高中数学专题数列求和问题市公开课一等奖市赛课获奖课件.pptx

专题,5,数列,第,24,练数列求和问题,题型分析,高考展望,数列求和是数列部分高考考察的两大重点之一，重要考察等差、等比数列的前n项和公式以及其他求和措施，尤其是错位相减法、裂项相消法是高考的热点内容，常与通项公式相结合考察，有时也与函数、方程、不等式等知识交汇，综合命题.,常,考题型精析,高考题,型精练,题型一分组转化法求和,题型二错位相减法求和,题型三裂项相消法求和,常考题型精析,题型一分组转化法求和,例1等比数列an中，a1，a2，a3分别是下表第一、二、三行中的某一种数，且a1，a2，a3中的任何两个数不在下表的同一列.,第一列,第二列,第三列,第一行,3,2,10,第二行,6,4,14,第三行,9,8,18,(1)求数列an的通项公式；,解当a13时，不合题意；,当a12时，当且仅当a26，a318时，符合题意；,当a110时，不合题意.,因此a12，a26，a318.因此公比q3.,故an23n1(nN*).,(2)若数列bn满足：bnan(1)nln an，求数列bn的前n项和Sn.,解由于bnan(1)nln an,23n1(1)nln(23n1),23n1(1)nln 2(n1)ln 3,23n1(1)n(ln 2ln 3)(1)nnln 3，,因此Sn2(133n1)111(1)n(ln 2ln 3)123(1)nnln 3.,因此当n为偶数时，,当,n,为奇数时，,点评分组求和常见的措施：(1)根据等差、等比数列分组，即分组后，每一组也许是等差数列或等比数列.(2)根据正号、负号分组.(3)根据数列的周期性分组.(4)根据奇数项、偶数项分组.,变式训练,1,已知等差数列,a,n,的前,n,项和为,S,n,，,n,N,*,，,a,3,5,，,S,10,100.,(1),求数列,a,n,的通项公式,；,解,设等差数列,a,n,的公差为,d,，,因此an2n1.,(2),设,b,n,2,a,n,2,n,，求数列,b,n,的前,n,项和,T,n,.,因此Tnb1b2bn,题型二错位相减法求和,例2(山东)设数列an的前n项和为Sn.已知2Sn3n3.,(1)求an的通项公式；,解由于2Sn3n3，因此2a133，故a13，,当n1时，2Sn13n13，,此时2an2Sn2Sn13n3n123n1，,即an3n1，,(2)若数列bn满足anbnlog3an，求bn的前n项和Tn.,解由于anbnlog3an，,当,n,1,时，,b,n,3,1,n,log,3,3,n,1,(,n,1)3,1,n,.,当,n,1,时，,T,n,b,1,b,2,b,3,b,n,因此3Tn1130231(n1)32n，,点评错位相减法的关注点：,(1)合用题型：等差数列an乘以等比数列bn对应项“anbn”型数列求和.,(2)环节：,求和时先乘以数列bn的公比.,把两个和的形式错位相减.,整顿成果形式.,变式训练,2,(,四川,),设等差数列,a,n,的公差为,d,，点,(,a,n,，,b,n,),在函数,f,(,x,),2,x,的图象上,(,n,N,*,).,(1),若,a,1,2,，点,(,a,8,4,b,7,),在函数,f,(,x,),的图象上，求数列,a,n,的前,n,项和,S,n,；,解,由已知，得,b,7,，,b,8,4,b,7,，,有,4,2,a,7,.,解得,d,a,8,a,7,2,.,(2),若,a,1,1,，函数,f,(,x,),的图象在点,(,a,2,，,b,2,),处的切线在,x,轴上的截距为,2,，,求数列,的前,n,项和,T,n解,函数,f,(,x,),2,x,在,(,a,2,，,b,2,),处的切线方程,为,y,2,(,2 ln,2)(,x,a,2,),，,因此da2a11，从而ann，bn2n.,a,2,a,2,题型三裂项相消法求和,例,3,在公差不为,0,的等差数列,a,n,中，,a,1,，,a,4,，,a,8,成等比数列,.,(1),已知数列,a,n,的前,10,项和为,45,，求数列,a,n,的通项公式；,解,设数列,a,n,的公差为,d,，,由,a,1,，,a,4,，,a,8,成等比数列可得,a,1,9,d,.,则数列,b,n,的前,n,项和为,故数列,a,n,的公差,d,1,或,1.,(2)运用裂项相消法求和时，应注意抵消后并不一定只剩第一项和最终一项，也也许前面剩两项，背面也剩两项.,变式训练,3,(,大纲全国,),等差数列,a,n,的前,n,项和为,S,n,，已知,a,1,10,，,a,2,为整数，且,S,n,S,4,.,(1),求,a,n,的通项公式；,解,由,a,1,10,，,a,2,为整数，,知等差数列,a,n,的公差,d,为整数,.,又,S,n,S,4,，故,a,4,0,，,a,5,0,，,于是,10,3,d,0,10,4,d,0,.,数列,a,n,的通项公式为,a,n,13,3,n,.,于是,T,n,b,1,b,2,b,n,高考题型精练,1.(,浙江,),已知,a,n,是等差数列，公差,d,不为零，前,n,项和是,S,n,，若,a,3,，,a,4,，,a,8,成等比数列，则,(,),A.,a,1,d,0,，,dS,4,0,B.,a,1,d,0,，,dS,4,0,C.,a,1,d,0,，,dS,4,0,D.,a,1,d,0,，,dS,4,0,1,2,3,4,5,6,7,8,9,10,11,12,解析,a,3,，,a,4,，,a,8,成等比数列，,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,答案,B,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,2.(,课标全国,),等差数列,a,n,的公差为,2,，若,a,2,，,a4,，,a,8,成等比数列，则,a,n,的前,n,项和,S,n,等于,(,),解析,由,a,2,，,a,4,，,a,8,成等比数列，得,即,(,a,1,6),2,(,a,1,2)(,a,1,14),，,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,a,1,2.,2,n,n,2,n,n,(,n,1).,答案,A,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,3.,若数列,a,n,的通项公式为,a,n,，,则其前,n,项和,S,n,为,(,),高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,因此Sna1a2an,高考题型精练,1,23,4,5,6,7,8,9,10,11,12,故选,D.,答案,D,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,答案,A,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,5.,设等差数列,a,n,的前,n,项和为,S,n,，,S,m,1,2,，,S,m,0,，,S,m,1,3,，则,m,等于,(,),A.3,B.4,C.5,D.6,解析,a,m,2,，,a,m,1,3,，故,d,1,，,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,由于amam15，,故amam12a1(2m1)d(m1)2m15，,即m5.,答案C,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,答案,B,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,7.,在等比数列,a,n,中，,a,1,3,，,a,4,81,，若数列,b,n,满足,b,n,log,3,a,n,，则,数列,的,前,n,项和,S,n,_.,解析,设等比数列,a,n,的公比为,q,，,因此ana1qn133n13n，故bnlog3ann，,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,8.,数列,a,n,满足,a,n,1,(,1),n,a,n,2,n,1,，则,a,n,的前,60,项和为,_.,解析,a,n,1,(,1),n,a,n,2,n,1,，,a,2,1,a,1,，,a,3,2,a,1,，,a,4,7,a,1,，,a,5,a,1,，,a,6,9,a,1,，,a,7,2,a,1,，,a,8,15,a,1,，,a,9,a,1,，,a,10,17,a,1,，,a,11,2,a,1,，,a,12,23,a,1,，,，,a,57,a,1,，,a58,113,a,1,，,a,59,2,a,1,，,a,60,119,a,1,，,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,a,1,a,2,a,60,(,a,1,a,2,a,3,a,4,),(,a,5,a,6,a,7,a,8,),(,a,57,a,58,a,59,a,60,),10,26,42,234,答案,1 830,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,9.在数列an中，a11，当n2时，其前n项和Sn满足,(1)求Sn的体现式；,a,n,S,n,S,n,1,(,n,2),，,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,即,2,S,n,1,S,n,S,n,1,S,n,，,由题意得,S,n,1,S,n,0,，,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,10.(,课标全国,),已知,a,n,是递增的等差数列，,a,2,，,a,4,是方程,x,2,5,x,6,0,的根,.,(1),求,a,n,的通项公式,；,解,方程,x,2,5,x,6,0,的两根为,2,3,，,由题意得,a,2,2,，,a,4,3.,设数列,a,n,的公差为,d,，则,a,4a,2,2,d,，,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,两式相减得,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,11,.(,天津,),已知数列,a,n,满足,a,n,2,qa,n,(,q,为实数，且,q,1),，,n,N,*,，,a,1,1,，,a,2,2,，且,a,2,a,3,，,a,3,a,4,，,a,4,a,5,成等差数列,.,(1),求,q,的值和,a,n,的通项公式,；,解,由,已知，,有,(a,3,a,4,),(,a,2,a,3,),(,a,4,a,5,),(,a,3,a,4,),，,即,a,4,a,2,a,5,a,3,，,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,因此a2(q1)a3(q1)，又由于q1，,故a3a22，由a3a1q，得q2.,由递推公式得,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,设,b,n,的前,n,项和为,S,n,，,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,上述两式相减得：,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,12.(,安徽,),设,n,N,*,，,x,n,是曲线,y,x,2,n,2,1,在点,(1,2),处的切线与,x,轴交点的横坐标,.,(1),求数列,x,n,的通项公式,；,解,y,(,x,2,n,2,1),(2,n,2),x,2,n,1,，曲线,y,x,2,n,2,1,在点,(1,2),处的切线斜率为,2,n,2,，,从而切线方程为,y,2,(2,n,2)(,x,1).,令,y,0,，,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,证明由题设和(1)中的计算成果知,高考题型精练,1,2,3,4,5,6,7,8,9,10,11,12,