高考数学复习第十章计数原理10.3二项式定理理市赛课公开课一等奖省名师优质课获奖课件.pptx

10.3,二项式定理,1/61,基础知识自主学习,课时作业,题型分类深度剖析,内容索引,2/61,基础知识自主学习,3/61,1.,二项式定理,知识梳理,二项式定理,(,a,b,),n,_,_(,n,N,*,),二项展开式,通项公式,T,r,1,_,，它表示第,项,二项式系数,二项展开式中各项系数 (r0,1,2，n),r,1,a,n,r,b,r,4/61,2.,二项式系数性质,1,2,n,1,5/61,二项展开式形式上特点,(1),项数为,.,(2),各项次数都等于二项式幂指数,n,，即,a,与,b,指数和为,n,.,(3),字母,a,按,排列，从第一项开始，次数由,n,逐项减,1,直到零字母,b,按,排列，从第一项起，次数由零逐项增,1,直到,n,.,知识拓展,n,1,降幂,升幂,6/61,判断以下结论是否正确,(,请在括号中打,“”,或,“”,),(1),a,n,r,b,r,是二项展开式第,r,项,.(,),(2),二项展开式中，系数最大项为中间一项或中间两项,.(,),(3)(,a,b,),n,展开式中某一项二项式系数与,a,，,b,无关,.(,),(4),在,(1,x,),9,展开式中系数最大项是第五、第六两项,.(,),(5),若,(3,x,1),7,a,7,x,7,a,6,x,6,a,1,x,a,0,，则,a,7,a,6,a,1,值为,128.(,),思索辨析,7/61,考点自测,(,x,y,),n,展开式中第,m,项系数为,1.(,教材改编,)(,x,y,),n,二项展开式中，第,m,项系数是,_.,答案,解析,8/61,由题意可知，含,x,4,项为,15,x,4,.,2.(,四川改编,),设,i,为虚数单位，则,(,x,i),6,展开式中含,x,4,项为,_.,答案,解析,15,x,4,9/61,63,答案,解析,10/61,4.(,苏州模拟,)(1,x,),8,(1,y,),4,展开式中,x,2,y,2,系数是,_.,答案,解析,168,11/61,题型分类深度剖析,12/61,题型一二项展开式,命题点,1,求二项展开式中特定项或指定项系数,例,1,(1)(,全国乙卷,)(2,x,),5,展开式中，,x,3,系数是,_.(,用数字填写答案,),10,答案,解析,x,3,系数是,10.,13/61,(2)(,课标全国,改编,)(,x,2,x,y,),5,展开式中，,x,5,y,2,系数为,_.,答案,解析,30,方法一利用二项展开式通项公式求解,.,(,x,2,x,y,),5,(,x,2,x,),y,5,，,方法二利用组合知识求解,.,14/61,命题点,2,已知二项展开式某项系数求参数,例,2,(1)(,课标全国,)(,a,x,)(1,x,),4,展开式中,x,奇数次幂项系数之和为,32,，则,a,_.,3,答案,解析,设,(,a,x)(1,x,),4,a,0,a,1,x,a,2,x,2,a,3,x,3,a,4,x,4,a,5,x,5,，,令,x,1,，得,16(,a,1),a,0,a,1,a,2,a,3,a,4,a,5,，,令,x,1,，得,0,a,0,a,1,a,2,a,3,a,4,a,5,.,，得,16(,a,1),2(,a,1,a,3,a,5,),，,即展开式中,x,奇数次幂项系数之和为,a,1,a,3,a,5,8(,a,1),，,所以,8(,a,1),32,，解得,a,3.,15/61,答案,解析,2,16/61,求二项展开式中特定项，普通是利用通项公式进行，化简通项公式后，令字母指数符合要求,(,求常数项时，指数为零；求有理项时，指数为整数等,),，解出项数,r,1,，代回通项公式即可,.,思维升华,17/61,跟踪训练,1,(1)(,连云港模拟,)(,x,)(1,),4,展开式中,x,系数是,_.,答案,解析,3,18/61,(2)(,x,a,),10,展开式中，,x,7,系数为,15,，则,a,_.(,用数字填写答案,),答案,解析,19/61,题型二二项式系数和或各项系数和问题,例,3,在,(2,x,3,y,),10,展开式中，求：,(1),二项式系数和；,解答,设,(2,x,3,y,),10,a,0,x,10,a,1,x,9,y,a,2,x,8,y,2,a,10,y,10,，,(*),各项系数和为,a,0,a,1,a,10,，奇数项系数和为,a,0,a,2,a,10,，偶数项系数和为,a,1,a,3,a,5,a,9,，,x,奇次项系数和为,a,1,a,3,a,5,a,9,，,x,偶次项系数和为,a,0,a,2,a,4,a,10,.,因为,(*),是恒等式，故可用,“,赋值法,”,求出相关系数和,.,20/61,(2),各项系数和；,解答,令,x,y,1,，各项系数和为,(2,3),10,(,1),10,1.,(3),奇数项二项式系数和与偶数项二项式系数和；,解答,21/61,(4),奇数项系数和与偶数项系数和；,解答,22/61,令,x,y,1,，得到,a,0,a,1,a,2,a,10,1,，,令,x,1,，,y,1(,或,x,1,，,y,1),，,得,a,0,a,1,a,2,a,3,a,10,5,10,，,得,2(,a,0,a,2,a,10,),1,5,10,，,得,2(,a,1,a,3,a,9,),1,5,10,，,23/61,(5),x,奇次项系数和与,x,偶次项系数和,.,解答,24/61,(1),“,赋值法,”,普遍适合用于恒等式，是一个主要方法，对形如,(,ax,b,),n,，,(,ax,2,bx,c,),m,(,a,，,b,R,),式子求其展开式各项系数之和，惯用赋值法，只需令,x,1,即可；对形如,(,ax,by,),n,(,a,，,b,R,),式子求其展开式各项系数之和，只需令,x,y,1,即可,.,(2),若,f,(,x,),a,0,a,1,x,a,2,x,2,a,n,x,n,，则,f,(,x,),展开式中各项系数之和为,f,(1),，奇数项系数之和为,a,0,a,2,a,4,，偶数项系数之和为,a,1,a,3,a,5,.,思维升华,25/61,跟踪训练,2,(1)(,淮安,月考,),设,m,为正整数，,(,x,y,),2,m,展开式二项式系数最大值为,a,，,(,x,y,),2,m,1,展开式二项式系数最大值为,b,，若,13,a,7,b,，则,m,_.,答案,解析,6,26/61,经检验符合题意,.,27/61,当,x,0,时，左边,1,，右边,a,0,，,a,0,1.,解答,28/61,题型三二项式定理应用,例,4,(1),设,a,Z,且,0,a,13,，若,51,2 016,a,能被,13,整除，则,a,_.,答案,解析,12,29/61,(2)1.02,8,近似值是,_.(,准确到小数点后三位,),答案,解析,1.172,30/61,(1),整除问题和求近似值是二项式定理中两类常见应用问题，整除问题中要关注展开式最终几项，而求近似值则应关注展开式前几项,.,(2),二项式定理应用基本思绪是正用或逆用二项式定理，注意选择适当形式,.,思维升华,31/61,1,前,10,项均能被,88,整除，,余数是,1.,答案,解析,32/61,原式,46,n,5,n,a,4(5,1),n,5,n,a,(2),已知,2,n,2,3,n,5,n,a,能被,25,整除，求正整数,a,最小值,.,显然正整数,a,最小值为,4.,解答,33/61,现场纠错,纠错心得,和二项展开式相关问题，要分清所求是展开式中项系数还是二项式系数，是系数和还是二项式系数和,.,错解展示,典例,(1)(,江苏镇江中学质检,),若,(,),n,展开式各项系数绝对值之和为,1 024,，则展开式中含,x,项系数为,_.,(2),已知,(,x,m,),7,a,0,a,1,x,a,2,x,2,a,7,x,7,展开式中,x,4,系数是,35,，则,a,1,a,2,a,7,_.,二项展开式系数与二项式系数,现场纠错系列,13,34/61,答案,(1)5,(2)2,7,1,返回,35/61,故展开式中含,x,项系数为,15.,36/61,(2),(,x,m,),7,a,0,a,1,x,a,2,x,2,a,7,x,7,，,令,x,0,，,a,0,(,m,),7,.,m,1.,a,0,(,m,),7,1.,在,(,x,m,),7,a,0,a,1,x,a,2,x,2,a,7,x,7,中，,令,x,1,，得,0,1,a,1,a,2,a,7,，,即,a,1,a,2,a,3,a,7,1.,答案,(1),15,(2)1,返回,37/61,课时作业,38/61,1在,x,2,(1,x,),6,展开式中，含,x,4,项系数为,_.,答案,解析,15,1,2,3,4,5,6,7,8,9,10,11,12,13,14,39/61,2.(,湖南改编,),已知,展开式中含,项系数为,30,，则,a,_.,答案,解析,6,1,2,3,4,5,6,7,8,9,10,11,12,13,14,40/61,3.(4,x,2,x,),6,(,x,R,),展开式中常数项是,_.,答案,解析,15,12,x,3,rx,0,恒成立，,r,4,，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,41/61,4.(,湖北改编,),已知,(1,x,),n,展开式中第,4,项与第,8,项二项式系数相等，则奇数项二项式系数和为,_.,答案,解析,512,则奇数项二项式系数和为,2,n,1,2,9,512.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,42/61,5.,若在,(,x,1),4,(,ax,1),展开式中，,x,4,系数为,15,，则,a,值为,_.,4,答案,解析,(,x,1),4,(,ax,1),(,x,4,4,x,3,6,x,2,4,x,1)(,ax,1),，,x,4,系数为,4,a,1,15,，,a,4.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,43/61,6.,若,(1,x,),(1,x,),2,(1,x,),n,a,0,a,1,(1,x,),a,2,(1,x,),2,a,n,(1,x,),n,，则,a,0,a,1,a,2,a,3,(,1),n,a,n,_.,答案,解析,在展开式中，令,x,2,，得,3,3,2,3,3,3,n,a,0,a,1,a,2,a,3,(,1),n,a,n,，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,44/61,7.(,扬州模拟,),已知,(1,2,x,),8,展开式二项式系数最大值为,a,，系数最大值为,b,，则,_.,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,45/61,8.(,北京,),在,(1,2,x,),6,展开式中，,x,2,系数为,_.(,用数字作答,),答案,解析,即,x,2,系数为,60.,60,1,2,3,4,5,6,7,8,9,10,11,12,13,14,46/61,答案,解析,9.(,天津,),8,展开式中,x,7,系数为,_.(,用数字作答,),56,当,16,3,r,7,时，,r,3,，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,47/61,10.,若将函数,f,(,x,),x,5,表示为,f,(,x,),a,0,a,1,(1,x,),a,2,(1,x,),2,a,5,(1,x,),5,，其中,a,0,，,a,1,，,a,2,，a,5,为实数，则,a,3,_.,答案,解析,f,(,x,),x,5,(1,x,1),5,，,它通项为,T,r,1,(1,x,),5,r,(,1),r,，,T,3,C(1,x,),3,(,1),2,10(1,x,),3,，,a,3,10.,10,1,2,3,4,5,6,7,8,9,10,11,12,13,14,48/61,11.(,苏锡常联考,),已知,(,ax,1),5,a,0,a,1,x,a,2,x,2,a,3,x,3,a,4,x,4,32,x,5,，则二项式,(,ax,1),5,展开后各项系数之和为,_.,(,ax,1),5,a,0,a,1,x,a,2,x,2,a,3,x,3,a,4,x,4,32,x,5,，,x,5,系数为,a,5,32,，解得,a,2.,在,(2,x,1),5,a,0,a,1,x,a,2,x,2,a,3,x,3,a,4,x,4,32,x,5,中，,令,x,1,可得二项式,(2,x,1),5,展开后各项系数之和为,1.,1,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,49/61,12.,已知,(1,2,x,),7,a,0,a,1,x,a,2,x,2,a,7,x,7,.,求：,(1),a,1,a,2,a,7,；,解答,令,x,1,，则,a,0,a,1,a,2,a,3,a,4,a,5,a,6,a,7,1.,令,x,1,，则a,0,a,1,a,2,a,3,a,4,a,5,a,6,a,7,3,7,.,a,0,1,，,a,1,a,2,a,3,a,7,2.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,50/61,(2),a,1,a,3,a,5,a,7,；,解答,(,)2,，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,51/61,(3),a,0,a,2,a,4,a,6,；,解答,(,)2,，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,52/61,(4)|,a,0,|,|,a,1,|,|,a,2,|,|,a,7,|.,解答,1,2,3,4,5,6,7,8,9,10,11,12,13,14,53/61,方法一,(1,2,x,),7,展开式中，,a,0,、,a,2,、,a,4,、,a,6,大于零，而,a,1,、,a,3,、,a,5,、,a,7,小于零，,|,a,0,|,|,a,1,|,|,a,2,|,|,a,7,|,(,a,0,a,2,a,4,a,6,),(,a,1,a,3,a,5,a,7,),1 093,(,1 094),2 187.,方法二,|,a,0,|,|,a,1,|,|,a,2,|,|,a,7,|,，,即,(1,2,x,),7,展开式中各项系数和，令,x,1,，,|,a,0,|,|,a,1,|,|,a,2,|,|,a,7,|,3,7,2 187.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,54/61,13.,求证：,1,2,2,2,2,5,n,1,(,n,N,*,),能被,31,整除,.,证实,1,2,3,4,5,6,7,8,9,10,11,12,13,14,55/61,2,5,n,1,32,n,1,(31,1),n,1,原式能被,31,整除,.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,56/61,*14.,若,(),n,展开式中前三项系数成等差数列，求：,(1),展开式中全部,x,有理项；,解答,1,2,3,4,5,6,7,8,9,10,11,12,13,14,57/61,设展开式中有理项为,T,r,1,，,r,为,4,倍数，又,0,r,8,，,r,0,4,8.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,58/61,1,2,3,4,5,6,7,8,9,10,11,12,13,14,59/61,(2),展开式中系数最大项,.,解答,1,2,3,4,5,6,7,8,9,10,11,12,13,14,60/61,设展开式中,T,r,1,项系数最大，,故展开式中系数最大项为,1,2,3,4,5,6,7,8,9,10,11,12,13,14,61/61,