新高考數(shù)學(xué)一輪復(fù)習(xí)計算題精練專題11 二項式的計算（解析版）

二項式定理的相關(guān)計算1．已知SKIPIF1<0展開式的二項式系數(shù)之和為128，則SKIPIF1<0__________.【解答】根據(jù)展開式的二項式系數(shù)之和為SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，故答案為：SKIPIF1<0.2．若SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0，則SKIPIF1<0__________.【解答】二項式SKIPIF1<0展開式的通項為SKIPIF1<0所以SKIPIF1<0的展開式中含SKIPIF1<0的項為SKIPIF1<0，所以SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<03．SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為______（用數(shù)字表示）.【解答】SKIPIF1<0的通項為SKIPIF1<0,令SKIPIF1<0，所以展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0,故答案為：2104．SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)是______.【解答】二項式SKIPIF1<0中，SKIPIF1<0，當(dāng)SKIPIF1<0中取x時，這一項為SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0中取y時，這一項為SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0故答案為：SKIPIF1<05．若SKIPIF1<0的展開式中所有項的系數(shù)和為SKIPIF1<0，則展開式中SKIPIF1<0的系數(shù)為__________．【解答】令SKIPIF1<0，得SKIPIF1<0，解得SKIPIF1<0，進(jìn)而可得SKIPIF1<0的展開式為SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，故SKIPIF1<0的系數(shù)為SKIPIF1<0．故答案為：SKIPIF1<06．SKIPIF1<0的展開式中，SKIPIF1<0項的系數(shù)為__________．【解答】由二項式展開式通項為SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0項的系數(shù)為SKIPIF1<0.故答案為：SKIPIF1<07．已知SKIPIF1<0，則SKIPIF1<0__________．【解答】依題意SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0.因為SKIPIF1<0可以得出SKIPIF1<0,SKIPIF1<0，故SKIPIF1<0.故答案為：SKIPIF1<0.8．已知二項式SKIPIF1<0的常數(shù)項為SKIPIF1<0，則SKIPIF1<0______________.【解答】由題意可知SKIPIF1<0，則其通項為SKIPIF1<0，而SKIPIF1<0的通項為SKIPIF1<0，令SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0，不合題意，由二項式SKIPIF1<0的常數(shù)項為SKIPIF1<0，可得SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，故答案為：SKIPIF1<09．在SKIPIF1<0的展開式中x的系數(shù)為______．【解答】SKIPIF1<0的展開式中x的項為SKIPIF1<0，所以展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0．故答案為：SKIPIF1<0.10．SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為______【解答】SKIPIF1<0展開式的通項為SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，所以展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0.故答案為：SKIPIF1<0.11．已知常數(shù)SKIPIF1<0，SKIPIF1<0的二項展開式中SKIPIF1<0項的系數(shù)是SKIPIF1<0，則SKIPIF1<0的值為_____________．【解答】由已知SKIPIF1<0，則其展開式的通項為SKIPIF1<0，又其二項展開式中SKIPIF1<0項的系數(shù)是SKIPIF1<0，則令SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，故答案為：SKIPIF1<0.12．若SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為60，則實數(shù)SKIPIF1<0________．【解答】∵SKIPIF1<0的展開式中含SKIPIF1<0的項為SKIPIF1<0，由已知SKIPIF1<0的系數(shù)為SKIPIF1<0，∴SKIPIF1<0．故答案為：SKIPIF1<0.13．SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)是______.（用數(shù)字作答）【解答】SKIPIF1<0，而SKIPIF1<0的通項為SKIPIF1<0，SKIPIF1<0，故展開式中SKIPIF1<0的系數(shù)是SKIPIF1<0,故答案為：SKIPIF1<0.14．在SKIPIF1<0的展開式中，SKIPIF1<0的系數(shù)為____________．（結(jié)果填數(shù)字）【解答】設(shè)SKIPIF1<0的展開式通項為SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0的系數(shù)為SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0的系數(shù)為SKIPIF1<0；所以SKIPIF1<0的系數(shù)為SKIPIF1<0.故答案為：3215．SKIPIF1<0展開式中含SKIPIF1<0項的系數(shù)為______.【解答】SKIPIF1<0展開式的通項公式為SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，所以含SKIPIF1<0項為SKIPIF1<0，所以SKIPIF1<0展開式中含SKIPIF1<0項的系數(shù)為14.故答案為：14.16．SKIPIF1<0展開式的常數(shù)項為___________.（用最簡分?jǐn)?shù)表示）【解答】SKIPIF1<0展開式通項公式SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0展開式的常數(shù)項是SKIPIF1<0.故答案為：SKIPIF1<017．SKIPIF1<0的展開式中含SKIPIF1<0的項與含SKIPIF1<0的項系數(shù)相等，則SKIPIF1<0___________.【解答】由SKIPIF1<0的展開式的通項為SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0；令SKIPIF1<0，可得SKIPIF1<0，因為展開式中含SKIPIF1<0的項與含SKIPIF1<0的項系數(shù)相等，可得SKIPIF1<0，又因為SKIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<0.18．已知SKIPIF1<0，則SKIPIF1<0的值等于______.【解答】令SKIPIF1<0，則SKIPIF1<0；令SKIPIF1<0，則SKIPIF1<0，上述兩式相加得SKIPIF1<0，故SKIPIF1<0；故答案為：SKIPIF1<0.19．已知SKIPIF1<0，則SKIPIF1<0___________．（用數(shù)字作答）【解答】因為SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0；令SKIPIF1<0，得SKIPIF1<0；又SKIPIF1<0，二項式SKIPIF1<0的通項公式為SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0．故答案為：SKIPIF1<020．SKIPIF1<0展開式中SKIPIF1<0項的系數(shù)為________.【解答】因為SKIPIF1<0的二項展開式為SKIPIF1<0，所以SKIPIF1<0項為SKIPIF1<0，即展開式中SKIPIF1<0項的系數(shù)為12.故答案為：12.21．已知a>0，若SKIPIF1<0，且SKIPIF1<0，則a=______.【解答】因為SKIPIF1<0，又SKIPIF1<0，展開式通項為SKIPIF1<0，SKIPIF1<0對應(yīng)SKIPIF1<0的系數(shù)，故得到SKIPIF1<0，解得SKIPIF1<0，其系數(shù)為SKIPIF1<0或SKIPIF1<0.又a>0，故實數(shù)a的值為2.故答案為：2.22．若SKIPIF1<0的展開式中各項系數(shù)之和為SKIPIF1<0，則展開式中SKIPIF1<0的系數(shù)為______.【解答】因為SKIPIF1<0的展開式中各項系數(shù)之和為SKIPIF1<0,令SKIPIF1<0,得SKIPIF1<0,所以SKIPIF1<06.因為SKIPIF1<0展開式的通頂公式為SKIPIF1<0,令SKIPIF1<0,得SKIPIF1<0;令SKIPIF1<0,得SKIPIF1<0,所以展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0.故答案為：SKIPIF1<023．SKIPIF1<0的展開式中含SKIPIF1<0項的系數(shù)為_________．【解答】解：SKIPIF1<0展開式的通項為SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，所以展開式中常數(shù)項為SKIPIF1<0．故答案為：SKIPIF1<024．SKIPIF1<0的展開式中，含SKIPIF1<0的項的系數(shù)是__________．【解答】由題意可知SKIPIF1<0中SKIPIF1<0的系數(shù)為SKIPIF1<0,SKIPIF1<0的系數(shù)為SKIPIF1<0,故SKIPIF1<0的展開式中，含SKIPIF1<0的項的系數(shù)是SKIPIF1<0，故答案為：1425．SKIPIF1<0展開式的常數(shù)項是__________．（用數(shù)字作答）【解答】SKIPIF1<0展開式的通項公式是SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0展開式的常數(shù)項為SKIPIF1<0.故答案為：2426．若SKIPIF1<0展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0，則SKIPIF1<0______．【解答】SKIPIF1<0的通項為：SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，解得：SKIPIF1<0.故答案為：SKIPIF1<0.27．已知SKIPIF1<0的展開式中各項系數(shù)的和為243，則這個展開式中SKIPIF1<0項的系數(shù)是__________．【解答】在SKIPIF1<0中令SKIPIF1<0得展開式中各項系數(shù)的和為SKIPIF1<0，求出SKIPIF1<0．SKIPIF1<0的展開式的通項SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0．故答案為：80．28．在SKIPIF1<0的展開式中，含SKIPIF1<0的項的系數(shù)為__________.【解答】在SKIPIF1<0展開式中，第SKIPIF1<0項為SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，得含有SKIPIF1<0的項的系數(shù)為SKIPIF1<0；故答案為：135.29．二項式SKIPIF1<0的展開式中的SKIPIF1<0項的系數(shù)為___________.【解答】SKIPIF1<0展開式的通項為SKIPIF1<0，SKIPIF1<0，所以當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，所以二項式SKIPIF1<0的展開式中含SKIPIF1<0項的系數(shù)為SKIPIF1<0.故答案為：SKIPIF1<0.30．在SKIPIF1<0的展開式中，SKIPIF1<0的系數(shù)為________．【解答】因為SKIPIF1<0的展開式的通項公式為SKIPIF1<0，即SKIPIF1<0，所以由SKIPIF1<0，得到SKIPIF1<0，故SKIPIF1<0的系數(shù)為SKIPIF1<0.故答案為：SKIPIF1<0.31．SKIPIF1<0的展開式中常數(shù)項為______.【解答】SKIPIF1<0的展開式中通項為SKIPIF1<0,所以要使SKIPIF1<0展開式中出現(xiàn)常數(shù)項，需SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0（舍去），所以常數(shù)項為SKIPIF1<0，故答案為：280.32．在二項式SKIPIF1<0的展開式中，SKIPIF1<0項的二項式系數(shù)為__________．【解答】因為SKIPIF1<0，SKIPIF1<0，1，2，…，6.令SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0項的二項式系數(shù)為SKIPIF1<0.故答案為：2033．SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為__________.（用數(shù)字作答）【解答】由題意得SKIPIF1<0，因為SKIPIF1<0的展開式的通項為SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0，故答案為：SKIPIF1<0.34．SKIPIF1<0的展開式中x的系數(shù)為___________.【解答】SKIPIF1<0的展開式的通項公式為SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，所以展開式中x的系數(shù)為SKIPIF1<0.故答案為:SKIPIF1<0.35．SKIPIF1<0的二項展開式中的常數(shù)項為______．【解答】二項式SKIPIF1<0展開式的通項為SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，所以展開式中常數(shù)項為SKIPIF1<0.故答案為：SKIPIF1<036．已知二項式SKIPIF1<0的展開式中SKIPIF1<0的系數(shù)為SKIPIF1<0，則該二項展開式中的常數(shù)項為___________.【解答】SKIPIF1<0的展開式的通項SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，∴SKIPIF1<0，解得SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，∴該二項展開式中的常數(shù)項為SKIPIF1<0.故答案為：SKIPIF1<0.37．SKIPIF1<0的展開式中的常數(shù)項為______.【解答】二項式SKIPIF1<0展開式的通項為SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0常數(shù)項為SKIPIF1<0.故答案為：SKIPIF1<0．38．已知SKIPIF1<0的展開式中常數(shù)項為20，則實數(shù)m的值為______．【解答】展開式的通項為SKIPIF1<0，令SKIPIF1<0解得SKIPIF1<0，∴SKIPIF1<0．∴SKIPIF1<0．故答案為：139．SKIPIF1<0的展開式中的常數(shù)項為______．【解答】SKIPIF1<0的展開式的通項公式為SKIPIF1<0.令SKIPIF1<0，令SKIPIF1<0.則SKIPIF1<0的展開式中的常數(shù)項為SKIPIF1<0.故答案為：SKIPIF1<040．二項式SKIPIF1<0的展開式中，常數(shù)項為_______________（用數(shù)值表示）．【解答】由二項式定理可得SKIPIF1<0，顯然其常數(shù)項為第三項即SKIPIF1<0，故答案為：2441．在SKIPIF1<0的展開式中，常數(shù)項為______________.（結(jié)果用數(shù)字表示）【解答】SKIPIF1<0展開式通項為：SKIPIF1<0，令SKIPIF1<0，解得：SKIPIF1<0，SKIPIF1<0，即常數(shù)項為SKIPIF1<0.故答案為：SKIPIF1<0.42．在SKIPIF1<0的展開式中，SKIPIF1<0項的系數(shù)是______．【解答】SKIPIF1<0展開式的通項公式為SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，所以含SKIPIF1<0項的系數(shù)為SKIPIF1<0，故答案為：SKIPIF1<0．43．SKIPIF1<0展開式中的常數(shù)項為__________．【解答】SKIPIF1<0展開式通項為SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，所以常數(shù)項為SKIPIF1<0．故答案為：SKIPIF1<0.44．二項式SKIPIF1<0的常數(shù)項為__________．【解答】SKIPIF1<0的展開式的通項公式為SKIPIF1<0，而SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0；令SKIPIF1<0，得SKIPIF1<0（舍）.所以SKIPIF1<0的展開式中的常數(shù)項為SKIPIF1<0．故答案為：SKIPIF1<045．若在SKIPIF1<0的展開式中，SKIPIF1<0的系數(shù)為__________．（用數(shù)字作答）【解答】SKIPIF1<0的展開式通項為SKIPIF1<0，令SKIPIF1<0