新高考數(shù)學(xué)一輪復(fù)習(xí)提升練習(xí)考向45 二項(xiàng)式定理 (含解析)

考向45二項(xiàng)式定理1．（2021·山東·高考真題）SKIPIF1<0的二項(xiàng)展開式中，所有項(xiàng)的二項(xiàng)式系數(shù)之和是（）A．0 B．SKIPIF1<0 C．SKIPIF1<0 D．32【答案】D【分析】根據(jù)SKIPIF1<0的二項(xiàng)展開式系數(shù)之和為SKIPIF1<0求解即可【詳解】SKIPIF1<0的二項(xiàng)展開式中所有項(xiàng)的二項(xiàng)式系數(shù)之和為SKIPIF1<0故選：D2．（2021·湖南·高考真題）SKIPIF1<0的展開式中常數(shù)項(xiàng)是______．（用數(shù)字作答）【答案】15【分析】寫出二項(xiàng)展開式的通項(xiàng)，由SKIPIF1<0的指數(shù)為0求得SKIPIF1<0值，則答案可求．【詳解】解：由SKIPIF1<0．取SKIPIF1<0，得SKIPIF1<0．SKIPIF1<0SKIPIF1<0展開式中常數(shù)項(xiàng)為SKIPIF1<0．故答案為：15．1.求二項(xiàng)展開式的特定項(xiàng)問題,實(shí)質(zhì)是考查通項(xiàng)的特點(diǎn),一般需要建立方程求k,再將k的值代回通項(xiàng)求解,注意k的取值范圍（）.（1）第項(xiàng)：：此時(shí)k+1=m,直接代入通項(xiàng).（2）常數(shù)項(xiàng)：即這項(xiàng)中不含“變?cè)?令通項(xiàng)中“變?cè)钡膬缰笖?shù)為0建立方程.（3）有理項(xiàng)：令通項(xiàng)中“變?cè)钡膬缰笖?shù)為整數(shù)建立方程.2．解題技巧：（1）形如(ax＋b)n，(ax2＋bx＋c)m(a，b，c∈R)的式子求其展開式的各項(xiàng)系數(shù)之和，常用賦值法，只需令x＝1即可．（2）對(duì)形如(ax＋by)n(a，b∈R)的式子求其展開式各項(xiàng)系數(shù)之和，只需令x＝y(tǒng)＝1即可．（3）若f(x)＝a0＋a1x＋a2x2＋…＋anxn，則f(x)展開式中各項(xiàng)系數(shù)之和為f(1)，奇數(shù)項(xiàng)系數(shù)之和為a0＋a2＋a4＋…＝SKIPIF1<0，偶數(shù)項(xiàng)系數(shù)之和為a1＋a3＋a5＋…＝SKIPIF1<0.1.二項(xiàng)式定理(a+b)n=an+an-1b+…+an-rbr+…+bn(n∈N*)2.二項(xiàng)展開式的通項(xiàng)Tr+1=an-rbr,它表示第r+1項(xiàng)3.二項(xiàng)式系數(shù)，，…，【知識(shí)拓展】1.=1，=1，=+.

2.=(0≤m≤n).

3.二項(xiàng)式系數(shù)先增后減中間項(xiàng)最大.當(dāng)n為偶數(shù)時(shí),第+1項(xiàng)的二項(xiàng)式系數(shù)最大,最大值為;當(dāng)n為奇數(shù)時(shí),第項(xiàng)和第項(xiàng)的二項(xiàng)式系數(shù)最大,最大值為或.4.各二項(xiàng)式系數(shù)和:+++…+=2n，+++…=+++…=2n-1.

1．（2021·云南大理·模擬預(yù)測(cè)（理））二項(xiàng)式SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)是SKIPIF1<0，則SKIPIF1<0（）A．SKIPIF1<0 B．1 C．SKIPIF1<0 D．SKIPIF1<02．（2021·廣西南寧·模擬預(yù)測(cè)（理））已知SKIPIF1<0的展開式中各項(xiàng)系數(shù)的和為3，則該展開式中常數(shù)項(xiàng)為（）A．80 B．160 C．240 D．3203．（2021·浙江嘉興·模擬預(yù)測(cè)）已知多項(xiàng)式SKIPIF1<0，則SKIPIF1<0______，SKIPIF1<0______.4．（2021·上?！つM預(yù)測(cè)）SKIPIF1<0二項(xiàng)展開式中的x的有理項(xiàng)的系數(shù)和為______1．（2021·上?！つM預(yù)測(cè)）二項(xiàng)式SKIPIF1<0的展開式中，其中是有理項(xiàng)的項(xiàng)數(shù)共有（）A．4項(xiàng) B．7項(xiàng) C．5項(xiàng) D．6項(xiàng)2．（2021·遼寧·撫順市第二中學(xué)模擬預(yù)測(cè)）SKIPIF1<0的展開式中，第二項(xiàng)為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2021·吉林長(zhǎng)春·一模（理））SKIPIF1<0展開式中，SKIPIF1<0的系數(shù)是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2021·全國(guó)·模擬預(yù)測(cè)）在SKIPIF1<0的二項(xiàng)展開式中，SKIPIF1<0的系數(shù)為（）A．40 B．20 C．-40 D．-205．（2021·全國(guó)·模擬預(yù)測(cè)）SKIPIF1<0展開式中SKIPIF1<0的系數(shù)是（）A．10 B．SKIPIF1<0 C．5 D．SKIPIF1<06．（2021·浙江·模擬預(yù)測(cè)）SKIPIF1<0的展開式中的常數(shù)項(xiàng)為SKIPIF1<032，則實(shí)數(shù)a的值為________；展開式中含SKIPIF1<0項(xiàng)的系數(shù)為________.7．（2021·全國(guó)·模擬預(yù)測(cè)）若二項(xiàng)式SKIPIF1<0展開式的各項(xiàng)系數(shù)和為81，則展開式中的常數(shù)項(xiàng)是___________.8．（2021·上?！つM預(yù)測(cè)）在SKIPIF1<0的展開式中，SKIPIF1<0與SKIPIF1<0項(xiàng)的系數(shù)和為___________.（結(jié)果用數(shù)值表示）9．（2021·全國(guó)·模擬預(yù)測(cè)（理））已知二項(xiàng)式SKIPIF1<0的展開式中，常數(shù)項(xiàng)為SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0___________.10．（2021·全國(guó)·模擬預(yù)測(cè)（理））已知SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0，SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0，SKIPIF1<0，則非零常數(shù)SKIPIF1<0的值為________．11．（2021·甘肅·嘉峪關(guān)市第一中學(xué)三模（理））若SKIPIF1<0展開式的二項(xiàng)式系數(shù)之和為64，則展開式中的常數(shù)項(xiàng)是________．12．（2021·全國(guó)·模擬預(yù)測(cè)）已知SKIPIF1<0的展開式的二項(xiàng)式系數(shù)和為128，若SKIPIF1<0SKIPIF1<0，則SKIPIF1<0________．1．（2020·山東·高考真題）在SKIPIF1<0的二項(xiàng)展開式中，第SKIPIF1<0項(xiàng)的二項(xiàng)式系數(shù)是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2021·江蘇·高考真題）已知SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為40，則SKIPIF1<0等于（）A．5 B．6 C．7 D．83．（2020·北京·高考真題）在SKIPIF1<0的展開式中，SKIPIF1<0的系數(shù)為（）．A．SKIPIF1<0 B．5 C．SKIPIF1<0 D．104．（2020·全國(guó)·高考真題（理））SKIPIF1<0的展開式中x3y3的系數(shù)為（）A．5 B．10C．15 D．205．（2019·全國(guó)·高考真題（理））（1+2x2）（1+x）4的展開式中x3的系數(shù)為A．12 B．16 C．20 D．246．（2021·浙江·高考真題）已知多項(xiàng)式SKIPIF1<0，則SKIPIF1<0___________，SKIPIF1<0___________.7．（2020·浙江·高考真題）設(shè)SKIPIF1<0，則SKIPIF1<0________；SKIPIF1<0________．8．（2021·天津·高考真題）在SKIPIF1<0的展開式中，SKIPIF1<0的系數(shù)是__________．9．（2020·天津·高考真題）在SKIPIF1<0的展開式中，SKIPIF1<0的系數(shù)是_________．10．（2019·江蘇·高考真題）設(shè)SKIPIF1<0.已知SKIPIF1<0.（1）求n的值；（2）設(shè)SKIPIF1<0，其中SKIPIF1<0，求SKIPIF1<0的值.1．【答案】B【分析】根據(jù)多項(xiàng)式乘法法則及排列組合知識(shí)即可求解.【詳解】解：SKIPIF1<0可以看作8個(gè)因式SKIPIF1<0的乘積，根據(jù)多項(xiàng)式乘法法則，展開式中SKIPIF1<0項(xiàng)需要從8個(gè)因式中取7個(gè)SKIPIF1<0和1個(gè)SKIPIF1<0相乘得到，所以由排列組合的知識(shí)有展開式中SKIPIF1<0的系數(shù)SKIPIF1<0，解得SKIPIF1<0，故選：B.2．【答案】D【分析】令SKIPIF1<0解得SKIPIF1<0，再求得SKIPIF1<0展開式的通項(xiàng)公式求解.【詳解】令SKIPIF1<0得SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0展開式的通項(xiàng)為SKIPIF1<0，則SKIPIF1<0展開式中常數(shù)項(xiàng)為SKIPIF1<0．故選：D3．【答案】SKIPIF1<0SKIPIF1<0【分析】設(shè)SKIPIF1<0，利用賦值法可得出SKIPIF1<0，求得SKIPIF1<0，利用賦值法可得出SKIPIF1<0的值.【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0，因?yàn)镾KIPIF1<0，所以，SKIPIF1<0，因此，SKIPIF1<0.故答案為：SKIPIF1<0；SKIPIF1<0.4．【答案】255【分析】易得SKIPIF1<0展開式的通項(xiàng)為SKIPIF1<0，再由SKIPIF1<0為有理數(shù)求解.【詳解】SKIPIF1<0展開式的通項(xiàng)為SKIPIF1<0，若SKIPIF1<0為有理數(shù)，則SKIPIF1<0，所以x的有理項(xiàng)的系數(shù)和為SKIPIF1<0，故答案為：2551．【答案】D【分析】根據(jù)二項(xiàng)展開式的通項(xiàng)公式，由SKIPIF1<0的指數(shù)值為整數(shù)即可解出．【詳解】二項(xiàng)式SKIPIF1<0的展開式中，通項(xiàng)公式為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0時(shí)滿足題意，共6項(xiàng)．故選：D.2．【答案】C【分析】先表示出展開式的通項(xiàng)，再令r=1可求得.【詳解】SKIPIF1<0,第二項(xiàng)是SKIPIF1<0，即SKIPIF1<0=SKIPIF1<0故選：C3．【答案】B【分析】寫出展開式的通項(xiàng)公式SKIPIF1<0，令SKIPIF1<0，即得解【詳解】SKIPIF1<0展開式的通項(xiàng)為SKIPIF1<0，令SKIPIF1<0，故SKIPIF1<0，故選：B.4．【答案】A【分析】由二項(xiàng)式得到展開式通項(xiàng)，進(jìn)而確定SKIPIF1<0的系數(shù).【詳解】SKIPIF1<0的展開式的通項(xiàng)SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，故SKIPIF1<0的系數(shù)為SKIPIF1<0，故選：A.5．【答案】B【分析】前一個(gè)括號(hào)內(nèi)有SKIPIF1<0與SKIPIF1<0兩項(xiàng)，SKIPIF1<0，SKIPIF1<0，所以分兩種情況討論得解.【詳解】前一個(gè)括號(hào)內(nèi)有SKIPIF1<0與SKIPIF1<0兩項(xiàng)，SKIPIF1<0，SKIPIF1<0展開式第SKIPIF1<0項(xiàng)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0展開式SKIPIF1<0系數(shù)為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0時(shí)，SKIPIF1<0不能出現(xiàn)SKIPIF1<0∴SKIPIF1<0的系數(shù)為SKIPIF1<0．故選：B．6．【答案】SKIPIF1<0SKIPIF1<0【分析】先求出SKIPIF1<0的展開式的通項(xiàng)公式為SKIPIF1<0，由SKIPIF1<0，可得SKIPIF1<0，從而可由題意可得SKIPIF1<0，可求出a的值，含SKIPIF1<0項(xiàng)的系數(shù)由SKIPIF1<0展開式的常數(shù)項(xiàng)加上二次項(xiàng)系數(shù)【詳解】因?yàn)镾KIPIF1<0的展開式的通項(xiàng)公式為SKIPIF1<0，所以SKIPIF1<0的展開式中的常數(shù)項(xiàng)為SKIPIF1<0，解得SKIPIF1<0.所以SKIPIF1<0的展開式中含SKIPIF1<0項(xiàng)的系數(shù)為SKIPIF1<0.故答案為：SKIPIF1<0，SKIPIF1<07．【答案】32【分析】利用賦值法求得SKIPIF1<0，結(jié)合二項(xiàng)式展開式的通項(xiàng)公式求得展開式中的常數(shù)項(xiàng).【詳解】令SKIPIF1<0得SKIPIF1<0，二項(xiàng)式SKIPIF1<0展開式的通項(xiàng)公式為SKIPIF1<0，由SKIPIF1<0解得SKIPIF1<0，所以展開式中的常數(shù)項(xiàng)為SKIPIF1<0.故答案為：SKIPIF1<08．【答案】SKIPIF1<0【分析】根據(jù)二項(xiàng)展開式的通項(xiàng)公式以及多項(xiàng)式的乘法原理即可解出．【詳解】因?yàn)镾KIPIF1<0展開式的通項(xiàng)公式為SKIPIF1<0，所以SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0，SKIPIF1<0項(xiàng)的系數(shù)為SKIPIF1<0，即SKIPIF1<0與SKIPIF1<0項(xiàng)的系數(shù)和為SKIPIF1<0．故答案為：SKIPIF1<0．9．【答案】2【分析】寫出二項(xiàng)式的展開式公式SKIPIF1<0，令SKIPIF1<0，結(jié)合題意即可求出參數(shù)a【詳解】二項(xiàng)式SKIPIF1<0的展開式通項(xiàng)公式為SKIPIF1<0令SKIPIF1<0，解得SKIPIF1<0，因?yàn)槌?shù)項(xiàng)為14，所以SKIPIF1<0，解得SKIPIF1<0，故答案為：210．【答案】SKIPIF1<0【分析】根據(jù)題設(shè)二項(xiàng)式分別寫出SKIPIF1<0的系數(shù)SKIPIF1<0、SKIPIF1<0，由已知等量關(guān)系列方程求參數(shù)SKIPIF1<0的值即可.【詳解】SKIPIF1<0的展開式中含SKIPIF1<0的項(xiàng)為：SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0的展開式中含SKIPIF1<0的項(xiàng)為：SKIPIF1<0，∴SKIPIF1<0，由SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0．故答案為：SKIPIF1<011．【答案】60【分析】根據(jù)二項(xiàng)式系數(shù)之和，可求得n值，求得SKIPIF1<0展開式的通項(xiàng)公式，令SKIPIF1<0，求得k值，計(jì)算即可得答案.【詳解】根據(jù)二項(xiàng)式系數(shù)之和為64，可得SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0展開式的通項(xiàng)公式為SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0，所以展開式中的常數(shù)項(xiàng)為SKIPIF1<0.故答案為：6012．【答案】SKIPIF1<0【分析】根據(jù)二項(xiàng)式系數(shù)和，可求得n值，設(shè)SKIPIF1<0，則SKIPIF1<0，所求即為SKIPIF1<0，根據(jù)展開式的通項(xiàng)公式，即可求得SKIPIF1<0，即可得答案.【詳解】由SKIPIF1<0的展開式的二項(xiàng)式系數(shù)和為128，則SKIPIF1<0，∴SKIPIF1<0．設(shè)SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0SKIPIF1<0，∴SKIPIF1<0．故答案為：SKIPIF1<01．【答案】A【分析】本題可通過二項(xiàng)式系數(shù)的定義得出結(jié)果.【詳解】第SKIPIF1<0項(xiàng)的二項(xiàng)式系數(shù)為SKIPIF1<0，故選：A.2．【答案】A【分析】寫出x2項(xiàng)，進(jìn)一步即可解出.【詳解】SKIPIF1<0，所以SKIPIF1<0.故選：A.3．【答案】C【分析】首先寫出展開式的通項(xiàng)公式，然后結(jié)合通項(xiàng)公式確定SKIPIF1<0的系數(shù)即可.【詳解】SKIPIF1<0展開式的通項(xiàng)公式為：SKIPIF1<0，令SKIPIF1<0可得：SKIPIF1<0，則SKIPIF1<0的系數(shù)為：SKIPIF1<0.故選：C.【點(diǎn)睛】二項(xiàng)式定理的核心是通項(xiàng)公式，求解此類問題可以分兩步完成：第一步根據(jù)所給出的條件(特定項(xiàng))和通項(xiàng)公式，建立方程來(lái)確定指數(shù)(求解時(shí)要注意二項(xiàng)式系數(shù)中n和r的隱含條件，即n，r均為非負(fù)整數(shù)，且n≥r，如常數(shù)項(xiàng)指數(shù)為零、有理項(xiàng)指數(shù)為整數(shù)等)；第二步是根據(jù)所求的指數(shù)，再求所求解的項(xiàng)．4．【答案】C【分析】求得SKIPIF1<0展開式的通項(xiàng)公式為SKIPIF1<0（SKIPIF1<0且SKIPIF1<0），即可求得SKIPIF1<0與SKIPIF1<0展開式的乘積為SKIPIF1<0或SKIPIF1<0形式，對(duì)SKIPIF1<0分別賦值為3，1即可求得SKIPIF1<0的系數(shù)，問題得解.【詳解】SKIPIF1<0展開式的通項(xiàng)公式為SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）所以SKIPIF1<0的各項(xiàng)與SKIPIF1<0展開式的通項(xiàng)的乘積可表示為：SKIPIF1<0和SKIPIF1<0在SKIPIF1<0中，令SKIPIF1<0，可得：SKIPIF1<0，該項(xiàng)中SKIPIF1<0的系數(shù)為SKIPIF1<0，在SKIPIF1<0中，令SKIPIF1<0，可得：SKIPIF1<0，該項(xiàng)中SKIPIF1<0的系數(shù)為SKIPIF1<0所以SKIPIF1<0的系數(shù)為SKIPIF1<0故選：C【點(diǎn)睛】本題主要考查了二項(xiàng)式定理及其展開式的通項(xiàng)公式，還考查了賦值法、轉(zhuǎn)化能力及分析能力，屬于中檔題.5．【答案】A【分析】本題利用二項(xiàng)展開式通項(xiàng)公式求展開式指定項(xiàng)的系數(shù)．【詳解】由題意得x3的系數(shù)為SKIPIF1<0，故選A．【點(diǎn)睛】本題主要考查二項(xiàng)式定理，利用展開式通項(xiàng)公式求展開式指定項(xiàng)的系數(shù)．6．【答案】SKIPIF1<0;SKIPIF1<0.【分析】根據(jù)二項(xiàng)展開式定理，分別求出SKIPIF1<0的展開式，即可得出結(jié)論.【詳解】SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<0.7．【答案】SKIPIF1<0SKIPIF1<0【分析】利用二項(xiàng)式展開式的通項(xiàng)公式計(jì)算即可.【詳解】SKIPIF1<0的通項(xiàng)為SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0；SKIPIF1<0.故答案為：SKIPIF1<0；SKIPIF1<0.【點(diǎn)晴】本題主要考查利用二項(xiàng)式定理求指定項(xiàng)的系數(shù)問題，考查學(xué)生的數(shù)學(xué)運(yùn)算能力，是一道基礎(chǔ)題.8．【答案】160【分析】求出二項(xiàng)式的展開式通項(xiàng)，