高考数学复习计数原理10.2二项式定理理.pptx

Click to edit Master title style,Click to edit Master text styles,Second level,Third level,Fourth level,Fifth level,*,*,*,10.2 二项式定理,高考理数,第1页,考点二项式定理应用,1.二项式定理,(,a,+,b,),n,=,a,n,+,a,n,-1,b,1,+,+,a,n,-,r,b,r,+,+,b,n,(,n,N,*,).,2.几个基本概念,(1)二项展开式:二项式定理中公式右边多项式叫做(,a,+,b,),n,二项展,开式.,(2)项数:二项展开式中共有,n,+1,项.,(3)二项式系数:在二项展开式中各项系数,(,r,=0,1,2,n,)叫做,二项式系数,.,(4)通项:在二项展开式中,a,n,-,r,b,r,叫做二项展开式通项,用,T,r,+1,表示,即,知识清单,第2页,通项为展开式第,r,+1项:,T,r,+1,=,a,n,-,r,b,r,(,r,=0,1,n,).,3.在二项式定理中,假如设,a,=1,b,=,x,则得到公式:(1+,x,),n,=1+,x,+,x,2,+,x,3,+,+,x,n,.若,a,=1,b,=-,x,则得到公式:(1-,x,),n,=1+(-1),1,x,+,x,2,+,+(-1),n,x,n,.,4.二项式系数性质,(1)对称性,与首末两端“等距离”两个二项式系数,相等,实际上,这一性,质可直接由公式,=,得到.,(2)增减性,=,当,k,时,二项式系数逐步增大,由对称性知后半,部分是逐步减小.,第3页,(3)最大值,当,n,为偶数时,中间一项,二项式系数最大,最大值为,;,当,n,为奇数时,中间两项,二项式系数相等,且同时,取得最大值,最大值为,或,.,5.各二项式系数和:(,a,+,b,),n,展开式各个二项式系数和等于2,n,即,+,+,+,+,=2,n,.,二项展开式中,奇数项二项式系数和等于偶数项二项式系数,和,即,+,+,+,=,+,+,+,=2,n,-1,.,第4页,求展开式中指定项或特定项,求二项展开式特定项问题,实质是考查通项,T,k,+1,=,a,n,-,k,b,k,特点,普通,需要建立方程求,k,再将,k,值代回通项求解,注意,k,取值范围(,k,=0,1,2,n,).,(1)第,m,项:此时,k,+1=,m,直接代入通项;,(2)常数项:即这项中不含“变元”,令通项中“变元”幂指数为0建立,方程;,(3)有理项:令通项中“变元”幂指数为整数建立方程.,特定项系数问题及相关参数值求解等都可依据上述方法求解.,方法,1,方法技巧,第5页,解题导引,例1(安徽合肥二模,15)在,展开式中,常数项为,.,第6页,解析,展开式中通项为,T,r,+1,=,(-1),4-,r,(,r,=0,1,2,3,4).,当,r,=0时,T,1,=1,当,r,0时,通项为,T,k,+1,=,x,r,-,k,=(-1),k,x,r,-2,k,(,k,=0,r,),令,r,-2,k,=0,即,r,=2,k,.,r,=2,k,=1;,r,=4,k,=2.,常数项=1-,+,1=-5.,答案-5,第7页,1.二项式系数与项系数是不一样两个概念,二项式系数是指,它只与各项项数相关,而与,a,b,值无关;而项系数是指该项中除,变量外常数部分,它不但与各项项数相关,也与,a,b,值相关,如(,a,+,bx,),n,展开式中,第,k,+1项二项式系数是,而项系数是,a,n,-,k,b,k,.,2.形如(,ax,+,b,),n,(,ax,2,+,bx,+,c,),m,(,a,b,c,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;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.,3.普通地,若,f,(,x,)=,a,0,+,a,1,x,+,a,2,x,2,+,+,a,n,x,n,则,f,(,x,)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,+,=,二项式系数与项系数,方法,2,第8页,.,例2(1)(湖南三湘名校联盟三模,7)在(,x,2,-4),展开式中,x,5,系数为,(D),A.36B.-144C.60D.-60,(2)(辽宁试验中学四模)若(1-,x,),5,=,a,0,+,a,1,x,+,a,2,x,2,+,a,3,x,3,+,a,4,x,4,+,a,5,x,5,则|,a,0,|-|,a,1,|,+|,a,2,|-|,a,3,|+|,a,4,|-|,a,5,|=,(A),A.0B.1C.32D.-1,第9页,解题导引,第10页,解析(1)(,x,2,-4),=(,x,2,-4)(,x,9,+,x,7,+,x,5,+,x,3,+,x,+,+,x,-9,),展开式中,x,5,系数为,-4,=84-144=-60,故选D.,(2),T,r,+1,=,(-,x,),r,=(-1),r,x,r,(,r,=0,1,2,3,4,5),当,r,为奇数时,a,r,0,|,a,0,|-|,a,1,|+|,a,2,|-|,a,3,|+|,a,4,|-|,a,5,|=,a,0,+,a,1,+,a,2,+,a,3,+,a,4,+,a,5,.,对(1-,x,),5,=,a,0,+,a,1,x,+,a,2,x,2,+,a,3,x,3,+,a,4,x,4,+,a,5,x,5,令,x,=1,可得,a,0,+,a,1,+,a,2,+,a,3,+,a,4,+,a,5,=(1-1),5,=0.故选A.,第11页,