數(shù)學(xué)同步優(yōu)化指導(dǎo)（人教選修23）ppt課件131二項(xiàng)式定理

第一章計(jì)數(shù)原理1.3二項(xiàng)式定理1.3.1二項(xiàng)式定理第一章計(jì)數(shù)原理1.3二項(xiàng)式定理數(shù)學(xué)同步優(yōu)化指導(dǎo)（人教選修23）ppt課件131二項(xiàng)式定理1．能用計(jì)數(shù)原理證明二項(xiàng)式定理．(難點(diǎn))2．掌握二項(xiàng)式定理及其展開(kāi)式的通項(xiàng)公式．(重點(diǎn))3．會(huì)用二項(xiàng)式定理解決與二項(xiàng)展開(kāi)式有關(guān)的簡(jiǎn)單問(wèn)題．(重點(diǎn))數(shù)學(xué)同步優(yōu)化指導(dǎo)（人教選修23）ppt課件131二項(xiàng)式定理n－k

n－kn＋1nn0b0nk＋1n＋1nn0b0nk＋13．在(a＋b)n的展開(kāi)式中，取a＝1，b＝x，則(1＋x)n＝________________________________在解題中是很有用的，要認(rèn)真體會(huì)，熟練掌握．?dāng)?shù)學(xué)同步優(yōu)化指導(dǎo)（人教選修23）ppt課件131二項(xiàng)式定理1．設(shè)S＝(x－1)3＋3(x－1)2＋3(x－1)＋1，則S等于()A．(x－1)3 B．(x－2)3C．x3 D．(x＋1)3解析：S＝[(x－1)＋1]3＝x3.答案：C1．設(shè)S＝(x－1)3＋3(x－1)2＋3(x－1)＋1，則數(shù)學(xué)同步優(yōu)化指導(dǎo)（人教選修23）ppt課件131二項(xiàng)式定理數(shù)學(xué)同步優(yōu)化指導(dǎo)（人教選修23）ppt課件131二項(xiàng)式定理4．已知(ax＋1)7(a≠0)的展開(kāi)式中，x3的系數(shù)是x2的系數(shù)與x4的系數(shù)的等差中項(xiàng)，則a的值為_(kāi)_______．4．已知(ax＋1)7(a≠0)的展開(kāi)式中，x3的系數(shù)是x2數(shù)學(xué)同步優(yōu)化指導(dǎo)（人教選修23）ppt課件131二項(xiàng)式定理1.二項(xiàng)展開(kāi)式的特點(diǎn)(1)展開(kāi)式共有n＋1項(xiàng)．(2)各項(xiàng)的次數(shù)和都等于二項(xiàng)式的冪指數(shù)n.(3)字母a的冪指數(shù)按降冪排列，從第一項(xiàng)開(kāi)始，次數(shù)由n逐項(xiàng)減1直到為0，字母b的冪指數(shù)按升冪排列，從第一項(xiàng)開(kāi)始，次數(shù)由0逐項(xiàng)加1直到為n.二項(xiàng)式定理及其通項(xiàng)公式1.二項(xiàng)展開(kāi)式的特點(diǎn)二項(xiàng)式定理及其通項(xiàng)公式數(shù)學(xué)同步優(yōu)化指導(dǎo)（人教選修23）ppt課件131二項(xiàng)式定理數(shù)學(xué)同步優(yōu)化指導(dǎo)（人教選修23）ppt課件131二項(xiàng)式定理(a＋b)n是n個(gè)(a＋b)相乘，每個(gè)(a＋b)在相乘時(shí)有兩種選擇，選a或b.而且每個(gè)(a＋b)中的a或b選定后才能得到展開(kāi)式的一項(xiàng)．由分步計(jì)數(shù)原理可知展開(kāi)式共有2n項(xiàng)(包括同類項(xiàng))，其中每一項(xiàng)都是an－kbk的形式，k＝0,1，…，n；對(duì)于每一項(xiàng)an－kbk，它是由n－k個(gè)(a＋b)選了a，k個(gè)(a＋b)選了b得到的，它出現(xiàn)的次數(shù)相當(dāng)于從n個(gè)(a＋b)中取k個(gè)b的組合數(shù)，將它們合并同類項(xiàng)，就得二項(xiàng)展開(kāi)式，這就是二項(xiàng)式定理．二項(xiàng)式定理的證明(a＋b)n是n個(gè)(a＋b)相乘，每個(gè)(a＋b)在相乘時(shí)有兩【想一想】(1)(a＋b)n展開(kāi)式中各項(xiàng)前的系數(shù)代表著什么？【想一想】(1)(a＋b)n展開(kāi)式中各項(xiàng)前的系數(shù)代表著什么二項(xiàng)式定理的正用、逆用二項(xiàng)式定理的正用、逆用數(shù)學(xué)同步優(yōu)化指導(dǎo)（人教選修23）ppt課件131二項(xiàng)式定理1．第(2)小題屬公式的“逆用”，首先轉(zhuǎn)化為展開(kāi)式的形式，適當(dāng)?shù)剡M(jìn)行配湊項(xiàng)處理．2．對(duì)于較復(fù)雜的二項(xiàng)式，有時(shí)先化簡(jiǎn)再展開(kāi)更簡(jiǎn)捷；要搞清楚二項(xiàng)展開(kāi)式中的項(xiàng)以及該項(xiàng)的系數(shù)與二項(xiàng)式系數(shù)的區(qū)別．逆用二項(xiàng)式定理可將多項(xiàng)式化簡(jiǎn)，對(duì)于這類問(wèn)題的求解，要熟悉公式的特點(diǎn)、項(xiàng)數(shù)、各項(xiàng)冪指數(shù)的規(guī)律以及各項(xiàng)的系數(shù)．1．第(2)小題屬公式的“逆用”，首先轉(zhuǎn)化為展開(kāi)式的形式，適數(shù)學(xué)同步優(yōu)化指導(dǎo)（人教選修23）ppt課件131二項(xiàng)式定理求二項(xiàng)式展開(kāi)式中的特定項(xiàng)求二項(xiàng)式展開(kāi)式中的特定項(xiàng)數(shù)學(xué)同步優(yōu)化指導(dǎo)（人教選修23）ppt課件131二項(xiàng)式定理數(shù)學(xué)同步優(yōu)化指導(dǎo)（人教選修23）ppt課件131二項(xiàng)式定理數(shù)學(xué)同步優(yōu)化指導(dǎo)（人教選修23）ppt課件131二項(xiàng)式定理數(shù)學(xué)同步優(yōu)化指導(dǎo)（人教選修23）ppt課件131二項(xiàng)式定理數(shù)學(xué)同步優(yōu)化指導(dǎo)（人教選修23）ppt課件131二項(xiàng)式定理展開(kāi)式通項(xiàng)的應(yīng)用展開(kāi)式通項(xiàng)的應(yīng)用[思路探究]

首先由“前三項(xiàng)系數(shù)成等差數(shù)列”，得到關(guān)于n的方程，解得n的值，然后根據(jù)題目的要求解答每一問(wèn)．每問(wèn)都與二項(xiàng)展開(kāi)式的通項(xiàng)公式有關(guān)．[思路探究]首先由“前三項(xiàng)系數(shù)成等差數(shù)列”，得到關(guān)于n的方數(shù)學(xué)同步優(yōu)化指導(dǎo)（人教選修23）ppt課件131二項(xiàng)式定理數(shù)學(xué)同步優(yōu)化指導(dǎo)（人教選修23）ppt課件131二項(xiàng)式定理利用二項(xiàng)式的通項(xiàng)公式求二項(xiàng)展開(kāi)式中具有某種特性的項(xiàng)是關(guān)于二項(xiàng)式定理的一類典型題型．常見(jiàn)的有求二項(xiàng)展開(kāi)式中的第r項(xiàng)、常數(shù)項(xiàng)、含某字母的r次方的項(xiàng)……等等．其通常解法就是據(jù)通項(xiàng)公式確定Tk＋1中k的值或取值范圍以滿足題設(shè)的條件．利用二項(xiàng)式的通項(xiàng)公式求二項(xiàng)展開(kāi)式中具有某種特性的項(xiàng)是關(guān)于二項(xiàng)數(shù)學(xué)同步優(yōu)化指導(dǎo)（人教選修23）ppt課件131二項(xiàng)式定理答案：C答案：C數(shù)學(xué)同步優(yōu)化指導(dǎo)（人教選修23）ppt課件131二項(xiàng)式定理數(shù)學(xué)同步優(yōu)化指導(dǎo)（人教選修23）ppt課件131二項(xiàng)式定理活頁(yè)作業(yè)(七)活頁(yè)作業(yè)(七)謝謝觀看！謝謝觀看！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