2017-2018学年高中数学人教A版选修2-3教学案：1.3.1二项式定理Word版含解析.pdf

131二项式定理预习课本P29 31，思考并完成以下问题1二项式定理是什么？2通项公式又是什么？3二项式定理有何结构特征，二项展开式中某项的二项式系数与某项的系数有区别吗？新知初探 二项式定理二项式定理(ab)nC0nan C1nan1b,Cknankbk,Cnnbn二项展开式公式右边的式子二项式系数Ckn(k0,1,2，,，n)二项展开式的通项Tk1Cknankbk点睛 应用通项公式要注意四点(1)Tk1是展开式中的第k1 项，而不是第k 项；(2)公式中 a，b 的指数和为n，且 a，b不能随便颠倒位置；(3)要将通项中的系数和字母分离开，以便于解决问题；(4)对二项式(ab)n展开式的通项公式要特别注意符号问题小试身手 1判断下列命题是否正确(正确的打“”，错误的打“”)(1)(ab)n展开式中共有n 项()(2)二项式(a b)n与(ba)n展开式中第r 1 项相同()(3)Cknankbk是(a b)n展开式中的第k 项()答案：(1)(2)(3)2 x1x5的展开式中含x3项的二项式系数为()A 10B10 C 5 D5 答案：D 3 x22x35展开式中的常数项为()A 80 B 80 C 40 D 40 答案：C 4(12x)5的展开式的第3 项的系数为 _，第三项的二项式系数为_答案：4010 二项式定理的应用典例(1)求3 x1x4的展开式；(2)化简：(x 1)55(x1)410(x1)310(x1)2 5(x1)解(1)法一：3 x1x4C04(3x)4C14(3x)31xC24(3 x)21x2C34 3 x1x3C441x481x2108x5412x1x2法二：3 x1x43x14x21x2(81x4108x354x212x1)81x2108x5412x1x2(2)原式 C05(x1)5C15(x1)4C25(x1)3C35(x1)2 C45(x1)C55(x1)01(x1)151 x51运用二项式定理的解题策略(1)正用：求形式简单的二项展开式时可直接由二项式定理展开，展开时注意二项展开文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10式的特点：前一个字母是降幂，后一个字母是升幂形如(ab)n的展开式中会出现正负间隔的情况对较繁杂的式子，先化简再用二项式定理展开(2)逆用：逆用二项式定理可将多项式化简，对于这类问题的求解，要熟悉公式的特点、项数、各项幂指数的规律以及各项的系数活学活用 1化简(x1)44(x1)36(x1)2 4(x1)1 的结果为()A x4B(x1)4C(x1)4Dx41 解析：选 A(x1)44(x1)36(x1)2 4(x1)1C04(x1)4C14(x1)3(1)1C24(x 1)2(1)2C34(x1)(1)3C44(x 1)0(1)4(x1)14x4，故选 A2 设 n为自然数，化简 C0n 2nC1n 2n1,(1)k Ckn 2nk,(1)n Cnn _解：原式 C0n 2n(1)0C1n2n1(1)1,(1)k Ckn2nk,(1)n Cnn 20(21)n1答案：1 二项式系数与项的系数问题典例(1)求二项式2x1x6的展开式中第6 项的二项式系数和第6 项的系数；(2)求 x1x9的展开式中x3的系数解(1)由已知得二项展开式的通项为Tr1Cr6(2x)6r1xr 26rCr6(1)r x33r2，T6 12 x92第 6 项的二项式系数为C566，第 6 项的系数为C56(1)5 2 12(2)设展开式中的第r1 项为含 x3的项，则Tr1Cr9x9r1xr(1)r Cr9 x92r，令 92r3，得 r3，即展开式中第四项含x3，其系数为(1)3 C39 84一题多变 1 变设问 本例问题(1)条件不变，问题改为“求第四项的二项式系数和第四项的系数”文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10解：由通项 Tr1(1)r Cr6 26r x332r，知第四项的二项式系数为C3620，第四项的系数为C36(1)3 23 1602变设问 本例问题(2)条件不变，问题改为“求展开式中x5的系数”，该如何求解解：设展开式中第r1 项为含 x5的项，则Tr1(1)r Cr9 x92r，令 92r5，得 r2即展开式中的第3 项含 x5，且系数为C2936求某项的二项式系数或展开式中含xr的项的系数，主要是利用通项公式求出相应的项，特别要注意某项二项式系数与系数两者的区别与展开式中的特定项有关的问题题点一：求展开式中的特定项1(四川高考)设 i 为虚数单位，则(xi)6的展开式中含x4的项为()A 15x4B15x4C 20ix4D20ix4解析：选 A二项式的通项为Tr1Cr6x6rir，由 6r 4 得 r2故 T3C26x4i2 15x4故选 A2(12x)3(13x)5的展开式中x 的系数是 _解析：(12x)3(13x)5的展开式的通项为2rCr3(1)sCs5x3r2s6(其中 r0,1,2,3；s0,1,2,3,4,5)，令3r2s61，得 3r2s 6，所以r0，s3或r2，s0.所以 x 的系数是 C354C232答案：2 题点二：由二项展开式某项的系数求参数问题3(山东高考)若ax21x5的展开式中x5的系数是 80，则实数 a_解析：Tr1Cr5(ax2)5r1xrCr5 a5rx1052r令 1052r5，解得 r2又展开式中x5的系数为 80，则有 C25 a3 80，解得 a 2答案：2 文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10求展开式中特定项的方法求展开式特定项的关键是抓住其通项公式，求解时先准确写出通项，再把系数和字母分离，根据题目中所指定的字母的指数所具有的特征，列出方程或不等式即可求解有理项问题的解法，要保证字母的指数一定为整数层级一学业水平达标1(x2)n的展开式共有12 项，则 n 等于()A 9B10 C 11 D8 解析：选 C(ab)n的展开式共有n1 项，而(x2)n的展开式共有12 项，n11 故选 C2(1i)10(i 为虚数单位)的二项展开式中第七项为()A 210 B210 C 120i D 210i 解析：选 A由通项公式得T7C610(i)6 C610 2103已知x1x7的展开式的第4 项等于 5，则 x等于()A17B17C 7 D 7 解析：选 BT4C37x41x35，x174若二项式x2xn的展开式中第5 项是常数项，则自然数n 的值可能为()A 6 B10 C 12 D15 解析：选 C T5C4n(x)n42x424 C4nxn122是常数项，n1220，n125(湖南高考)12x2y5的展开式中x2y3的系数是()A 20 B 5 C 5 D20 解析：选 A由二项展开式的通项可得，第四项 T4C3512x2(2y)3 20 x2y3，故 x2y3的系数为 20，选 A文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G106(全国卷)(2xx)5的展开式中，x3的系数是 _(用数字填写答案)解析：(2xx)5展开式的通项为Tr1 Cr5(2x)5r(x)r 25r Cr5 x5r2令 5r2 3，得 r4故 x3的系数为254 C452C4510答案：10 7若(1 2x)6的展开式中的第2 项大于它的相邻两项，则x 的取值范围是_解析：由T2T1，T2T3，得C162x1，C162xC262x2.解得112x15答案：112，158若(x a)10的展开式中，x7的系数为15，则 a_(用数字填写答案)解析：二项展开式的通项公式为Tr1Cr10 x10rar，当 10r7 时，r3，T4 C310a3x7，则 C310a315，故 a12答案：129若二项式xax6(a 0)的展开式中x3的系数为A，常数项为B，且 B4A，求 a的值解：Tr1Cr6x6raxr(a)rCr6x63r2，令 63r23，则 r2，得 AC26 a2 15a2；令 63r20，则 r4，得 BC46 a4 15a4由 B4A 可得 a24，又 a0，所以 a210已知m，nN*，f(x)(1x)m(1 x)n展开式中x 的系数为19，求 x2的系数的最小值及此时展开式中x7的系数解：由题设 mn19，m，nN*m1，n18，m2，n17，,，m18，n1.x2的系数 C2mC2n12(m2m)12(n2n)m219m171 m19223234当 m9 或 10 时，x2的系数取最小值81，此时 x7的系数为C79 C710156文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10层级二应试能力达标1在(1 x3)(1x)10的展开式中x5的系数是()A 297B 252 C 297 D207 解析：选 Dx5应是(1x)10中含 x5项与含 x2项其系数为C510C210(1)2072使3x1xxn(nN*)的展开式中含有常数项的最小的n 为()A 4 B5 C 6 D7 解析：选 B由二项式定理得，Tr1Crn(3x)nr1x xrCrn3nrxn52r，令 n52r0，当 r2 时，n5，此时 n 最小3(13x)n(其中 nN 且 n 6)的展开式中，若x5与 x6的系数相等，则n()A 6 B7 C 8 D9 解析：选 B二项式(13x)n的展开式的通项是Tr1Crn1nr(3x)r Crn 3r xr 依题意得C5n 35C6n 36，即n n1 n2n3n45！3n n1n2n 3 n 4 n56！(n6)，得 n74在 x21xn的展开式中，常数项为15，则 n 的一个值可以是()A 3 B4 C 5 D6 解析：选 D通项 Tr1Crn(x2)nr1xr(1)rCrnx2n3r，常数项是15，则 2n3r，且Crn15，验证 n6 时，r4 合题意，故选D5x x2x7的展开式中，x4的系数是 _(用数字作答)解析：x4的系数，即x2x7展开式中x3的系数，Tr1Cr7 x7r2xr(2)r Cr7 x72r，令 72r3 得，r2，所求系数为(2)2C27 84答案：84 文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM9U8N9G5G10文档编码:CT3A6A1D2E2 HG6T5I2Z5T2 ZM