新高考數(shù)學(xué)二輪復(fù)習(xí)重點突破訓(xùn)練第13講 二項式定理（含解析）

第13講二項式定理真題展示2022新高考一卷第13題SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0（用數(shù)字作答）．【思路分析】由題意依次求出SKIPIF1<0中SKIPIF1<0，SKIPIF1<0項的系數(shù)，求和即可．【解析】SKIPIF1<0的通項公式為SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0．故答案為：SKIPIF1<0．【試題評價】本題考查二項式定理的應(yīng)用，考查運算求解能力，是基礎(chǔ)題．知識要點整理知識點一二項式定理(a＋b)n＝Ceq\o\al(0,n)an＋Ceq\o\al(1,n)an－1b＋Ceq\o\al(2,n)an－2b2＋…＋Ceq\o\al(k,n)an－kbk＋…＋Ceq\o\al(n,n)bn(n∈N*)．(1)這個公式叫做二項式定理．(2)展開式：等號右邊的多項式叫做(a＋b)n的二項展開式，展開式中一共有n＋1項．(3)二項式系數(shù)：各項的系數(shù)Ceq\o\al(k,n)(k∈{0,1,2，…，n})叫做二項式系數(shù)．知識點二二項展開式的通項(a＋b)n展開式的第k＋1項叫做二項展開式的通項，記作Tk＋1＝Ceq\o\al(k,n)an－kbk.知識點三二項式系數(shù)的性質(zhì)對稱性在(a＋b)n的展開式中，與首末兩端“等距離”的兩個二項式系數(shù)相等，即Ceq\o\al(m,n)＝Ceq\o\al(n－m,n)增減性與最大值增減性：當k<eq\f(n＋1,2)時，二項式系數(shù)是逐漸增大的；當k>eq\f(n＋1,2)時，二項式系數(shù)是逐漸減小的．最大值：當n為偶數(shù)時，中間一項的二項式系數(shù)SKIPIF1<0最大；當n為奇數(shù)時，中間兩項的二項式系數(shù)SKIPIF1<0，SKIPIF1<0相等，且同時取得最大值各二項式系數(shù)的和(1)Ceq\o\al(0,n)＋Ceq\o\al(1,n)＋Ceq\o\al(2,n)＋…＋Ceq\o\al(n,n)＝2n；(2)Ceq\o\al(0,n)＋Ceq\o\al(2,n)＋Ceq\o\al(4,n)＋…＝Ceq\o\al(1,n)＋Ceq\o\al(3,n)＋Ceq\o\al(5,n)＋…＝2n－1三年真題一、單選題1．若SKIPIF1<0，則SKIPIF1<0（

）A．40 B．41 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0，故選：B.2．在SKIPIF1<0的二項展開式中，第SKIPIF1<0項的二項式系數(shù)是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】第SKIPIF1<0項的二項式系數(shù)為SKIPIF1<0，故選：A.3．在SKIPIF1<0的展開式中，SKIPIF1<0的系數(shù)為（

）．A．SKIPIF1<0 B．5 C．SKIPIF1<0 D．10【答案】C【詳解】SKIPIF1<0展開式的通項公式為：SKIPIF1<0，令SKIPIF1<0可得：SKIPIF1<0，則SKIPIF1<0的系數(shù)為：SKIPIF1<0.故選：C.4．SKIPIF1<0的展開式中x3y3的系數(shù)為（

）A．5 B．10C．15 D．20【答案】C【詳解】SKIPIF1<0展開式的通項公式為SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）所以SKIPIF1<0的各項與SKIPIF1<0展開式的通項的乘積可表示為：SKIPIF1<0和SKIPIF1<0在SKIPIF1<0中，令SKIPIF1<0，可得：SKIPIF1<0，該項中SKIPIF1<0的系數(shù)為SKIPIF1<0，在SKIPIF1<0中，令SKIPIF1<0，可得：SKIPIF1<0，該項中SKIPIF1<0的系數(shù)為SKIPIF1<0所以SKIPIF1<0的系數(shù)為SKIPIF1<0故選：C5．（1+2x2）（1+x）4的展開式中x3的系數(shù)為A．12 B．16 C．20 D．24【答案】A【詳解】由題意得x3的系數(shù)為SKIPIF1<0，故選A．二、填空題6．在SKIPIF1<0的展開式中，SKIPIF1<0的系數(shù)是__________．【答案】160【詳解】SKIPIF1<0的展開式的通項為SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0的系數(shù)是SKIPIF1<0.故答案為：160.7．在SKIPIF1<0的展開式中，常數(shù)項為__________．【答案】SKIPIF1<0【詳解】的展開式的通項令SKIPIF1<0，解得，故常數(shù)項為．故答案為：SKIPIF1<0.8．SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為________________（用數(shù)字作答）．【答案】-28【分析】SKIPIF1<0可化為SKIPIF1<0，結(jié)合二項式展開式的通項公式求解.【詳解】因為SKIPIF1<0，所以SKIPIF1<0的展開式中含SKIPIF1<0的項為SKIPIF1<0，SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為-28故答案為：-289．已知多項式SKIPIF1<0，則SKIPIF1<0__________，SKIPIF1<0___________．【答案】

SKIPIF1<0

SKIPIF1<0【詳解】含SKIPIF1<0的項為：SKIPIF1<0，故SKIPIF1<0；令SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0，故答案為：SKIPIF1<0；SKIPIF1<0．10．SKIPIF1<0的展開式中的常數(shù)項為______.【答案】SKIPIF1<0【詳解】由題意SKIPIF1<0的展開式的通項為SKIPIF1<0，令SKIPIF1<0即SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0的展開式中的常數(shù)項為SKIPIF1<0.故答案為：SKIPIF1<0.11．在SKIPIF1<0的展開式中，SKIPIF1<0的系數(shù)是_________．【答案】10【詳解】因為SKIPIF1<0的展開式的通項公式為SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0．所以SKIPIF1<0的系數(shù)為SKIPIF1<0．故答案為：SKIPIF1<0．12．SKIPIF1<0的展開式中常數(shù)項是__________（用數(shù)字作答）．【答案】SKIPIF1<0【詳解】SKIPIF1<0SKIPIF1<0其二項式展開通項：SKIPIF1<0SKIPIF1<0SKIPIF1<0當SKIPIF1<0，解得SKIPIF1<0SKIPIF1<0SKIPIF1<0的展開式中常數(shù)項是：SKIPIF1<0.故答案為：SKIPIF1<0.13．SKIPIF1<0展開式中的常數(shù)項為________.【答案】SKIPIF1<0【詳解】SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，所以的常數(shù)項為SKIPIF1<0.【點睛】本題考查二項式定理的應(yīng)用，牢記常數(shù)項是由指數(shù)冪為0求得的．三、雙空題14．已知多項式SKIPIF1<0，則SKIPIF1<0___________，SKIPIF1<0___________.【答案】

SKIPIF1<0;

SKIPIF1<0.【詳解】SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<0.15．設(shè)SKIPIF1<0，則SKIPIF1<0________；SKIPIF1<0________．【答案】

SKIPIF1<0

SKIPIF1<0【詳解】SKIPIF1<0的通項為SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0；SKIPIF1<0.故答案為：SKIPIF1<0；SKIPIF1<0.16．在二項式SKIPIF1<0的展開式中，常數(shù)項是________；系數(shù)為有理數(shù)的項的個數(shù)是_______.【答案】

SKIPIF1<0

SKIPIF1<0【詳解】SKIPIF1<0的通項為SKIPIF1<0可得常數(shù)項為SKIPIF1<0，因系數(shù)為有理數(shù)，SKIPIF1<0，有SKIPIF1<0共5個項【點睛】此類問題解法比較明確，首要的是要準確記憶通項公式，特別是“冪指數(shù)”不能記混，其次，計算要細心，確保結(jié)果正確.四、解答題17．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.【答案】（1）SKIPIF1<0；（2）-32.【詳解】（1）因為SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0．因為SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0．（2）由（1）知，SKIPIF1<0．SKIPIF1<0SKIPIF1<0SKIPIF1<0．解法一：因為SKIPIF1<0，所以SKIPIF1<0，從而SKIPIF1<0．解法二：SKIPIF1<0SKIPIF1<0．因為SKIPIF1<0，所以SKIPIF1<0．因此SKIPIF1<0．三年模擬一、單選題1．SKIPIF1<0的展開式中的常數(shù)項是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．20【答案】B【詳解】SKIPIF1<0展開式的通項為SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，故SKIPIF1<0展開式的常數(shù)項是SKIPIF1<0.故選：B．2．“SKIPIF1<0”是“SKIPIF1<0的二項展開式中存在常數(shù)項”的（

）A．充分非必要條件 B．必要非充分條件C．充要條件 D．既非充分也非必要條件【答案】A【詳解】二項式SKIPIF1<0的通項為SKIPIF1<0，SKIPIF1<0的二項展開式中存在常數(shù)項SKIPIF1<0為正偶數(shù)，SKIPIF1<0為正偶數(shù)，n為正偶數(shù)推不出SKIPIF1<0∴SKIPIF1<0是SKIPIF1<0的二項展開式中存在常數(shù)項的充分不必要條件．故選：A3．SKIPIF1<0的展開式中x項的系數(shù)為（

）A．568 B．－160 C．400 D．120【答案】D【詳解】因為SKIPIF1<0，又SKIPIF1<0的展開式的通項為SKIPIF1<0，SKIPIF1<0且SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0的展開式的通項為SKIPIF1<0SKIPIF1<0且SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，則x項的系數(shù)為SKIPIF1<0，故選：D．4．已知SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為10，則實數(shù)a的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．2【答案】B【詳解】SKIPIF1<0的展開式的通項公式為SKIPIF1<0，SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，解得SKIPIF1<0，故選：B.5．二項式SKIPIF1<0展開式中SKIPIF1<0的系數(shù)為（

）A．120 B．135 C．140 D．100【答案】B【詳解】SKIPIF1<0的展開式通項公式為SKIPIF1<0，其中SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故二項式SKIPIF1<0中SKIPIF1<0的四次方項為SKIPIF1<0，即展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0.故選：B6．若二項式SKIPIF1<0SKIPIF1<0的展開式中只有第7項的二項式系數(shù)最大，若展開式的有理項中第SKIPIF1<0項的系數(shù)最大，則SKIPIF1<0（

）A．5 B．6 C．7 D．8【答案】A【詳解】由已知可得，SKIPIF1<0.根據(jù)二項式定理，知展開式的通項為SKIPIF1<0，顯然當SKIPIF1<0是偶數(shù)時，該項為有理項，SKIPIF1<0時，SKIPIF1<0；SKIPIF1<0時，SKIPIF1<0；SKIPIF1<0時，SKIPIF1<0；SKIPIF1<0時，SKIPIF1<0；SKIPIF1<0時，SKIPIF1<0；SKIPIF1<0時，SKIPIF1<0；SKIPIF1<0時，SKIPIF1<0.經(jīng)比較可得，SKIPIF1<0，即SKIPIF1<0時系數(shù)最大，即展開式的有理項中第5項的系數(shù)最大．故選：A.7．SKIPIF1<0展開式中SKIPIF1<0的系數(shù)為（

）A．SKIPIF1<0 B．21 C．SKIPIF1<0 D．35【答案】A【詳解】因為SKIPIF1<0展開式的通項公式為SKIPIF1<0，所以當SKIPIF1<0時，含有SKIPIF1<0的項，此時SKIPIF1<0，故SKIPIF1<0的系數(shù)為SKIPIF1<0．故選：A二、填空題8．在SKIPIF1<0的展開式中，SKIPIF1<0的系數(shù)為__________.【答案】SKIPIF1<0【詳解】由題意可知，把二項式SKIPIF1<0看成由SKIPIF1<0和SKIPIF1<0兩項構(gòu)成，展開式中含SKIPIF1<0的項為SKIPIF1<0，再將SKIPIF1<0展開可得含SKIPIF1<0的項為SKIPIF1<0即可知SKIPIF1<0的系數(shù)為SKIPIF1<0.故答案為：SKIPIF1<09．在二項式SKIPIF1<0的展開式中，系數(shù)最大的項的系數(shù)為__________（結(jié)果用數(shù)值表示）.【答案】462【詳解】二項式SKIPIF1<0的展開式的通項公式為SKIPIF1<0，所以當SKIPIF1<0或SKIPIF1<0時，其系數(shù)最大，則最大系數(shù)為SKIPIF1<0，故答案為：462.10．已知SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)是20，則實數(shù)SKIPIF1<0__________.【答案】2【詳解】解：因為SKIPIF1<0則展開式中SKIPIF1<0的系數(shù)是SKIPIF1<0，求得SKIPIF1<0.故答案為：2.11．SKIPIF1<0的二項展開式中SKIPIF1<0的系數(shù)為______.【答案】80【詳解】SKIPIF1<0的二項展開式中含SKIPIF1<0的項為SKIPIF1<0，所以SKIPIF1<0的系數(shù)為SKIPIF1<0.故答案為：SKIPIF1<012．若SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為10，則SKIPIF1<0______．【答案】SKIPIF1<0【詳解】SKIPIF1<0的通項SKIPIF1<0，所以SKIPIF1<0的展開式中SKIPIF1<0項為SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0.故答案為：SKIPIF1<013．在SKIPIF1<0的二項展開式中，SKIPIF1<0項的系數(shù)是___________.【答案】SKIPIF1<0【詳解】二項式SKIPIF1<0的通項為SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0項的系數(shù)是SKIPIF1<0.故答案為：SKIPIF1<0.14．已知SKIPIF1<0（n是正整數(shù)），SKIPIF1<0，則SKIPIF1<0________．【答案】243【詳解】因為SKIPIF1<0，所以SKIPIF1<0，解得，SKIPIF1<0.令SKIPIF1<0得，SKIPIF1<0，故SKIPIF1<0，故答案為：243.15．在SKIPIF1<0展開式中，含有SKIPIF1<0項的系數(shù)為______．【答案】SKIPIF1<0【詳解】因為SKIPIF1<0的展開式通項為SKIPIF1<0，由題意可知，在SKIPIF1<0展開式中，含有SKIPIF1<0項的系數(shù)為SKIPIF1<0.故答案為：SKIPIF1<0.16．已知常數(shù)SKIPIF1<0，在SKIPIF1<0的二項展開式中，SKIPIF1<0項的系數(shù)等于SKIPIF1<0，則SKIPIF1<0_______.【答案】SKIPIF1<0【詳解】根據(jù)已知條件SKIPIF1<0是二項式展開式的某一項，故得SKIPIF1<0.由SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0.得SKIPIF1<0，根據(jù)已知可得SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0.故答案為：SKIPIF1<0.17．在SKIPIF1<0的二項展開式中SKIPIF1<0項的系數(shù)為______．【答案】35【詳解】由SKIPIF1<0，令SKIPIF1<0，解得：SKIPIF1<0.SKIPIF1<0.得SKIPIF1<0項的系數(shù)為SKIPIF1<0.故答案為：SKIPIF1<018．在SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為__________.【答案】SKIPIF1<0【詳解】SKIPIF1<0展開式的通項公式為SKIPIF1<0，所以SKIPIF1<0的展開式中含SKIPIF1<0的項等于SKIPIF1<0，故答案為:SKIPIF1<0.19．若SKIPIF1<0的展開式的二項式系數(shù)和為32，則展開式中x的系數(shù)為______.【答案】40【詳解】依題意SKIPIF1<0的展開式的二項式系數(shù)和為SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0.二項式SKIPIF1<0展開式的通項公式為SKIPIF1<0SKIPIF1<0.令SKIPIF1<0，所以x的系數(shù)為SKIPIF1<0.故答案為：SKIPIF1<020．已知SKIPIF1<0的展開式中，僅有第5項的二項式系數(shù)最大，則展開式中有理項的個數(shù)為___________．【答案】2【詳解】SKIPIF1<0的展開式有SKIPIF1<0項,因為僅有第5項的二項式系數(shù)最大,所以SKIPIF1<0SKIPIF1<0當SKIPIF1<0時,SKIPIF1<0,當SKIPIF1<0時,SKIPIF1<0,符合題意所以展開式中有理項的個數(shù)為2故答案為：221．已知SKIPIF1<0且滿足SKIPIF1<0能被8整除，則符合條件的一個SKIPIF1<0的值為___________.【答案】5（答案不唯一）【詳解】由已知得SKIPIF1<0SKIPIF1<0，由已知SKIPIF1<0且滿足SKIPIF1<0能被8整除，則SKIPIF1<0是8的整數(shù)倍，所以SKIPIF1<0（SKIPIF1<0），則符合條件的一個SKIPIF1<0的值為5.故答案為：SKIPIF1<0（答案不唯一）22．SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為______（用數(shù)字作答）．【答案】-800【詳解】由題意知，在SKIPIF1<0的展開式中取第4項，即SKIPIF1<0，SKIPIF1<0的展開式中取第2項，即SKIPIF1<0，故SKIPIF1<0的系數(shù)為SKIPIF1<0．故答案為：-80023．己知SKIPIF1<0，則SKIPIF1<0________.（用數(shù)字作案）【答案】34【詳解】令SKIPIF1<0，得SKIPIF1<0；令SKIPIF1<0，得SKIP