新高考數(shù)學(xué)一輪復(fù)習(xí)講義 第51講 二項(xiàng)式定理（含解析）

第51講二項(xiàng)式定理（精講）題型目錄一覽①二項(xiàng)展開式中的特定項(xiàng)問題②二項(xiàng)式系數(shù)問題③二項(xiàng)展開式中各項(xiàng)系數(shù)的和問題④三項(xiàng)展開式的問題⑤兩個(gè)二項(xiàng)式乘積展開式的系數(shù)⑥賦值法一、知識(shí)點(diǎn)梳理一、知識(shí)點(diǎn)梳理一、二項(xiàng)式展開式的特定項(xiàng)、特定項(xiàng)的系數(shù)問題1.二項(xiàng)式定理一般地，對(duì)于任意正整數(shù)，都有：SKIPIF1<0，這個(gè)公式所表示的定理叫做二項(xiàng)式定理，等號(hào)右邊的多項(xiàng)式叫做的二項(xiàng)展開式．式中的SKIPIF1<0做二項(xiàng)展開式的通項(xiàng)，用SKIPIF1<0表示，即通項(xiàng)為展開式的第SKIPIF1<0項(xiàng)：SKIPIF1<0，其中的系數(shù)SKIPIF1<0（r=0，1，2，…，n）叫做二項(xiàng)式系數(shù)，2.二項(xiàng)式SKIPIF1<0的展開式的特點(diǎn)：①項(xiàng)數(shù)：共有SKIPIF1<0項(xiàng)，比二項(xiàng)式的次數(shù)大1；②二項(xiàng)式系數(shù)：第SKIPIF1<0項(xiàng)的二項(xiàng)式系數(shù)為SKIPIF1<0，最大二項(xiàng)式系數(shù)項(xiàng)居中；③次數(shù)：各項(xiàng)的次數(shù)都等于二項(xiàng)式的冪指數(shù)SKIPIF1<0．字母SKIPIF1<0降冪排列，次數(shù)由SKIPIF1<0到SKIPIF1<0；字母SKIPIF1<0升冪排列，次數(shù)從SKIPIF1<0到SKIPIF1<0，每一項(xiàng)中，SKIPIF1<0，SKIPIF1<0次數(shù)和均為SKIPIF1<0；④項(xiàng)的系數(shù)：二項(xiàng)式系數(shù)依次是SKIPIF1<0，項(xiàng)的系數(shù)是SKIPIF1<0與SKIPIF1<0的系數(shù)（包括二項(xiàng)式系數(shù)）．3.兩個(gè)常用的二項(xiàng)展開式：①（）②4.二項(xiàng)展開式的通項(xiàng)公式二項(xiàng)展開式的通項(xiàng)：SKIPIF1<0SKIPIF1<0公式特點(diǎn)：①它表示二項(xiàng)展開式的第SKIPIF1<0項(xiàng)，該項(xiàng)的二項(xiàng)式系數(shù)是；②字母SKIPIF1<0的次數(shù)和組合數(shù)的上標(biāo)相同；③SKIPIF1<0與SKIPIF1<0的次數(shù)之和為SKIPIF1<0．注意：①二項(xiàng)式SKIPIF1<0的二項(xiàng)展開式的第r+1項(xiàng)和SKIPIF1<0的二項(xiàng)展開式的第r+1項(xiàng)是有區(qū)別的，應(yīng)用二項(xiàng)式定理時(shí)，其中的SKIPIF1<0和SKIPIF1<0是不能隨便交換位置的．②通項(xiàng)是針對(duì)在SKIPIF1<0這個(gè)標(biāo)準(zhǔn)形式下而言的，如SKIPIF1<0的二項(xiàng)展開式的通項(xiàng)是（只需把SKIPIF1<0看成SKIPIF1<0代入二項(xiàng)式定理）．二、二項(xiàng)式展開式中的最值問題1.二項(xiàng)式系數(shù)的性質(zhì)=1\*GB3①每一行兩端都是SKIPIF1<0，即SKIPIF1<0；其余每個(gè)數(shù)都等于它“肩上”兩個(gè)數(shù)的和，即SKIPIF1<0．=2\*GB3②對(duì)稱性每一行中，與首末兩端“等距離”的兩個(gè)二項(xiàng)式系數(shù)相等，即SKIPIF1<0．=3\*GB3③二項(xiàng)式系數(shù)和令SKIPIF1<0，則二項(xiàng)式系數(shù)的和為SKIPIF1<0，變形式SKIPIF1<0．=4\*GB3④奇數(shù)項(xiàng)的二項(xiàng)式系數(shù)和等于偶數(shù)項(xiàng)的二項(xiàng)式系數(shù)和在二項(xiàng)式定理中，令SKIPIF1<0，則SKIPIF1<0，從而得到：SKIPIF1<0．=5\*GB3⑤最大值：如果二項(xiàng)式的冪指數(shù)SKIPIF1<0是偶數(shù)，則中間一項(xiàng)SKIPIF1<0的二項(xiàng)式系數(shù)SKIPIF1<0最大；如果二項(xiàng)式的冪指數(shù)SKIPIF1<0是奇數(shù)，則中間兩項(xiàng)SKIPIF1<0，SKIPIF1<0的二項(xiàng)式系數(shù)SKIPIF1<0，SKIPIF1<0相等且最大．2.系數(shù)的最大項(xiàng)求SKIPIF1<0展開式中最大的項(xiàng)，一般采用待定系數(shù)法．設(shè)展開式中各項(xiàng)系數(shù)分別為SKIPIF1<0，設(shè)第SKIPIF1<0項(xiàng)系數(shù)最大，應(yīng)有SKIPIF1<0，從而解出SKIPIF1<0來．三、二項(xiàng)式展開式中系數(shù)和有關(guān)問題常用賦值舉例：（1）設(shè)，二項(xiàng)式定理是一個(gè)恒等式，即對(duì)SKIPIF1<0，SKIPIF1<0的一切值都成立，我們可以根據(jù)具體問題的需要靈活選取SKIPIF1<0，SKIPIF1<0的值．①令，可得：②令SKIPIF1<0，可得：，即：（假設(shè)為偶數(shù)），再結(jié)合①可得：．（2）若SKIPIF1<0，則①常數(shù)項(xiàng)：令SKIPIF1<0，得SKIPIF1<0．②各項(xiàng)系數(shù)和：令SKIPIF1<0，得SKIPIF1<0．注意：常見的賦值為令SKIPIF1<0，SKIPIF1<0或SKIPIF1<0，然后通過加減運(yùn)算即可得到相應(yīng)的結(jié)果．【常用結(jié)論】奇數(shù)項(xiàng)的系數(shù)和與偶數(shù)項(xiàng)的系數(shù)和①5當(dāng)SKIPIF1<0為偶數(shù)時(shí)，奇數(shù)項(xiàng)的系數(shù)和為SKIPIF1<0；偶數(shù)項(xiàng)的系數(shù)和為SKIPIF1<0．（可簡記為：SKIPIF1<0為偶數(shù)，奇數(shù)項(xiàng)的系數(shù)和用“中點(diǎn)公式”，奇偶交錯(cuò)搭配）②當(dāng)SKIPIF1<0為奇數(shù)時(shí)，奇數(shù)項(xiàng)的系數(shù)和為SKIPIF1<0；偶數(shù)項(xiàng)的系數(shù)和為SKIPIF1<0．（可簡記為：SKIPIF1<0為奇數(shù)，偶數(shù)項(xiàng)的系數(shù)和用“中點(diǎn)公式”，奇偶交錯(cuò)搭配）若SKIPIF1<0，同理可得．二、題型分類精講二、題型分類精講題型一二項(xiàng)展開式中的特定項(xiàng)問題策略方法形如(a＋b)n的展開式問題二項(xiàng)展開式中的特定項(xiàng)，是指展開式中的某一項(xiàng)，如第n項(xiàng)、常數(shù)項(xiàng)、有理項(xiàng)等，求解二項(xiàng)展開式中的特定項(xiàng)的關(guān)鍵點(diǎn)如下：①求通項(xiàng)，利用(a＋b)n的展開式的通項(xiàng)公式Tr＋1＝Ceq\o\al(r,n)an－rbr(r＝0,1,2，…，n)求通項(xiàng)．②列方程(組)或不等式(組)，利用二項(xiàng)展開式的通項(xiàng)及特定項(xiàng)的特征，列出方程(組)或不等式(組)．③求特定項(xiàng)，先由方程(組)或不等式(組)求得相關(guān)參數(shù)，再根據(jù)要求寫出特定項(xiàng)．【典例1】（單選題）已知SKIPIF1<0的展開式中的常數(shù)項(xiàng)為SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0（

）A．2 B．-2 C．8 D．-8【答案】B【分析】直接利用二項(xiàng)式定理計(jì)算得到答案.【詳解】SKIPIF1<0展開式的通項(xiàng)為：SKIPIF1<0，取SKIPIF1<0得到常數(shù)項(xiàng)為SKIPIF1<0，解得SKIPIF1<0.故選：B【典例2】（單選題）二項(xiàng)式SKIPIF1<0展開式中含x項(xiàng)的系數(shù)是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根據(jù)二項(xiàng)式定理寫出通項(xiàng)公式進(jìn)而求解.【詳解】二項(xiàng)式SKIPIF1<0的通項(xiàng)公式SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0.則二項(xiàng)式SKIPIF1<0展開式中含x項(xiàng)的系數(shù)是SKIPIF1<0.故選：C【題型訓(xùn)練】一、單選題1．在SKIPIF1<0的展開式中，第四項(xiàng)為（

）A．160 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】直接根據(jù)二項(xiàng)展開式的通項(xiàng)求第四項(xiàng)即可.【詳解】在SKIPIF1<0的展開式中，第四項(xiàng)為SKIPIF1<0.故選：D.2．SKIPIF1<0展開式中的常數(shù)項(xiàng)為－160，則a＝（

）A．－1 B．1 C．±1 D．2【答案】B【分析】寫出該二項(xiàng)展開式的通項(xiàng)公式，令x的冪指數(shù)等于0，求出r的值，即可求得常數(shù)項(xiàng)，再根據(jù)常數(shù)項(xiàng)等于-160求得實(shí)數(shù)a的值.【詳解】SKIPIF1<0的展開式通項(xiàng)為SKIPIF1<0，∴令SKIPIF1<0，解得SKIPIF1<0，∴SKIPIF1<0的展開式的常數(shù)項(xiàng)為SKIPIF1<0，∴SKIPIF1<0∴SKIPIF1<0故選：B.3．SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為（

）A．40 B．SKIPIF1<0 C．80 D．SKIPIF1<0【答案】A【分析】首先寫出展開式的通項(xiàng)，再代入計(jì)算可得；【詳解】SKIPIF1<0的展開式的通項(xiàng)SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0項(xiàng)的系數(shù)為SKIPIF1<0，故選：A4．已知SKIPIF1<0的展開式中的常數(shù)項(xiàng)是672，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．2 D．1【答案】C【分析】寫出二項(xiàng)式通項(xiàng)SKIPIF1<0，整理后讓SKIPIF1<0的次數(shù)為SKIPIF1<0，得出SKIPIF1<0的值，再根據(jù)題意常數(shù)項(xiàng)的系數(shù)列出等式方程即可得出SKIPIF1<0的值.【詳解】展開式的通項(xiàng)為SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，∴常數(shù)項(xiàng)是SKIPIF1<0，故SKIPIF1<0．故選：C5．二項(xiàng)式SKIPIF1<0的展開式中的常數(shù)項(xiàng)為（

）A．1792 B．－1792 C．1120 D．－1120【答案】C【分析】根據(jù)二項(xiàng)式定理展開式求解即可.【詳解】因?yàn)镾KIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，所以二項(xiàng)式展開式中的常數(shù)項(xiàng)為SKIPIF1<0．故選：C.6．若SKIPIF1<0展開式中含有常數(shù)項(xiàng)，則n的最小值是（

）A．2 B．3 C．12 D．10【答案】A【分析】根據(jù)通項(xiàng)公式可求出結(jié)果.【詳解】SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0時(shí)，SKIPIF1<0取最小值SKIPIF1<0.故選：A7．SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)是126，則SKIPIF1<0（

）A．2 B．4 C．1 D．3【答案】C【分析】求出展開式通項(xiàng)，令SKIPIF1<0，解出SKIPIF1<0，即可得出答案.【詳解】展開式通項(xiàng)為SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，因?yàn)镾KIPIF1<0的系數(shù)是126，所以SKIPIF1<0，解得SKIPIF1<0，故選：C.二、填空題8．二項(xiàng)式SKIPIF1<0的展開式中的常數(shù)項(xiàng)為.【答案】240【分析】利用二項(xiàng)式展開式的通項(xiàng)公式求解即得.【詳解】二項(xiàng)式SKIPIF1<0的展開式通項(xiàng)為SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，所以所求常數(shù)項(xiàng)為SKIPIF1<0.故答案為：2409．SKIPIF1<0的展開式的第8項(xiàng)的系數(shù)為（結(jié)果用數(shù)值表示）.【答案】960【分析】根據(jù)二項(xiàng)式定理求出展開式中的第8項(xiàng)，由此即可求解．【詳解】因?yàn)?，SKIPIF1<0展開式的第8項(xiàng)為SKIPIF1<0，所以，SKIPIF1<0的展開式的第8項(xiàng)的系數(shù)為960.故答案為：96010．二項(xiàng)式SKIPIF1<0的展開式中的常數(shù)項(xiàng)是.【答案】SKIPIF1<0【分析】利用二項(xiàng)式SKIPIF1<0的通項(xiàng)公式SKIPIF1<0，即可求出結(jié)果.【詳解】二項(xiàng)式SKIPIF1<0的通項(xiàng)公式為SKIPIF1<0，由SKIPIF1<0，得到SKIPIF1<0，所以二項(xiàng)式SKIPIF1<0的展開式中的常數(shù)項(xiàng)是SKIPIF1<0，故答案為：SKIPIF1<0.11．若在SKIPIF1<0的展開式中，第4項(xiàng)是常數(shù)項(xiàng)，則SKIPIF1<0．【答案】12【分析】寫出二項(xiàng)展開式的通項(xiàng)公式，再根據(jù)題意可得到SKIPIF1<0，即可求得答案【詳解】設(shè)展開式中第SKIPIF1<0項(xiàng)為SKIPIF1<0，則SKIPIF1<0，又展開式中第4項(xiàng)是常數(shù)項(xiàng)，∴SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0故答案為：1212．設(shè)常數(shù)SKIPIF1<0，若SKIPIF1<0的二項(xiàng)展開式中SKIPIF1<0的系數(shù)為144，則SKIPIF1<0．【答案】2【分析】利用公式SKIPIF1<0，令SKIPIF1<0即可求值．【詳解】解：SKIPIF1<0．令SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0．故答案為：2．【點(diǎn)睛】本題考查了二項(xiàng)式定理的應(yīng)用、組合數(shù)的計(jì)算公式，考查了推理能力與計(jì)算能力，屬于基礎(chǔ)題．13．已知SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)是SKIPIF1<0，則SKIPIF1<0.【答案】SKIPIF1<0【分析】結(jié)合二項(xiàng)展開式通項(xiàng)，根據(jù)SKIPIF1<0的系數(shù)可構(gòu)造方程求得結(jié)果.【詳解】因?yàn)镾KIPIF1<0展開式通項(xiàng)為SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0，即SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0.故答案為：SKIPIF1<0.題型二二項(xiàng)式系數(shù)問題策略方法二項(xiàng)式系數(shù)和：(a＋b)n的展開式中二項(xiàng)式系數(shù)的和為Ceq\o\al(0,n)＋Ceq\o\al(1,n)＋…＋Ceq\o\al(n,n)＝2n．【典例1】（單選題）SKIPIF1<0展開式中的各二項(xiàng)式系數(shù)之和為SKIPIF1<0，則SKIPIF1<0的值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】利用二項(xiàng)式的系數(shù)和可得出關(guān)于SKIPIF1<0的等式，解之即可.【詳解】SKIPIF1<0展開式中的各二項(xiàng)式系數(shù)之和為SKIPIF1<0，解得SKIPIF1<0.故選：A.【典例2】（單選題）．若SKIPIF1<0二項(xiàng)展開式中的各項(xiàng)的二項(xiàng)式系數(shù)只有第SKIPIF1<0項(xiàng)最大，則展開式的常數(shù)項(xiàng)的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】根據(jù)二項(xiàng)式系數(shù)的性質(zhì)得到SKIPIF1<0，再寫出展開式的通項(xiàng)，即可求出常數(shù)項(xiàng).【詳解】因?yàn)槎?xiàng)展開式中的各項(xiàng)的二項(xiàng)式系數(shù)只有第SKIPIF1<0項(xiàng)最大，所以SKIPIF1<0，則SKIPIF1<0展開式的通項(xiàng)為SKIPIF1<0（SKIPIF1<0且SKIPIF1<0），令SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，即展開式中常數(shù)項(xiàng)為SKIPIF1<0.故選：D【題型訓(xùn)練】一、單選題1．SKIPIF1<0的展開式中二項(xiàng)式系數(shù)最大的項(xiàng)是（

）A．160 B．240 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根據(jù)二項(xiàng)式系數(shù)的性質(zhì)求解即可.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0的展開式中二項(xiàng)式系數(shù)最大為SKIPIF1<0，即展開式的第4項(xiàng)，即SKIPIF1<0.故選：C．2．已知二項(xiàng)式SKIPIF1<0的展開式中僅有第SKIPIF1<0項(xiàng)的二項(xiàng)式系數(shù)最大，則SKIPIF1<0為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】分析可知，二項(xiàng)式SKIPIF1<0的展開式共SKIPIF1<0項(xiàng)，即可求出SKIPIF1<0的值.【詳解】因?yàn)槎?xiàng)式SKIPIF1<0的展開式中僅有第SKIPIF1<0項(xiàng)的二項(xiàng)式系數(shù)最大，則二項(xiàng)式SKIPIF1<0的展開式共SKIPIF1<0項(xiàng)，即SKIPIF1<0，解得SKIPIF1<0.故選：A.3．已知SKIPIF1<0的展開式中第4項(xiàng)與第5項(xiàng)的二項(xiàng)式系數(shù)相等，則展開式中的SKIPIF1<0項(xiàng)的系數(shù)為（

）A．―4 B．84 C．―280 D．560【答案】B【分析】根據(jù)二項(xiàng)式系數(shù)的性質(zhì)求得SKIPIF1<0，再根據(jù)二項(xiàng)式展開的通項(xiàng)即可求得指定項(xiàng)的系數(shù).【詳解】因?yàn)镾KIPIF1<0的展開式中第4項(xiàng)與第5項(xiàng)的二項(xiàng)式系數(shù)相等，所以SKIPIF1<0．則SKIPIF1<0又因?yàn)镾KIPIF1<0的展開式的通項(xiàng)公式為SKIPIF1<0，令SKIPIF1<0，所以展開式中的SKIPIF1<0項(xiàng)的系數(shù)為SKIPIF1<0．故選：B.4．若SKIPIF1<0的展開式中所有項(xiàng)的二項(xiàng)式系數(shù)之和為16，則SKIPIF1<0的展開式中的常數(shù)項(xiàng)為（

）A．6 B．8 C．28 D．56【答案】C【分析】根據(jù)SKIPIF1<0的展開式中所有項(xiàng)的二項(xiàng)式系數(shù)之和求出n的值，從而寫出SKIPIF1<0的展開式的通項(xiàng)公式，再令x的指數(shù)為0，即可求解常數(shù)項(xiàng)．【詳解】由SKIPIF1<0的展開式中所有項(xiàng)的二項(xiàng)式系數(shù)之和為16，得SKIPIF1<0，所以SKIPIF1<0，則二項(xiàng)式SKIPIF1<0的展開式的通項(xiàng)公式為SKIPIF1<0（SKIPIF1<0且SKIPIF1<0），令SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0的展開式中的常數(shù)項(xiàng)為28，故選：C．5．若SKIPIF1<0的展開式中第3項(xiàng)與第9項(xiàng)的系數(shù)相等，則展開式中二項(xiàng)式系數(shù)最大的項(xiàng)為（

）A．第4項(xiàng) B．第5項(xiàng) C．第6項(xiàng) D．第7項(xiàng)【答案】C【分析】根據(jù)二項(xiàng)展開式可知SKIPIF1<0，計(jì)算出SKIPIF1<0，即可知二項(xiàng)式系數(shù)最大為SKIPIF1<0，即為第6項(xiàng).【詳解】由二項(xiàng)式定理可得第3項(xiàng)與第9項(xiàng)的系數(shù)分別為SKIPIF1<0和SKIPIF1<0，即SKIPIF1<0，由二項(xiàng)式系數(shù)性質(zhì)可得SKIPIF1<0；因此展開式中二項(xiàng)式系數(shù)最大的項(xiàng)為SKIPIF1<0，是第6項(xiàng).故選：C6．二項(xiàng)式SKIPIF1<0的展開式中，含SKIPIF1<0項(xiàng)的二項(xiàng)式系數(shù)為（

）A．84 B．56 C．35 D．21【答案】B【分析】易知展開式中，含SKIPIF1<0項(xiàng)的二項(xiàng)式系數(shù)為SKIPIF1<0，再利用組合數(shù)的性質(zhì)求解.【詳解】解：因?yàn)槎?xiàng)式為SKIPIF1<0，所以其展開式中，含SKIPIF1<0項(xiàng)的二項(xiàng)式系數(shù)為：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.故選：B二、填空題7．若SKIPIF1<0展開式的二項(xiàng)式系數(shù)和為64，則展開式中第三項(xiàng)的二項(xiàng)式系數(shù)為.【答案】SKIPIF1<0【分析】根據(jù)二項(xiàng)式系數(shù)和得到SKIPIF1<0，再計(jì)算第三項(xiàng)的二項(xiàng)式系數(shù)即可.【詳解】SKIPIF1<0展開式的二項(xiàng)式系數(shù)和為SKIPIF1<0，故SKIPIF1<0，展開式中第三項(xiàng)的二項(xiàng)式系數(shù)為SKIPIF1<0.故答案為：SKIPIF1<0.8．若SKIPIF1<0的展開式的奇數(shù)項(xiàng)的二項(xiàng)式系數(shù)和為16，則展開式中SKIPIF1<0的系數(shù)為.【答案】SKIPIF1<0【分析】由展開式的奇數(shù)項(xiàng)的二項(xiàng)式系數(shù)和為16可得SKIPIF1<0，則SKIPIF1<0展開式中第SKIPIF1<0項(xiàng)為SKIPIF1<0，令SKIPIF1<0可得答案.【詳解】因SKIPIF1<0的展開式的奇數(shù)項(xiàng)的二項(xiàng)式系數(shù)和為16，則SKIPIF1<0.則SKIPIF1<0展開式中第SKIPIF1<0項(xiàng)為SKIPIF1<0.令SKIPIF1<0可得SKIPIF1<0，則SKIPIF1<0的系數(shù)為SKIPIF1<0.故答案為：SKIPIF1<09．SKIPIF1<0的展開式中含SKIPIF1<0項(xiàng)的二項(xiàng)式系數(shù)為.【答案】SKIPIF1<0【分析】寫出二項(xiàng)式展開式的通項(xiàng)公式，確定第五項(xiàng)中k的值，再求二項(xiàng)式系數(shù)．【詳解】由題意知：通項(xiàng)為SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0的展開式中含SKIPIF1<0項(xiàng)為第六項(xiàng)，第六項(xiàng)的二項(xiàng)式系數(shù)為：SKIPIF1<0．故答案為：SKIPIF1<0．10．SKIPIF1<0的展開式中二項(xiàng)式系數(shù)最大的項(xiàng)是.【答案】SKIPIF1<0【分析】根據(jù)二項(xiàng)式系數(shù)的性質(zhì)即可知SKIPIF1<0最大，由二項(xiàng)式展開式的通項(xiàng)特征即可求解.【詳解】SKIPIF1<0的二項(xiàng)展開式有7項(xiàng)，其二項(xiàng)式系數(shù)為SKIPIF1<0，由組合數(shù)的性質(zhì)可知SKIPIF1<0最大，故由二項(xiàng)式定理得二項(xiàng)式系數(shù)最大的一項(xiàng)是SKIPIF1<0.故答案為：SKIPIF1<011．已知SKIPIF1<0的二項(xiàng)展開式中第3項(xiàng)與第10項(xiàng)的二項(xiàng)式系數(shù)相等，則展開式中含SKIPIF1<0的系數(shù)為.【答案】SKIPIF1<0【分析】根據(jù)題意求得SKIPIF1<0，得到二項(xiàng)式為SKIPIF1<0，結(jié)合展開式的通項(xiàng)，即可求解.【詳解】因?yàn)镾KIPIF1<0的二項(xiàng)展開式中第3項(xiàng)與第10項(xiàng)的二項(xiàng)式系數(shù)相等，可得SKIPIF1<0，即SKIPIF1<0，即二項(xiàng)式為SKIPIF1<0，其展開式的通項(xiàng)為SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0，即展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0.故答案為：SKIPIF1<0.12．已知SKIPIF1<0的展開式中第9項(xiàng)、第10項(xiàng)、第11項(xiàng)的二項(xiàng)式系數(shù)成等差數(shù)列，則SKIPIF1<0.【答案】14或23【分析】根據(jù)二項(xiàng)式系數(shù)的定義列出等式，解方程即可求得SKIPIF1<0或SKIPIF1<0.【詳解】由題意可得SKIPIF1<0成等差數(shù)列，則SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.故答案為：14或23題型三二項(xiàng)展開式中各項(xiàng)系數(shù)的和問題策略方法常用賦值法，參考題型六【典例1】（單選題）已知SKIPIF1<0的展開式中所有項(xiàng)的系數(shù)之和為256，則SKIPIF1<0（

）A．3 B．4 C．6 D．8【答案】B【分析】令SKIPIF1<0，得到SKIPIF1<0，即可求解.【詳解】由二項(xiàng)式SKIPIF1<0的展開式中所有項(xiàng)的系數(shù)之和為256，令SKIPIF1<0，可得SKIPIF1<0，解得SKIPIF1<0.故選：B.【題型訓(xùn)練】一、單選題1．若SKIPIF1<0的展開式中常數(shù)項(xiàng)等于SKIPIF1<0，則其展開式各項(xiàng)系數(shù)之和為（

）A．1 B．32 C．0 D．64【答案】C【分析】寫出二項(xiàng)式的通項(xiàng)，根據(jù)展開式中常數(shù)項(xiàng)等于SKIPIF1<0，則就出參數(shù)SKIPIF1<0，則賦值給SKIPIF1<0即可求出展開式各項(xiàng)系數(shù)之和.【詳解】因?yàn)镾KIPIF1<0的展開式中常數(shù)項(xiàng)等于SKIPIF1<0，所以由SKIPIF1<0，當(dāng)SKIPIF1<0，此時(shí)常數(shù)項(xiàng)為：SKIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0，其展開式各項(xiàng)系數(shù)之和為0，故選：C.2．已知SKIPIF1<0的展開式中所有項(xiàng)的系數(shù)和為512，則展開式中的常數(shù)項(xiàng)為（

）A．－756 B．756 C．－2268 D．2268【答案】D【分析】利用賦值法結(jié)合條件可得SKIPIF1<0，然后利用展開式的通項(xiàng)根據(jù)結(jié)合條件即得.【詳解】令SKIPIF1<0可得展開式中所有項(xiàng)的系數(shù)和為SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，所以展開式中的常數(shù)項(xiàng)為：SKIPIF1<0．故選：D．3．已知SKIPIF1<0的展開式中各項(xiàng)系數(shù)之和為0，則展開式中SKIPIF1<0的系數(shù)為（

）A．28 B．-28 C．45 D．-45【答案】A【分析】根據(jù)展開式各項(xiàng)系數(shù)之和可得SKIPIF1<0的值，從而可得展開式的通項(xiàng)，進(jìn)而可得SKIPIF1<0的系數(shù).【詳解】SKIPIF1<0的展開式中各項(xiàng)系數(shù)之和為0所以令SKIPIF1<0得SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0的通項(xiàng)為SKIPIF1<0所以展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0.故選：A.4．已知SKIPIF1<0的二項(xiàng)展開式中，第SKIPIF1<0項(xiàng)與第SKIPIF1<0項(xiàng)的系數(shù)相等，則所有項(xiàng)的系數(shù)之和為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】利用二項(xiàng)式定理求得SKIPIF1<0的展開通項(xiàng)，從而利用SKIPIF1<0與SKIPIF1<0的系數(shù)相等得到關(guān)于SKIPIF1<0的方程，進(jìn)而求得SKIPIF1<0的值，由此得解.【詳解】因?yàn)镾KIPIF1<0的展開通項(xiàng)為SKIPIF1<0又因?yàn)榈赟KIPIF1<0項(xiàng)與第SKIPIF1<0項(xiàng)的系數(shù)相等，所以SKIPIF1<0，由二項(xiàng)式系數(shù)的性質(zhì)知SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0的二項(xiàng)展開式中所有項(xiàng)的系數(shù)之和為SKIPIF1<0.故選：C.二、填空題5．已知SKIPIF1<0的展開式中，各項(xiàng)系數(shù)之和為SKIPIF1<0，則二項(xiàng)式系數(shù)之和為.【答案】SKIPIF1<0【分析】令SKIPIF1<0，結(jié)合二項(xiàng)式SKIPIF1<0各項(xiàng)系數(shù)和可求得SKIPIF1<0的值，進(jìn)而可求得該二項(xiàng)式系數(shù)之和.【詳解】因?yàn)镾KIPIF1<0的展開式中，各項(xiàng)系數(shù)之和為SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0，解得SKIPIF1<0，因此，二項(xiàng)式系數(shù)之和為SKIPIF1<0.故答案為：SKIPIF1<0.6．在SKIPIF1<0的展開式中，二項(xiàng)式系數(shù)和是16，則展開式中各項(xiàng)系數(shù)的和為．【答案】16【分析】由二項(xiàng)式系數(shù)的性質(zhì)可求SKIPIF1<0，再利用賦值法求各項(xiàng)系數(shù)和.【詳解】因?yàn)槎?xiàng)式SKIPIF1<0的展開式中，所有二項(xiàng)式系數(shù)的和是16，所以SKIPIF1<0，故SKIPIF1<0，取SKIPIF1<0可得二項(xiàng)式SKIPIF1<0的展開式中各項(xiàng)系數(shù)和為SKIPIF1<0，即16.故答案為：16.7．已知SKIPIF1<0的展開式中各項(xiàng)系數(shù)和為243，則展開式中常數(shù)項(xiàng)為．【答案】80【分析】根據(jù)題意，由各項(xiàng)系數(shù)之和可得SKIPIF1<0，再由二項(xiàng)式展開式的通項(xiàng)公式即可得到結(jié)果.【詳解】由題意，令SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0的展開式第SKIPIF1<0項(xiàng)SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<08．已知SKIPIF1<0的二項(xiàng)展開式的各項(xiàng)系數(shù)和為32，則二項(xiàng)展開式中SKIPIF1<0的系數(shù)為．【答案】10【分析】由二項(xiàng)展開式的各項(xiàng)系數(shù)和為32，求出SKIPIF1<0，用通項(xiàng)公式求解即可.【詳解】因?yàn)镾KIPIF1<0的二項(xiàng)展開式的各項(xiàng)系數(shù)和為32，令SKIPIF1<0得：SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0.通項(xiàng)公式為：SKIPIF1<0，令SKIPIF1<0，得：SKIPIF1<0，所以SKIPIF1<0的系數(shù)為：SKIPIF1<0.故答案為：109．在SKIPIF1<0的展開式中，各項(xiàng)系數(shù)和與二項(xiàng)式系數(shù)和的比值為SKIPIF1<0，則二項(xiàng)展開式中的常數(shù)項(xiàng)為.【答案】240【分析】由已知求得SKIPIF1<0，再根據(jù)二項(xiàng)式通項(xiàng)公式的展開式求出常數(shù)項(xiàng)即可.【詳解】SKIPIF1<0的展開式中，二項(xiàng)式系數(shù)和為SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0的展開式中，各項(xiàng)系數(shù)和為SKIPIF1<0，由題意可得SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0的展開式的通項(xiàng)為SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，故展開式的常數(shù)項(xiàng)為SKIPIF1<0，故答案為：24010．在SKIPIF1<0的二項(xiàng)式中，所有的二項(xiàng)式系數(shù)之和為64，則各項(xiàng)的系數(shù)的絕對(duì)值之和為．【答案】729【分析】根據(jù)二項(xiàng)式系數(shù)之和求出n的值，進(jìn)而設(shè)出各項(xiàng)的系數(shù)，然后采用賦值法即可求得答案.【詳解】由題意SKIPIF1<0的二項(xiàng)式中，所有的二項(xiàng)式系數(shù)之和為64，即SKIPIF1<0，設(shè)SKIPIF1<0的各項(xiàng)的系數(shù)為SKIPIF1<0，則各項(xiàng)的系數(shù)的絕對(duì)值之和為SKIPIF1<0，即為SKIPIF1<0中各項(xiàng)的系數(shù)的和，令SKIPIF1<0，SKIPIF1<0，即各項(xiàng)的系數(shù)的絕對(duì)值之和為SKIPIF1<0，故答案為：729題型四三項(xiàng)展開式的問題策略方法求三項(xiàng)展開式中某些特定項(xiàng)的系數(shù)的方法(1)通過變形先把三項(xiàng)式轉(zhuǎn)化為二項(xiàng)式，再用二項(xiàng)式定理求解．(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．【典例1】（單選題）SKIPIF1<0的展開式中，SKIPIF1<0的系數(shù)為（

）A．80 B．60 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】由題得SKIPIF1<0，再利用二項(xiàng)式的通項(xiàng)即可得到答案.【詳解】SKIPIF1<0，則其展開式通項(xiàng)為SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0的展開式中含SKIPIF1<0的項(xiàng)為SKIPIF1<0，所以SKIPIF1<0的系數(shù)為SKIPIF1<0，故選：D．【題型訓(xùn)練】一、單選題1．SKIPIF1<0的展開式共（

）A．10項(xiàng) B．15項(xiàng) C．20項(xiàng) D．21項(xiàng)【答案】B【分析】根據(jù)二項(xiàng)式定理的展開式項(xiàng)數(shù)即可得出結(jié)論.【詳解】∵SKIPIF1<0，由二項(xiàng)式定理可知，SKIPIF1<0展示式中共有SKIPIF1<0項(xiàng)，∴SKIPIF1<0的展開式共有SKIPIF1<0項(xiàng).故選：B．2．在SKIPIF1<0的展開式中，SKIPIF1<0的系數(shù)是（

）A．24 B．32 C．36 D．40【答案】D【分析】根據(jù)題意，SKIPIF1<0的項(xiàng)為SKIPIF1<0，化簡后即可求解.【詳解】根據(jù)題意，SKIPIF1<0的項(xiàng)為SKIPIF1<0，所以SKIPIF1<0的系數(shù)是SKIPIF1<0.故選：D．3．SKIPIF1<0的展開式中的常數(shù)項(xiàng)為（

）A．588 B．589 C．798 D．799【答案】B【分析】因?yàn)镾KIPIF1<0展開式中的項(xiàng)可以看作8個(gè)含有三個(gè)單項(xiàng)式SKIPIF1<0各取一個(gè)相乘而得，分析組合可能，結(jié)合組合數(shù)運(yùn)算求解.【詳解】因?yàn)镾KIPIF1<0展開式中的項(xiàng)可以看作8個(gè)含有三個(gè)單項(xiàng)式SKIPIF1<0中各取一個(gè)相乘而得，若得到常數(shù)項(xiàng)，則有：①8個(gè)1；②2個(gè)SKIPIF1<0，1個(gè)SKIPIF1<0，5個(gè)1；③4個(gè)SKIPIF1<0，2個(gè)SKIPIF1<0，2個(gè)1；所以展開式中的常數(shù)項(xiàng)為SKIPIF1<0.故選：B.4．SKIPIF1<0的展開式中含SKIPIF1<0的項(xiàng)的系數(shù)為（

）A．SKIPIF1<0 B．60 C．SKIPIF1<0 D．30【答案】A【分析】將SKIPIF1<0轉(zhuǎn)化為SKIPIF1<0，根據(jù)二項(xiàng)式定理求出含SKIPIF1<0的項(xiàng)，即可得出答案.【詳解】因?yàn)镾KIPIF1<0的展開式中含SKIPIF1<0的項(xiàng)為SKIPIF1<0，SKIPIF1<0的展開式中含SKIPIF1<0的項(xiàng)為SKIPIF1<0，所以SKIPIF1<0的展開式中含SKIPIF1<0的項(xiàng)的系數(shù)為SKIPIF1<0.故選：A.5．已知SKIPIF1<0展開式的各項(xiàng)系數(shù)之和為SKIPIF1<0，則展開式中SKIPIF1<0的系數(shù)為（

）A．270 B．SKIPIF1<0 C．330 D．SKIPIF1<0【答案】D【分析】令SKIPIF1<0，得SKIPIF1<0，得SKIPIF1<0.再根據(jù)二項(xiàng)展開式的通項(xiàng)公式即可求解.【詳解】令SKIPIF1<0，則SKIPIF1<0，得SKIPIF1<0.所以SKIPIF1<0SKIPIF1<0SKIPIF1<0，又因?yàn)橹挥蠸KIPIF1<0，SKIPIF1<0展開式中有含SKIPIF1<0的項(xiàng)，所以SKIPIF1<0的系數(shù)為SKIPIF1<0.故選：D6．已知SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】利用二項(xiàng)式定理可分別得到SKIPIF1<0和SKIPIF1<0展開式的通項(xiàng)，令SKIPIF1<0即可討論得到SKIPIF1<0的取值，結(jié)合展開式通項(xiàng)，利用SKIPIF1<0的系數(shù)構(gòu)造方程即可求得SKIPIF1<0的值.【詳解】SKIPIF1<0展開式的通項(xiàng)為：SKIPIF1<0，SKIPIF1<0且SKIPIF1<0；SKIPIF1<0展開式的通項(xiàng)為：SKIPIF1<0，SKIPIF1<0且SKIPIF1<0；令SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0，SKIPIF1<0的系數(shù)為SKIPIF1<0，解得：SKIPIF1<0.故選：A.7．SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為12，則SKIPIF1<0（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根據(jù)乘法的運(yùn)算法則，結(jié)合組合數(shù)的性質(zhì)、二倍角的余弦公式進(jìn)行求解即可.【詳解】SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)可以看成：6個(gè)因式SKIPIF1<0中選取5個(gè)因式提供SKIPIF1<0，余下一個(gè)因式中提供SKIPIF1<0或者6個(gè)因式SKIPIF1<0中選取4個(gè)因式提供SKIPIF1<0，余下兩個(gè)因式中均提供SKIPIF1<0，故SKIPIF1<0的系數(shù)為SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，故選：C二、填空題8．SKIPIF1<0展開式中，SKIPIF1<0項(xiàng)的系數(shù)為.【答案】SKIPIF1<0【分析】由二項(xiàng)式定理求解．【詳解】SKIPIF1<0，∵SKIPIF1<0的指數(shù)是3，∴得到SKIPIF1<0，∵SKIPIF1<0的指數(shù)是2，得到SKIPIF1<0，∴SKIPIF1<0項(xiàng)的系數(shù)為SKIPIF1<0.故答案為：SKIPIF1<09．在SKIPIF1<0的展開式中，SKIPIF1<0的系數(shù)為．【答案】SKIPIF1<0【分析】原多項(xiàng)式中寫出含SKIPIF1<0的項(xiàng)，然后再從SKIPIF1<0中寫出含SKIPIF1<0的項(xiàng)，即可得含SKIPIF1<0的系數(shù).【詳解】由含SKIPIF1<0的項(xiàng)中對(duì)應(yīng)SKIPIF1<0的指數(shù)分別為SKIPIF1<0，所以SKIPIF1<0，對(duì)于SKIPIF1<0中含SKIPIF1<0的項(xiàng)為SKIPIF1<0，所以含SKIPIF1<0的系數(shù)是SKIPIF1<0.故答案為：SKIPIF1<0.10．SKIPIF1<0的展開式中，含SKIPIF1<0的項(xiàng)的系數(shù)為．【答案】SKIPIF1<0【分析】SKIPIF1<0的展開式中，含SKIPIF1<0的項(xiàng)有以下兩類，第一類：4個(gè)因式中有1個(gè)取到SKIPIF1<0，其余3個(gè)都取到2；第二類：4個(gè)因式中有2個(gè)取到SKIPIF1<0，其余2個(gè)都取到2，結(jié)合組合數(shù)即可求解.【詳解】SKIPIF1<0的展開式中，含SKIPIF1<0的項(xiàng)有以下兩類：第一類：4個(gè)因式中有1個(gè)取到SKIPIF1<0，其余3個(gè)都取到2，即SKIPIF1<0第二類：4個(gè)因式中有2個(gè)取到SKIPIF1<0，其余2個(gè)都取到2，即SKIPIF1<0所以SKIPIF1<0的展開式中含SKIPIF1<0的項(xiàng)為SKIPIF1<0，故含SKIPIF1<0的項(xiàng)的系數(shù)為SKIPIF1<0.故答案為：SKIPIF1<011．SKIPIF1<0展開式中常數(shù)項(xiàng)是.（答案用數(shù)字作答）【答案】SKIPIF1<0【分析】根據(jù)二項(xiàng)式展開式的通項(xiàng)化簡得常數(shù)項(xiàng)滿足SKIPIF1<0，即可代入求解.【詳解】SKIPIF1<0的展開式的通項(xiàng)為SKIPIF1<0，SKIPIF1<0,令SKIPIF1<0,則SKIPIF1<0或SKIPIF1<0，或SKIPIF1<0，所以常數(shù)項(xiàng)為SKIPIF1<0，故答案為：SKIPIF1<012．已知二項(xiàng)式SKIPIF1<0的展開式中含SKIPIF1<0的項(xiàng)的系數(shù)為SKIPIF1<0，則SKIPIF1<0．【答案】2【分析】SKIPIF1<0表示有5個(gè)SKIPIF1<0因式相乘，根據(jù)SKIPIF1<0的來源分析即可求出答案.【詳解】SKIPIF1<0表示有5個(gè)SKIPIF1<0因式相乘，SKIPIF1<0來源如下：有1個(gè)SKIPIF1<0提供SKIPIF1<0，有3個(gè)SKIPIF1<0提供SKIPIF1<0，有1個(gè)SKIPIF1<0提供常數(shù)，此時(shí)SKIPIF1<0系數(shù)是SKIPIF1<0，即SKIPIF1<0，解得：SKIPIF1<0故答案為：SKIPIF1<0.13．在SKIPIF1<0的展開式中，SKIPIF1<0項(xiàng)的系數(shù)為.【答案】220【分析】根據(jù)給定條件，分析展開式中SKIPIF1<0項(xiàng)出現(xiàn)的情況，再列式計(jì)算作答.【詳解】SKIPIF1<0的展開式通項(xiàng)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0展開式中SKIPIF1<0的最高指數(shù)小于12，而SKIPIF1<0的指數(shù)小于等于SKIPIF1<0，因此SKIPIF1<0中SKIPIF1<0的指數(shù)是負(fù)整數(shù)，要得到SKIPIF1<0項(xiàng)，當(dāng)且僅當(dāng)SKIPIF1<0，所以展開式中SKIPIF1<0項(xiàng)的系數(shù)是SKIPIF1<0展開式中SKIPIF1<0項(xiàng)的系數(shù)SKIPIF1<0.故答案為：220題型五兩個(gè)二項(xiàng)式乘積展開式的系數(shù)策略方法求解形如(a＋b)n(c＋d)m的展開式問題的思路(1)若n，m中一個(gè)比較小，可考慮把它展開得到多個(gè)，如(a＋b)2(c＋d)m＝(a2＋2ab＋b2)(c＋d)m，然后展開分別求解．(2)觀察(a＋b)(c＋d)是否可以合并，如(1＋x)5(1－x)7＝[(1＋x)(1－x)]5(1－x)2＝(1－x2)5(1－x)2．(3)分別得到(a＋b)n，(c＋d)m的通項(xiàng)公式，綜合考慮．【典例1】（單選題）SKIPIF1<0的展開式中含SKIPIF1<0項(xiàng)的系數(shù)為（

）A．10 B．12 C．4 D．5【答案】A【分析】利用二項(xiàng)式定理的通項(xiàng)公式進(jìn)行分類討論即可求解.【詳解】SKIPIF1<0的二項(xiàng)展開式的通項(xiàng)為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的展開式中含SKIPIF1<0項(xiàng)為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的展開式中含SKIPIF1<0項(xiàng)為SKIPIF1<0；所以SKIPIF1<0的展開式中含SKIPIF1<0項(xiàng)的系數(shù)為SKIPIF1<0.故選：A.【題型訓(xùn)練】一、單選題1．SKIPIF1<0的展開式中，SKIPIF1<0的系數(shù)為（

）A．200 B．40 C．120 D．80【答案】B【分析】根據(jù)二項(xiàng)式定理先求通項(xiàng)，再根據(jù)項(xiàng)進(jìn)行分別求系數(shù)，最后求和.【詳解】SKIPIF1<0，而SKIPIF1<0展開式的通項(xiàng)為SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的系數(shù)為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的系數(shù)為SKIPIF1<0，所以SKIPIF1<0的系數(shù)為SKIPIF1<0，故選：B2．SKIPIF1<0的展開式中各項(xiàng)系數(shù)之和為SKIPIF1<0，則該展開式中常數(shù)項(xiàng)為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】取SKIPIF1<0代入計(jì)算得到SKIPIF1<0，確定SKIPIF1<0展開式的通項(xiàng)，分別取SKIPIF1<0和SKIPIF1<0計(jì)算得到答案.【詳解】SKIPIF1<0的展開式中各項(xiàng)系數(shù)之和為SKIPIF1<0，令SKIPIF1<0，可知SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0展開式的通項(xiàng)為SKIPIF1<0，分別取SKIPIF1<0和SKIPIF1<0得到常數(shù)項(xiàng)為：SKIPIF1<0，故選：C3．已知SKIPIF1<0展開式中SKIPIF1<0的系數(shù)為48，則實(shí)數(shù)SKIPIF1<0（

）A．1 B．SKIPIF1<0 C．2 D．SKIPIF1<0【答案】A【分析】根據(jù)二項(xiàng)式的通項(xiàng)公式進(jìn)行求解即可.【詳解】二項(xiàng)式SKIPIF1<0的通項(xiàng)公式為：SKIPIF1<0SKI