高考数学复习专题四数列推理与证明第2讲数列的求和问题文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,1/42,热点分类突破,真题押题精练,2/42,热点分类突破,3/42,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,.,4/42,例,1,(,届安徽省合肥市模拟,),已知等差数列,a,n,前,n,项和为,S,n,，且满足,S,4,24,，,S,7,63.,(1),求数列,a,n,通项公式；,解答,解,a,n,为等差数列，,5/42,解答,思维升华,(2),若,求数列,b,n,前,n,项和,T,n,.,6/42,解,2,2,n,1,(,1),n,(2,n,1),24,n,(,1),n,(2,n,1),，,7/42,8/42,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,.,9/42,(1),求证：数列,b,n,为等差数列；,则,b,n,1,b,n,2(,n,1),1,2,n,1,2.,所以数列,b,n,是以,1,为首项，,2,为公差等差数列,.,证实,10/42,(2),设,c,n,a,n,b,2,n,，求数列,c,n,前,n,项和,T,n,.,解答,11/42,解,由,(1),知，,b,2,n,4,n,1,，,则数列,b,2,n,是以,3,为首项，,4,为公差等差数列,.,12/42,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,.,13/42,例,2,(,河南省夏邑一高模拟,),已知,a,n,是等差数列，其前,n,项和为,S,n,，,b,n,是等比数列，且,a,1,b,1,2,，,a,4,b,4,27,，,S,4,b,4,10.,(1),求数列,a,n,与,b,n,通项公式；,解答,14/42,解,设等差数列,a,n,公差为,d,，等比数列,b,n,公比为,q,，,由,a,1,b,1,2,，得,a,4,2,3,d,，,b,4,2,q,3,，,S,4,8,6,d,，,故,a,n,3,n,1,，,b,n,2,n,(,n,N,*,).,15/42,(2),求,T,n,a,1,b,1,a,2,b,2,a,n,b,n,值,.,解,T,n,2,2,5,2,2,8,2,3,(3,n,1),2,n,，,2,T,n,2,2,2,5,2,3,8,2,4,(3,n,1),2,n,1,，,，得,T,n,2,2,3,2,2,3,2,3,3,2,n,(3,n,1),2,n,1,3,2,3,2,2,3,2,3,3,2,n,2,(3,n,1),2,n,1,(4,3,n,)2,n,1,8,，,T,n,8,(3,n,4)2,n,1,.,解答,思维升华,16/42,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,.,17/42,跟踪演练,2,(,江西省赣州市十四县,(,市,),联考,),等差数列,a,n,前,n,项和为,S,n,，已知,a,2,7,，,a,3,为整数，且,S,n,最大值为,S,5,.,(1),求,a,n,通项公式；,解答,解,由,a,2,7,，,a,3,为整数知，等差数列,a,n,公差,d,为整数,.,又,S,n,S,5,，故,a,5,0,，,a,6,0,，,所以,d,2,，,数列,a,n,通项公式为,a,n,11,2,n,(,n,N,*,).,18/42,(2),设,b,n,，求数列,b,n,前,n,项和,T,n,.,解答,19/42,20/42,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,.,21/42,例,3,(,届湖南省郴州市质量检测,),已知等差数列,a,n,前,n,项和为,S,n,(,n,N,*,),，,a,3,3,，且,S,n,a,n,a,n,1,，在等比数列,b,n,中，,b,1,2,，,b,3,a,15,1.,(1),求数列,a,n,及,b,n,通项公式；,解答,思维升华,22/42,解,S,n,a,n,a,n,1,，,a,3,3,，,a,1,a,1,a,2,，且,(,a,1,a,2,),a,2,a,3,3,a,2,，,a,2,，,a,1,a,2,a,3,3,，,数列,a,n,是等差数列，,a,1,a,3,2,a,2,，,即,2,a,2,a,1,3,，,由,得,a,1,1,，,a,2,2,，,a,n,n,，,2,，,b,1,4,，,b,3,16,，则,b,n,2,n,1,或,b,n,(,2),n,1,(,n,N,*,).,23/42,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,.,24/42,解答,思维升华,25/42,思维升华,惯用裂项公式,26/42,跟踪演练,3,(,吉林省吉林市普通高中调研,),已知等差数列,a,n,前,n,和为,S,n,，公差,d,0,，且,a,3,S,5,42,，,a,1,，,a,4,，,a,13,成等比数列,.,(1),求数列,a,n,通项公式；,解答,解,设数列,a,n,首项为,a,1,，因为等差数列,a,n,前,n,和为,S,n,，,a,3,S,5,42,，,a,1,，,a,4,，,a,13,成等比数列,.,又公差,d,0,，,所以,a,1,3,，,d,2,，,所以,a,n,a,1,(,n,1),d,2,n,1.,27/42,解答,则,T,n,b,1,b,2,b,3,b,n,28/42,真题押题精练,29/42,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,1,2,30/42,解析,设等差数列,a,n,公差为,d,，则,1,2,31/42,2.(,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,32/42,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,.,1,2,33/42,(2),求数列,a,2,n,b,2,n,1,前,n,项和,(,n,N,*,).,1,2,解答,34/42,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,，,1,2,35/42,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,38/42,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,，,39/42,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,，,1,2,40/42,解答,1,2,41/42,1,2,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,.,42/42,