(课标专用)天津市高考数学二轮复习 专题四 数列 4.2 数列的通项与求和课件.pptx

,专题四,4.2,数列的通项与求和,#,高频考点,探究突破,核心归纳,预测演练,4.2,数列的通项与求和,-,2,-,突破点一,突破点二,突破点三,由数列的递推关系求通项,【例,1,】,根据下列条件,确定数列,a,n,的通项公式,:,(1),数列,a,n,满足,a,1,+,3,a,2,+,+,(2,n-,1),a,n,=,2,n,;,(3),a,1,=,1,a,x+,1,=,3,a,n,+,2,.,分析推理,(1),根据式子结构特征,把,(2,n-,1),a,n,看作一个整体,则该问题就看作已知和,S,n,求通项的问题,根据项与和的关系式求解即可,;(2),根据递推关系以及对数运算,可以利用累加法求其通项,;(3),因为递推关系中两项的系数不同,所以应该通过变形构造等比数列求解通项,.,-,3,-,突破点一,突破点二,突破点三,解,:,(1),a,1,+,3,a,2,+,+,(2,n-,1),a,n,=,2,n,故当,n,2,时,a,1,+,3,a,2,+,+,(2,n-,3),=,2(,n-,1),两式相减得,(2,n-,1),a,n,=,2,-,4,-,突破点一,突破点二,突破点三,又,a,1,=,2,适合上式,故,a,n,=,2,+,ln,n,(,n,N,*,),.,-,5,-,突破点一,突破点二,突破点三,(3),方法一,(,直接变形,),由,a,n+,1,=,3,a,n,+,2,得,a,n+,1,+,1,=,3(,a,n,+,1),.,a,1,=,1,知,a,1,+,1,=,2,a,n,+,10,数列,a,n,+,1,是以,2,为首项,以,3,为公比的等比数列,.,则,a,n,+,1,=,2,3,n-,1,故,a,n,=,2,3,n-,1,-,1,.,方法二,(,待定系数法,),由已知,设,a,n+,1,+t=,3(,a,n,+t,),则整理得,a,n+,1,=,3,a,n,+,2,t.,由已知可得,2,t=,2,解得,t=,1,.,a,n+,1,+,1,=,3(,a,n,+,1),.,下同方法一,.,-,6,-,突破点一,突破点二,突破点三,【例,1,】,(3),中,若已知,a,n+,1,=,3,a,n,+,2,n-,1,呢,?,解法一,(,直接变形,),由,a,n+,1,=,3,a,n,+,2,n-,1,得,a,n+,1,+,(,n+,1),=,3(,a,n,+n,),.,令,b,n,=a,n,+n,则有,b,n+,1,=,3,b,n,.,-,7,-,突破点一,突破点二,突破点三,解法二,(,待定系数法,),因为,a,n+,1,与,a,n,的系数不相等,故可构造等比数列,.,设,a,n+,1,+,k,(,n+,1),+t,=,3(,a,n,+kn+t,),整理得,a,n+,1,=,3,a,n,+,2,kn+,2,t-k.,由已知,a,n+,1,=,3,a,n,+,2,n-,1,所以,a,n+,1,+,(,n+,1),=,3(,a,n,+n,),.,下同解法一,.,-,8,-,突破点一,突破点二,突破点三,规律方法,1,.,由递推关系求数列的通项的基本思想是转化,常用的方法,:,(1),a,n+,1,-a,n,=f,(,n,),型,采用迭加法,;,(3),a,n+,1,=pa,n,+q,(,p,0,p,1),型,转化为等比数列解决,;,(4),a,n+,1,=,(,a,n,0,p,q,为非零常数,),型,可用倒数法转化为等差数列解决,.,2,.,已知,S,n,求,a,n,的三个步骤,:,(1),先利用,a,1,=S,1,求出,a,1,;,(2),用,n-,1,替换,S,n,中的,n,得到一个新的关系,利用,a,n,=S,n,-,(,n,2),便可求出当,n,2,时,a,n,的表达式,;,(3),注意检验,n=,1,时的表达式是否可以与,n,2,的表达式合并,.,-,9,-,突破点一,突破点二,突破点三,即时巩固,1,根据下列条件,确定数列,a,n,的通项公式,:,(3),S,n,=,2,a,n,+n,.,-,10,-,突破点一,突破点二,突破点三,(3),S,n,=,2,a,n,+n,当,n=,1,时,a,1,=S,1,=,2,a,1,+,1,即,a,1,=-,1;,当,n,2,时,a,n,=S,n,-S,n-,1,=,(2,a,n,+n,),-,2,a,n-,1,+,(,n-,1),=,2,a,n,-,2,a,n-,1,+,1,即,a,n,=,2,a,n-,1,-,1,数列,a,n,-,1,为首项,a,1,-,1,=-,2,公比,q=,2,的等比数列,a,n,-,1,=-,22,n-,1,即,a,n,=,1,-,2,n,.,-,11,-,突破点一,突破点二,突破点三,裂项求和,法,(1),求,a,n,的通项公式,;,分析推理,(1),首先令,n=,1,求出首项,然后根据,S,n,与,a,n,的关系,将已知转化为数列的项之间的关系,化简后判断数列的性质,进而求其通项,;(2),根据第,(1),问所求,写出,b,n,的表达式,然后将其裂成两项之差,利用裂项相消法求和,.,-,12,-,突破点一,突破点二,突破点三,所以,a,n,是首项为,3,公差为,2,的等差数列,通项公式为,a,n,=,2,n+,1,.,-,13,-,突破点一,突破点二,突破点三,规律方法,裂项相消法的基本思想就是把通项,a,n,分拆成,a,n,=b,n+k,-b,n,(,k,N,*,),的形式,从而达到在求和时绝大多数项相消的目的,在解题时要善于根据这个基本思想变换数列,a,n,的通项公式,使之符合裂项相消的条件,.,-,14,-,突破点一,突破点二,突破点三,即时巩固,2,(2019,湖南岳阳二模,),已知数列,a,n,a,1,=,3,且,na,n+,1,-a,n,=,na,n,n,N,*,.,(1),求数列,a,n,的通项公式,;,(2),记,S,n,为数列,a,n,的前,n,项和,求,数列,的,前,n,项和,T,n,.,得,a,n,=,3,n,所以数列,a,n,的通项公式为,a,n,=,3,n,n,N,*,.,-,15,-,突破点一,突破点二,突破点三,-,16,-,突破点一,突破点二,突破点三,错位相减法求和,【例,3,】,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,的等比数列,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,的通项公式,;,(2),求数列,a,2,n,b,2,n-,1,的前,n,项和,(,n,N,*,),.,分析推理,(1),等比数列,b,n,已知首项,故可直接利用已知,“,b,2,+b,3,=,12”,列出公比所满足的方程求解,即可得到其通项公式,;,然后代入已知列出方程组求出等差数列,a,n,的首项与公差,进而求其通项,.,(2),首先根据第,(1),问写出数列,a,2,n,b,2,n-,1,的通项公式,根据等差、等比数列的性质可知,a,2,n,为等差数列,b,2,n-,1,为等比数列,故应利用错位相减法求和,.,-,17,-,突破点一,突破点二,突破点三,解,:,(1),设等差数列,a,n,的公差为,d,等比数列,b,n,的公比为,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,的通项公式为,a,n,=,3,n-,2,数列,b,n,的通项公式为,b,n,=,2,n,.,-,18,-,突破点一,突破点二,突破点三,(2),设数列,a,2,n,b,2,n-,1,的前,n,项和为,T,n,由,a,2,n,=,6,n-,2,b,2,n-,1,=,24,n-,1,有,a,2,n,b,2,n-,1,=,(3,n-,1)4,n,故,T,n,=,24,+,54,2,+,84,3,+,+,(3,n-,1)4,n,4,T,n,=,24,2,+,54,3,+,84,4,+,+,(3,n-,4)4,n,+,(3,n-,1)4,n+,1,上述两式相减,得,-,3,T,n,=,24,+,34,2,+,34,3,+,+,34,n,-,(3,n-,1)4,n+,1,-,19,-,突破点一,突破点二,突破点三,规律方法,错位相减法适用于求数列,a,n,b,n,的前,n,项和,其中,a,n,为等差数列,b,n,为等比数列,;,所谓,“,错位,”,就是要找,“,同类项,”,相减,.,要注意的是相减后得到部分等比数列的和,此时一定要查清其项数,.,-,20,-,突破点一,突破点二,突破点三,即时巩固,3,(2019,江西上饶二模,),设数列,a,n,的前,n,项和为,S,n,.,已知,a,1,=,1,a,n+,1,=,3,S,n,+,1,n,N,*,.,(1),求数列,a,n,的通项公式,;,(2),记,T,n,为数列,na,n,的前,n,项和,求,T,n,.,解,:,(1),由题意,a,n+,1,=,3,S,n,+,1,则当,n,2,时,a,n,=,3,S,n-,1,+,1,.,两式相减,得,a,n+,1,=,4,a,n,(,n,2),.,所以数列,a,n,是以首项为,1,公比为,4,的等比数列,所以数列,a,n,的通项公式是,a,n,=,4,n-,1,(,n,N,*,),.,-,21,-,突破点一,突破点二,突破点三,(2),因为,T,n,=a,1,+,2,a,2,+,3,a,3,+,+na,n,=,1,+,24,+,34,2,+,+n,4,n-,1,所以,4,T,n,=,41,+,24,2,+,34,3,+,+,(,n-,1),4,n-,1,+n,4,n,-,22,-,核心归纳,预测演练,-,23,-,核心归纳,预测演练,1,.,(2019,山东泰安上学期期末,),已知数列,a,n,中,a,1,=,1,a,n+,1,=,2,a,n,+,1(,n,N,*,),S,n,为其前,n,项和,则,S,5,的值为,(,),A.57B.61,C.62,D.63,A,解析,:,由条件可得,a,1,=,1,a,2,=,2,a,1,+,1,=,3,a,3,=,2,a,2,+,1,=,7,a,4,=,2,a,3,+,1,=,15,a,5,=,2,a,4,+,1,=,31,所以,S,5,=a,1,+a,2,+a,3,+a,4,+a,5,=,1,+,3,+,7,+,15,+,31,=,57,故选,A,.,-,24,-,核心归纳,预测演练,a,n,的通项公式为,(,),A.,a,n,=n,B.,a,n,=n,2,B,a,n,=n,2,n,N,*,.,故选,B,.,-,25,-,核心归纳,预测演练,3,.,已知数列,a,n,的各项都是正数,其前,n,项和,S,n,满足,2,S,n,=a,n,+,n,N,*,则数列,a,n,的通项公式为,.,解析,:,因为数列,a,n,的各项都是正数,其前,n,项和,S,n,满足,-,26,-,核心归纳,预测演练,4,.,(2019,山东淄博实验中学、淄博五中一诊,),已知等差数列,a,n,的公差,d,0,其前,n,项和为,S,n,且,S,5,=,20,a,3,a,5,a,8,成等比数列,.,(1),求数列,a,n,的通项公式,;,所以,(,a,1,+,4,d,),2,=,(,a,1,+,2,d,)(,a,1,+,7,d,),化简得,a,1,=,2,d.,联立,和,得,a,1,=,2,d=,1,所以,a,n,=n+,1,.,-,27,-,核心归纳,预测演练,-,28,-,核心归纳,预测演练,5,.,(2019,江西上饶二模,),已知首项为,1,的等比数列,a,n,满足,a,2,+a,4,=,3(,a,1,+a,3,),等差数列,b,n,满足,b,1,=a,2,b,4,=a,3,数列,b,n,的前,n,项和为,S,n,.,(1),求数列,a,n,b,n,的通项公式,;,解,:,(1),设,a,n,的通项公式为,a,n,=a,1,q,n-,1,a,1,q+a,1,q,3,=,3(,a,1,+a,1,q,2,),q=,3,.,a,1,=,1,a,n,=,3,n-,1,.,a,2,=,3,a,3,=,9,b,1,=,3,b,4,=,9,.,b,n,=,2,n+,1,.,-,29,-,核心归纳,预测演练,综上可知,c,n,=,(2,n+,1),3,n-,1,n,N,*,.,T,n,=,33,0,+,53,1,+,+,(2,n+,1)3,n-,1,3,T,n,=,33,1,+,53,2,+,+,(2,n+,1)3,n,两式相减,得,-,2,T,n,=,3,-,(2,n+,1)3,n,+,23,1,+,23,2,+,+,23,n-,1,T,n,=n,3,n,.,-,30,-,