高考数学复习第十章计数原理10.3二项式定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,1/54,基础知识自主学习,课时作业,题型分类深度剖析,内容索引,2/54,基础知识自主学习,3/54,1.,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,，它表示第,项,二项式系数,二项展开式中各项系数 (k0,1,2，n),k,1,4/54,2.,二项式系数性质,1,1,5/54,二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,6/54,判断以下结论是否正确,(,请在括号中打,“”,或,“”,),(1),a,n,k,b,k,是二项展开式第,k,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,7/54,考点自测,1.(,教材改编,)(,x,y),n,二项展开式中，第,m,项系数是,答案,解析,(,x,y,),n,展开式中第,m,项系数为,8/54,2.(,四川,),设,i,为虚数单位，则,(,x,i),6,展开式中含,x,4,项为,A.,15,x,4,B.15,x,4,C.,20i,x,4,D.20i,x,4,答案,解析,9/54,答案,解析,A.130 B.135 C.121 D.139,式中含,x,2,项系数为,10/54,4.,在,展开式中，只有第,5,项二项式系数最大，则展开式中常数项是,_.,答案,解析,7,11/54,题型分类深度剖析,12/54,题型一二项展开式,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,),答案,解析,命题点,1,求二项展开式中特定项或指定项系数,10,x,3,系数是,10.,13/54,(2)(,课标全国,)(,x,2,x,y,),5,展开式中，,x,5,y,2,系数为,A.10 B.20C.30 D.60,答案,解析,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,.,14/54,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,_.,答案,解析,命题点,2,已知二项展开式某项系数求参数,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,奇数次幂系数之和为,a,1,a,3,a,5,8(,a,1),，,所以,8(,a,1),32,，解得,a,3.,3,15/54,2,答案,解析,16/54,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,k,1,，代回通项公式即可,.,思维升华,17/54跟踪训练,1,(1)(,x,y,)(,x,y,),8,展开式中,x,2,y,7,系数为,_.(,用数字填写答案,),20,答案,解析,18/54,(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,19/54,题型二,二项式系数和或各项系数和问题,例,2,在,(2,x,3,y,),10,展开式中，求：,(1),二项式系数和；,(2),各项系数和；,(3),奇数项二项式系数和与偶数项二项式系数和；,(4),奇数项系数和与偶数项系数和；,(5),x,奇次项系数和与,x,偶次项系数和,.,解答,20/54,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,.,(2),令,x,y,1,，各项系数和为,(2,3),10,(,1),10,1.,21/54,(4),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,，,22/54,23/54,(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,24/54,跟踪训练,2,(1)(,北京海淀区模拟,),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,等于,A.5 B.6 C.7 D.8,答案,解析,经检验符合题意，故选,B.,25/54,解答,当,x,0,时，左边,1,，右边,a,0,，,a,0,1.,26/54,题型三二项式定理应用,例,4,(1),设,a,Z,且,0,a,13,，若,51,2 012,a,能被,13,整除，则,a,等于,A.0 B.1 C.11 D.12,答案,解析,27/54,(2)1.02,8,近似值是,_.(,准确到小数点后三位,),1.172,答案,解析,28/54,(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,29/54,A.,1 B.1 C.,87 D.87,前,10,项均能被,88,整除，,余数是,1.,答案,解析,30/54,(2),已知,2,n,2,3,n,5,n,a,能被,25,整除，求正整数,a,最小值,.,解答,原式,46,n,5,n,a,4(5,1),n,5,n,a,显然正整数,a,最小值为,4.,31/54,典例,(1)(,河北武邑中学期末,),若,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)(,河北邯郸一中调研,),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,15,错解展示,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,.,32/54,解析,答案,(1)5,(2)2,7,1,返回,33/54,解析,故展开式中含,x,项系数为,15.,34/54,答案,(1),15,(2)1,令,x,1,，得,0,1,a,1,a,2,a,7,，,即,a,1,a,2,a,3,a,7,1.,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，,返回,35/54,课时作业,36/54,1.,在,x,2,(1,x,),6,展开式中，含,x,4,项系数为,A.30 B.20 C.15 D.10,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,37/54,2.(,湖南,),已知,展开式中含,项系数为,30,，则,a,等于,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,38/54,3.(4,x,2,x,),6,(,x,R,),展开式中常数项是,A.,20 B.,15,C.15 D.20,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,12,x,3,kx,0,恒成立，,k,4,，,39/54,4.(,湖北,),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,A.2,9,B.2,10,C.2,11,D.2,12,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,40/54,5.,若在,(,x,1),4,(,ax,1),展开式中，,x,4,系数为,15,，则,a,值为,1,2,3,4,5,6,7,8,9,10,11,12,13,14,(,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.,答案,解析,41/54,6.,若,(1,x,),(1,x,),2,(1,x,),n,a,0a,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,等于,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,，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,42/54,7.,若,(,x,a,),2,(,1),5,展开式中常数项为,1,，则,a,值为,A.1 B.9,C.,1,或,9 D.1,或,9,1,2,3,4,5,6,7,8,9,10,11,12,13,14,答案,解析,依题意,a,2,10,a,10,1,，解得a,2,10,a,9,0,，即,a,1,或,a,9.,43/54,8.(,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,60,1,2,3,4,5,6,7,8,9,10,11,12,13,14,44/54,9.(,天津,),展开式中,x,7,系数为,_.(,用数字作答,),答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,56,45/54,10.,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,1,2,3,4,5,6,7,8,9,10,11,12,13,14,10,f,(,x,),x,5,(1,x,1),5,，,46/54,11.(1,x,),8,(1,y,),4,展开式中,x,2,y,2,系数是,_.,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,168,47/54,解答,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,；,(2),a,1,a,3,a,5,a,7,；,(3),a,0,a,2,a,4,a,64)|,a,0,|,|,a,1,|,|,a,2,|,|,a,7,|.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,48/54,1,2,3,4,5,6,7,8,9,10,11,12,13,14,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,.,(2)(,)2,，,(3)(,)2,，,49/54,(4),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,、,a7,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.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,50/54,13.,求证：,1,2,2,2,2,5,n,1,(,n,N,*,),能被,31,整除,.,证实1,2,3,4,5,6,7,8,9,10,11,12,13,14,2,5,n,1,32,n,1,(31,1),n,1,原式能被,31,整除,.,51/54,*14.,若,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；,1,2,3,4,5,6,7,8,9,10,11,12,13,14,解答,52/54,k,为,4,倍数，又,0,k,8,，,k,0,4,8.,故有理项为,1,2,3,4,5,6,7,8,9,10,11,12,13,14,53/54,(2),展开式中系数最大项,.,解答,设展开式中,T,k,1,项系数最大，,故展开式中系数最大项为,1,2,3,4,5,6,7,8,9,10,11,12,13,14,54/54,