并,.,4/53,例,1,(,山东省平阴县第一中学模拟,),已知数列,a,n,是等差数列，其前,n,项和为,S,n,，数列,b,n,是公比大于,0,等比数列，且,b,1,2,a,1,2,，,a,3,b,2,1,，,S,3,2,b,3,7.,(1),求数列,a,n,和,b,n,通项公式；,解答,5/53,解,设数列,a,n,公差为,d,，,b,n,公比为,q,，且,q,0,，,由题易知，,a,1,1,，,b,1,2,，,a,n,2,n,1,，,b,n,2,n,.,6/53,解答,思维升华,7/53,解,由,(1),知，,a,n,2,n,1,，,b,n,2,n,，,T,n,(,c,1,c,3,c,5,c,n,1,),(,c,2,c,4,c,n,),n,(,c,2,c,4,c,n,),，,令,H,n,c,2,c,4,c,6,c,n,，,8/53,以上两式相减，得,9/53,10/53,当,n,(,n,3),为奇数时，,n,1,为偶数，,经验证，,n,1,也适合上式,.,11/53,思维升华,在处理普通数列求和时，一定要注意使用转化思想,.,把普通数列求和转化为等差数列或等比数列进行求和，在求和时要分析清楚哪些项组成等差数列，哪些项组成等比数列，清楚正确地求解,.,在利用分组求和法求和时，因为数列各项是正负交替，所以普通需要对项数,n,进行讨论，最终再验证是否能够合并为一个公式,.,12/53,证实,13/53,14/53,得,na,n,1,2(,n,1),a,n,n,(,n,1),，,由,a,1,0,及递推关系，可知,a,n,0,，,15/53,16/53,(2),求数列,a,n,前,n,项和,S,n,.,解答,17/53,a,n,n,2,n,n,，,S,n,2,2,2,2,3,2,3,(,n,1)2,n,1,n2,n,1,2,3,(,n,1),n,，,设,T,n,2,2,2,2,3,2,3,(,n,1)2,n,1,n,2,n,，,则,2,T,n,2,2,2,2,3,3,2,4,(,n,1)2,n,n,2,n,1,，,由,，得,T,n,2,2,2,2,3,2,n,n,2,n,1,18/53,T,n,(,n,1)2,n,1,2,，,19/53,热点二错位相减法求和,错位相减法是在推导等比数列前,n,项和公式时所用方法，这种方法主要用于求数列,a,n,b,n,前,n,项和，其中,a,n,，,b,n,分别是等差数列和等比数列,.,20/53,(1),求数列,a,n,和,b,n,通项公式；,解答,21/53解,因为数列,a,n,为等差数列，,又因为,a,3,5,，所以,a,1,1,，所以,a,n,2,n,1.,所以,b,n,2,b,n,1,，,即数列,b,n,是首项为,1,，公比为,2,等比数列，,所以,b,n,(,2),n,1,.,22/53,(2),设,c,n,a,n,|,b,n,|,，求数列,c,n,前,n,项和,T,n,.,解,因为,c,n,a,n,|,b,n,|,(2,n,1)2,n,1,，,所以,T,n,1,1,3,2,5,2,2,(2,n,1)2,n,1,，,2,T,n,1,2,3,2,2,5,2,3,(2,n,1)2,n,，,两式相减，得,T,n,1,1,2,2,2,2,2,22,n,1,(2,n,1)2,n,1,2(2,2,2,2,n,1,),(2,n,1)2,n,1,2,n,1,4,(2,n,1)2,n,3,(3,2,n,)2,n,，,所以,T,n,3,(2,n,3)2,n,.,解答,思维升华,23/53,思维升华,(1),错位相减法适合用于求数列,a,n,b,n,前,n,项和，其中,a,n,为等差数列，,b,n,为等比数列,.,(2),所谓,“,错位,”,，就是要找,“,同类项,”,相减,.,要注意是相减后得到部分求等比数列和，此时一定要查清其项数,.,(3),为确保结果正确，可对得到和取,n,1,2,进行验证,.,24/53,(1),求数列,a,n,与,bn,通项公式；,解答,25/53,解,当,n,1,时，,a,1,1,，,当,n,2,时，,a,n,S,n,S,n,1,2,n,1(,n,N,*,),，,检验,a,1,1,，满足,a,n,2,n,1(,n,N,*,).,且,b,n,0,，,2,b,n,1,b,n,，,26/53,(2),设,c,n,a,n,b,n,，求数列,c,n,前,n,项和,T,n,.,解答,27/53,28/53,29/53,热点三裂项相消法求和,裂项相消法是指把数列和式中各项分别裂开后，一些项能够相互抵消从而求和方法，主要适合用于,(,其中,a,n,为等差数列,),等形式数列求和,.,30/53,例,3,(,届山东省青岛市二模,),在公差不为,0,等差数列,a,n,中，,a,3,a,6,，且,a,3,为,a,1,与,a,11,等比中项,.,(1),求数列,a,n,通项公式；,解,设数列,a,n,公差为,d,，,即,(,a,1,2,d,),2,a,1,(,a,1,10,d,),，,d,0,，由,解得,a,1,2,，,d,3.,数列,a,n,通项公式为,a,n,3,n,1.,(,a,1,d,),2,a,1,2,d,a,1,5,d,，,解答,思维升华,31/53,思维升华,裂项相消法基本思想就是把通项,a,n,分拆成,a,n,b,n,k,b,n,(,k,1,，,k,N,*,),形式，从而在求和时到达一些项相消目标在解题时要善于依据这个基本思想变换数列,a,n,通项公式，使之符合裂项相消条件,.,32/53,解答,思维升华,33/53,思维升华,惯用裂项公式,34/53,(1),求数列,a,n,通项公式；,解答,35/53,36/53,解答,37/53,38/53,39/53,真题押题精练,40/53,真题体验,1.(,全国,),等差数列,a,n,前,n,项和为,S,n,，,a,3,3,，,S,4,10,，则,_.,答案,解析,1,2,41/53,解析,设等差数列,a,n,公差为,d,，则,1,2,42/53,2.(,天津,),已知,a,n,为等差数列，前,n,项和为,S,n,(,n,N,*,),，,b,n,是首项为,2,等比数列，且公比大于,0,，,b,2,b,3,12,，,b,3,a,4,2,a,1,，,S,11,11,b,4,.,(1),求,a,n,和,b,n,通项公式；,1,2,解答,43/53,解,设等差数列,a,n,公差为,d,，等比数列,b,n,公比为,q,.,由已知,b,2,b,3,12,，得,b,1,(,q,q,2,),12,，而,b,1,2,，,所以,q,2,q,6,0.,又因为,q,0,，解得,q,2,，所以,b,n,2,n,.,由,b,3,a,4,2,a,1,，可得,3,d,a,1,8,，,由,S,11,11,b,4,，可得,a,1,5,d,16,，,联立,，解得,a,1,1,，,d,3,，由此可得,a,n,3,n,2.,所以数列,a,n,通项公式为,a,n,3,n,2,，数列,b,n,通项公式为,b,n,2,n,.,1,2,44/53,(2),求数列,a,2,n,b,2,n,1,前,n,项和,(,n,N,*,).,1,2,解答,45/53,解,设数列,a,2,n,b,2,n,1,前,n,项和为,T,n,，,由,a,2,n,6,n,2,，,b,2,n,1,2,4,n,1,，得,a,2,n,b,2,n,1,(3,n,1),4,n,，故,T,n,2,4,5,4,2,8,4,3,(3,n,1),4,n,，,4,T,n,2,4,2,5,4,3,8,4,4,(3,n,4),4,n,(3,n,1),4,n,1,，,，得,3,T,n,2,4,3,4,2,3,4,3,3,4,n,(3,n,1),4,n,1,(3,n,2),4,n,1,8,，,1,2,46/53,押题预测,答案,解析,押题依据,数列通项以及求和是高考重点考查内容，也是考试纲领中明确提出知识点，年年在考，年年有变，变是试题外壳，即在题设条件上有变革，有创新，但在变中有不变性，即解答问题惯用方法有规律可循,.,1,2,1.,已知数列,a,n,通项公式为,a,n,，其前,n,项和为,S,n,，若存在,M,Z,，满足对任意,n,N,*,，都有,S,n,0),，且,4,a,3,是,a,1,与,2,a,2等差中项,.,(1),求,a,n,通项公式；,解答,押题依据,错位相减法求和是高考重点和热点，本题先利用,a,n,，,S,n,关系求,a,n,，也是高考出题常见形式,.,1,2,押题依据,49/53,解,当,n,1,时，,S,1,a,(,S,1,a,1,1),，所以,a,1,a,，,当,n,2,时，,S,n,a,(,S,n,a,n,1),，,S,n,1,a,(,S,n,1,a,n,1,1),，,1,2,故,a,n,是首项,a,1,a,，公比为,a,等比数列，,所以,a,n,a,a,n,1,a,n,.,故,a,2,a,2,，,a,3,a,3,.,由,4,a,3,是,a,1,与,2,a,2,等差中项，可得,8,a,3,a,1,2,a,2,，,50/53,即,8,a,3,a,2,a,2,，,因为,a,0,，整理得,8,a,2,2,a,1,0,，,即,(2,a,1)(4,a,1),0,，,1,2,51/53,解答,1,2,52/53,1,2,所以,T,n,3,2,5,2,2,7,2,3,(2,n,1),2,n,1,(2,n,1),2,n,，,2,T,n,3,2,2,5,2,3,7,2,4,(2,n,1)2,n,(2,n,1),2,n,1,，,由,，得,T,n,3,2,2(2,2,2,3,2,n,),(2,n,1),2,n,1,2,2,n,2,(2,n,1),2,n,1,2,(2,n,1),2,n,1,，,所以,T,n,2,(2,n,1),2,n,1,.,53/53,