高考数学复习第六章数列与数学归纳法6.4数列求和市赛课公开课一等奖省名师优质课获奖课件.pptx

6.4,数列求和,1/68,基础知识自主学习,课时训练,题型分类深度剖析,内容索引,2/68,基础知识自主学习,3/68,知识梳理,4/68,3.,一些常见数列前,n,项和公式,n,2,n,(,n,1),5/68,数列求和惯用方法,(1),公式法,等差、等比数列或可化为等差、等比数列可直接使用公式求和,.,(2),分组转化法,把数列每一项分成两项或几项，使其转化为几个等差、等比数列，再求解,.,知识拓展,6/68,(3),裂项相消法,把数列通项拆成两项之差求和，正负相消剩下首尾若干项,.,常见裂项公式,(4),倒序相加法,把数列分别正着写和倒着写再相加，即等差数列求和公式推导过程推广,.,7/68,(5),错位相减法,主要用于一个等差数列与一个等比数列对应项相乘所得数列求和，即等比数列求和公式推导过程推广,.,(6),并项求和法,一个数列前,n,项和中，可两两结合求解，则称之为并项求和,.,形如,a,n,(,1),n,f,(,n,),类型，可采取两项合并求解,.,比如，,S,n,100,2,99,2,98,2,97,2,2,2,1,2,(100,99),(98,97),(2,1),5 050.,8/68,判断以下结论是否正确,(,请在括号中打,“”,或,“”,),思索辨析,(3),求,S,n,a,2,a,2,3,a,3,na,n,之和时，只要把上式等号两边同时乘以,a,即可依据错位相减法求得,.(,),9/68,(5),推导等差数列求和公式方法叫做倒序求和法，利用此法可求得,sin,2,1,sin,2,2,sin,2,3,sin,2,88,sin,2,89,44.5.(,),10/68,1.(,潍坊模拟,),设,a,n,是公差不为,0,等差数列，,a,1,2,，且,a,1,，,a,3,，,a,6,成等比数列，则,a,n,前,n,项和,S,n,等于,答案,解析,考点自测,11/68,设等差数列公差为,d,，则,a,1,2,，,a,3,2,2,d,，,a,6,2,5,d,.,即,(2,2,d,),2,2(2,5,d,),，整理得,2,d,2,d,0.,12/68,答案,解析,13/68,3.,数列,a,n,通项公式为,a,n,(,1),n,1,(4,n,3),，则它前,100,项之和,S,100,等于,A.200 B.,200 C.400 D.,400,答案,解析,S,100,(4,1,3),(4,2,3),(4,3,3),(4,100,3),4,(1,2),(3,4),(99,100),4,(,50),200.,14/68,1 008,答案,解析,观察此数列规律以下：,a,1,0,，,a,2,2,，,a,3,0,，,a,4,4.,故,S,4,a,1,a,2,a,3,a,4,2.,a,5,0,，,a,6,6,，,a,7,0,，,a,8,8,，,故,a,5,a,6,a,7,a,8,2,，,周期,T,4.,15/68,题型分类深度剖析,16/68,题型一分组转化法求和,当,n,1,时，,a,1,S,1,1,；,a,1,也满足,a,n,n,，,故数列,a,n,通项公式为,a,n,n,.,解答,17/68,(2),设,b,n,2,a,n,(,1),n,a,n,，求数列,b,n,前,2,n,项和,.,解答,由,(1),知,a,n,n,，故,b,n,2,n,(,1),n,n,.,记数列,b,n,前,2,n,项和为,T,2,n,，,则,T,2,n,(2,1,2,2,2,2,n,),(,1,2,3,4,2,n,).,记,A,2,1,2,2,2,2,n,，,B,1,2,3,4,2,n,，,故数列,b,n,前,2,n,项和,T,2,n,A,B,2,2,n,1,n,2.,18/68,由,(1),知,b,n,2,n,(,1),n,n,.,引申探究,例,1(2),中，求数列,b,n,前,n,项和,T,n,.,解答,19/68,分组转化法求和常见类型,(1),若,a,n,b,n,c,n,，且,b,n,，,c,n,为等差或等比数列，可采取分组求和法求,a,n,前,n,项和,.,思维升华,提醒：,一些数列求和是将数列转化为若干个可求和新数列和或差，从而求得原数列和，注意在含有字母数列中对字母讨论,.,20/68,S,n,2(1,3,3,n,1,),1,1,1,(,1)n,(ln 2,ln 3),1,2,3,(,1),n,n,ln 3,，,跟踪训练,1,已知数列,a,n,通项公式是,a,n,23,n,1,(,1),n,(ln 2,ln 3),(,1),n,n,ln 3,，求其前,n,项和,S,n,.,解答,21/68,题型二错位相减法求和,例,2,(,山东,),已知数列,a,n,前,n,项和,S,n,3,n,2,8,n,，,b,n,是等差数列，且,a,n,b,n,b,n,1,.,(1),求数列,b,n,通项公式；,解答,由题意知，当,n,2,时，,a,n,S,n,S,n,1,6,n,5,，,当,n,1,时，,a,1,S,1,11,，满足上式，所以,a,n,6,n,5.,22/68,解答,又,T,n,c,1,c,2,c,n,，,得,T,n,3,22,2,32,3,(,n,1)2,n,1,，,2,T,n,322,3,32,4,(,n,1)2,n,2,.,所以,T,n,3,n,2,n,2,.,23/68,错位相减法求和时注意点,(1),要善于识别题目类型，尤其是等比数列公比为负数情形；,(2),在写出,“,S,n,”,与,“,qS,n,”,表示式时应尤其注意将两式,“,错项对齐,”,方便下一步准确写出,“,S,n,qS,n,”,表示式；,(3),在应用错位相减法求和时，若等比数列公比为参数，应分公比等于,1,和不等于,1,两种情况求解,.,思维升华,24/68,跟踪训练,2,设等差数列,a,n,公差为,d,，前,n,项和为,S,n,，等比数列,b,n,公比为,q,，已知,b,1,a,1,，,b,2,2,，,q,d,，,S,10,100.,(1),求数列,a,n,，,b,n,通项公式；,解答,25/68,解答,由,d,1,，知,a,n,2,n,1,，,b,n,2,n,1,，,26/68,题型三裂项相消法求和,例,3,(,课标全国,),S,n,为数列,a,n,前,n,项和,.,已知,a,n,0,，,2,a,n,4,S,n,3.,(1),求,a,n,通项公式；,解答,27/68,由,a,n,0,，可得,a,n,1,a,n,2.,所以,a,n,是首项为,3,，公差为,2,等差数列，通项公式为,a,n,2,n,1.,28/68,解答,29/68,答案,解析,30/68,则,f,(,x,),31/68,思维升华,32/68,解答,33/68,即,2,S,n,1,S,n,S,n,1,S,n,，,由题意得,S,n,1,S,n,0,，,34/68,解答,35/68,题型四数列求和综合应用,(1),求数列,a,n,通项公式,a,n,；,解答,因为,a,n,是正项数列，所以,S,n,10.,所以,S,n,n,2,n,(,n,N,*,).,n,2,时，,a,n,S,n,S,n,1,2,n,，,n,1,时，,a,1,S,1,2,适合上式,.,所以,a,n,2,n,(,n,N,*,).,36/68,证实,37/68,数列和其它知识综合，可先确定数列项递推关系，求出数列通项或前,n,项和；也可经过放缩法适当变形后再求和，进而证实一些不等式,.,思维升华,38/68,(1),若,t,0,，求数列,a,n,通项公式；,解答,39/68,又,a,1,1,，所以,a,n,0,，,从而,ln,a,n,1,2ln,a,n,ln 2,，,所以,ln,a,n,1,ln 2,2(ln,a,n,ln 2),，,40/68,证实,41/68,所以,a,n,为递减数列,.,42/68,所以命题得证,.,43/68,四审结构定方案,审题路线图系列,规范解答,审题路线图,44/68,返回,45/68,当,n,1,时，上式也成立,.,46/68,T,n,4.14,分,返回,47/68,课时训练,48/68,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,49/68,2.(,西安模拟,),设等比数列,a,n,前,n,项和为,S,n,，已知,a,1,2 016,，且,a,n,2,a,n,1,a,n,2,0(,n,N,*,),，则,S,2 016,等于,A.0 B.2 016,C.2 015 D.2 014,答案,解析,a,n,2,a,n,1,a,n,2,0(,n,N,*,),，,a,n,2,a,n,q,a,n,q,2,0,，,q,为等比数列,a,n,公比，,即,q,2,2,q,1,0,，,q,1.,a,n,(,1),n,1,2 016,，,S,2 016,(,a,1,a,2,),(,a,3,a,4,),(,a,2 015,a,2 016,),0.,1,2,3,4,5,6,7,8,9,10,11,12,13,50/68,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,51/68,4.,在,数列,a,n,中，,若,a,n,1,(,1),n,a,n,2,n,1,，则数列,a,n,前,12,项和等于,A.76 B.78,C.80 D.82,答案,解析,由已知,a,n,1,(,1),n,a,n,2,n,1,，,得,an,2,(,1),n,1,a,n,1,2,n,1,，得,a,n,2,a,n,(,1),n,(2,n,1),(2,n,1),，,取,n,1,，,5,，,9,及,n,2,，,6,，,10,，,结果相加可得,S,12,a,1,a,2,a,3,a,4,a,11,a,12,78.,故选,B.,1,2,3,4,5,6,7,8,9,10,11,12,13,52/68,由题意，得,a,1,a,2,a,3,a,100,1,2,2,2,2,2,3,2,3,2,4,2,4,2,5,2,99,2,100,2,100,2,101,2,(1,2),(3,2),(4,3),(99,100),(101,100),(1,2,99,100),(2,3,100,101),50,101,50,103,100.,故选,B.,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,53/68,6.,设数列,a,n,通项公式为,a,n,2,n,7,，则,|,a,1,|,|,a,2,|,|,a,15,|,等于,A.153 B.210,C.135 D.120,答案,解析,从第,4,项开始大于,0,，,1,2,3,4,5,6,7,8,9,10,11,12,13,54/68,答案,解析,120,1,2,3,4,5,6,7,8,9,10,11,12,13,55/68,8.,在等差数列,a,n,中，,a,1,0,，,a,10,a,11,0,，若此数列前,10,项和,S,10,36,，前,18,项和,S,18,12,，则数列,|,a,n,|,前,18,项和,T,18,值是,_.,答案,解析,由,a,1,0,，,a,10,a,11,0,可知,d,0,，,a,10,0,，,a,11,0,，,T,18,a,1,a,10,a,11,a,18,S,10,(,S,18,S,10,),60.,60,1,2,3,4,5,6,7,8,9,10,11,12,13,56/68,答案,解析,则数列前,n,项和为,_.,1,2,3,4,5,6,7,8,9,10,11,12,13,57/68,1,2,3,4,5,6,7,8,9,10,1112,13,58/68,9,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,59/68,又,a,n,为正项数列，,a,n,1,a,n,1,0,，,即,a,n,1,a,n,1.,1,2,3,4,5,6,7,8,9,10,11,12,13,60/68,数列,a,n,是以,1,为首项，,1,为公差等差数列,.,a,n,n,，,T,1,，,T,2,，,T,3,，,，,T,100,中有理数个数为,9.,1,2,3,4,5,6,7,8,9,10,11,12,13,61/68,11.,已知数列,a,n,中，,a,1,3,，,a,2,5,，且,a,n,1,是等比数列,.,(1),求数列,a,n,通项公式；,解答,a,n,1,22,n,1,2,n,，,a,n,2,n,1.,1,2,3,4,5,6,7,8,9,10,11,12,13,62/68,b,n,na,n,n,2,n,n,，,故,T,n,b,1,b,2,b,3,b,n,(2,2,2,2,3,2,3,n,2,n,),(1,2,3,n,).,令,T,2,2,2,2,3,2,3,n,2,n,，,则,2,T,2,2,2,2,3,3,2,4,n,2,n,1,.,T,2(1,2,n,),n,2,n,1,2,(,n,1)2,n,1,.,(2),若,b,n,na,n,，求数列,b,n,前,n,项和,T,n,.,解答,1,2,3,4,56,7,8,9,10,11,12,13,63/68,解答,(1),求,a,n,通项公式；,设数列,a,n,公比为,q,.,解得,q,2,或,q,1.,所以,a,n,2,n,1,.,1,2,3,4,5,6,7,8,9,10,11,12,13,64/68,解答,1,2,3,4,5,6,7,8,9,10,11,12,13,65/68,解答,1,2,3,4,5,6,7,8,9,10,11,12,13,66/68,证实,1,2,3,4,5,6,7,8,9,10,11,12,13,67/68,由,c,n,1,c,n,2,n,1,，,得当,n,2,时，,c,n,c,1,(,c,2,c,1,),(,c,3,c,2,),(,c,n,c,n,1,),0,3,5,(2,n,1),n,2,1,(,n,1)(,n,1),，,原式得证,.,1,2,3,4,5,6,7,8,9,10,11,12,13,68/68,