202X屆高考數(shù)學(xué)一輪復(fù)習(xí)第九章計(jì)數(shù)原理9.3二項(xiàng)式定理課件新人教A版

1、9.3二項(xiàng)式定理 -2-知識(shí)梳理雙基自測(cè)1.二項(xiàng)式定理 r+1 -3-知識(shí)梳理雙基自測(cè)2.二項(xiàng)式系數(shù)的性質(zhì) -4-知識(shí)梳理雙基自測(cè)3.常用結(jié)論 2n 2n-1 2-5-知識(shí)梳理雙基自測(cè)3415-6-知識(shí)梳理雙基自測(cè)23415A-7-知識(shí)梳理雙基自測(cè)234153.已知(1+x)n的展開(kāi)式中第4項(xiàng)與第8項(xiàng)的二項(xiàng)式系數(shù)相等,則奇數(shù)項(xiàng)的二項(xiàng)式系數(shù)和為()A.212B.211C.210D.29D解析 由條件知 ,則n=10.故(1+x)10中二項(xiàng)式系數(shù)和為210,其中奇數(shù)項(xiàng)的二項(xiàng)式系數(shù)和為210-1=29.-8-知識(shí)梳理雙基自測(cè)234154.在(1-2x)6的展開(kāi)式中,x2的系數(shù)為.(用數(shù)字作答)60-

2、9-知識(shí)梳理雙基自測(cè)234155.已知(1+3x)n的展開(kāi)式中含有x2項(xiàng)的系數(shù)是54,則n=.4-10-考點(diǎn)1考點(diǎn)2考向一已知二項(xiàng)式求其特定項(xiàng)(或系數(shù))思考如何求二項(xiàng)展開(kāi)式的項(xiàng)或特定項(xiàng)的系數(shù)?已知特定項(xiàng)的系數(shù)如何求二項(xiàng)式中的參數(shù)?A-56 -11-考點(diǎn)1考點(diǎn)2考向二已知三項(xiàng)式求其特定項(xiàng)(或系數(shù))例2(1)在(x2+x+y)5的展開(kāi)式中,x5y2的系數(shù)為 ()A.10 B.20C.30D.60(2)在(x2-x+1)3展開(kāi)式中,x項(xiàng)的系數(shù)為()A.-3B.-1C.1D.3思考如何求三項(xiàng)式中某一特定項(xiàng)的系數(shù)?CA-12-考點(diǎn)1考點(diǎn)2-13-考點(diǎn)1考點(diǎn)2考向三求兩個(gè)因式之積的特定項(xiàng)系數(shù)例3(1) (

3、1+x)6展開(kāi)式中x2的系數(shù)為()A.15 B.20C.30 D.35(2)(x-y)(x+y)8的展開(kāi)式中x2y7的系數(shù)為.(用數(shù)字填寫(xiě)答案)思考如何求兩個(gè)因式之積的特定項(xiàng)系數(shù)?C-20-14-考點(diǎn)1考點(diǎn)2-15-考點(diǎn)1考點(diǎn)2-16-考點(diǎn)1考點(diǎn)2解題心得1.求二項(xiàng)展開(kāi)式中的項(xiàng)或項(xiàng)的系數(shù)的方法:求二項(xiàng)展先建立方程求k,再將k的值代回通項(xiàng)求解,注意k的取值范圍(k=0,1,2,n).特定項(xiàng)的系數(shù)問(wèn)題及相關(guān)參數(shù)值的求解等都可依據(jù)上述方法求解.2.求三項(xiàng)展開(kāi)式中某些特殊項(xiàng)的系數(shù)的方法:(1)通過(guò)變形先把三項(xiàng)式轉(zhuǎn)化為二項(xiàng)式,再用二項(xiàng)式定理去解;(2)兩次利用二項(xiàng)式定理的通項(xiàng)公式求解;(3)由二項(xiàng)式定理

4、的推證方法知,可用排列組合的基本原理去求,即把三項(xiàng)式看作幾個(gè)因式之積,要得到特定項(xiàng)看有多少種方法從這幾個(gè)因式中取因式中的量.3.求兩個(gè)因式之積的特定項(xiàng)系數(shù)也有兩種方法:(1)利用通項(xiàng)公式法;(2)用排列組合法.-17-考點(diǎn)1考點(diǎn)22 141 128-18-考點(diǎn)1考點(diǎn)2則取常數(shù)項(xiàng)時(shí)r=2m.由題可知r0,1,2,3,4,5,6,m0,1,2,3,4,5,6,則m的可能取值為0,1,2,3,對(duì)應(yīng)的r分別為0,2,4,6.當(dāng)m=0,r=0時(shí),常數(shù)項(xiàng)為1;當(dāng)m=1,r=2時(shí),常數(shù)項(xiàng)為30;當(dāng)m=2,r=4時(shí),常數(shù)項(xiàng)為90;當(dāng)m=3,r=6時(shí),常數(shù)項(xiàng)為20;故常數(shù)項(xiàng)為1+30+90+20=141.-19

5、-考點(diǎn)1考點(diǎn)2-20-考點(diǎn)1考點(diǎn)2B-21-考點(diǎn)1考點(diǎn)2 -8 064 -15 360 x4 -22-考點(diǎn)1考點(diǎn)2-23-考點(diǎn)1考點(diǎn)2考向三求二項(xiàng)式展開(kāi)式中系數(shù)的和例6(a+x)(1+x)4的展開(kāi)式中x的奇數(shù)次冪項(xiàng)的系數(shù)之和為32,則a=.思考求二項(xiàng)式系數(shù)和的常用方法是什么?3=x4+4x3+6x2+4x+1,(a+x)(1+x)4的奇數(shù)次冪項(xiàng)的系數(shù)為4a+4a+1+6+1=32.a=3.(方法二)設(shè)(a+x)(1+x)4=b0+b1x+b2x2+b3x3+b4x4+b5x5.令x=1,得16(a+1)=b0+b1+b2+b3+b4+b5,令x=-1,得0=b0-b1+b2-b3+b4-b5,由-,得16(a+1)=2(b1+b3+b5).即8(a+1)=32,解得a=3.-24-考點(diǎn)1考點(diǎn)2-25-考點(diǎn)1考點(diǎn)23.求二項(xiàng)式系數(shù)和的常用方法是賦值法:(1)“賦值法”普遍適用于恒等式,對(duì)形如(ax+b)n,(ax2+bx+c)m(a,bR)的式子,求其展