2022新高考總復(fù)習(xí)《數(shù)學(xué)》(人教)第十章　計(jì)數(shù)原理課時(shí)規(guī)范練54　二項(xiàng)式定理

1、 課時(shí)規(guī)范練54二項(xiàng)式定理基礎(chǔ)鞏固組1.1-1x3(1+x)7展開(kāi)式中x3的系數(shù)為() A.-7B.28C.35D.422.若x4(x+4)8=a0+a1(x+3)+a2(x+3)2+a12(x+3)12,則log2(a1+a3+a11)=()A.4B.8C.12D.113.若二項(xiàng)式x-1xn的展開(kāi)式中第m項(xiàng)為常數(shù)項(xiàng),則m,n應(yīng)滿足()A.2n=3mB.2n=3(m-1)C.2n=3(m+1)D.2n=m4.3x-1x6的展開(kāi)式中有理項(xiàng)的項(xiàng)數(shù)為()A.0B.1C.2D.35.設(shè)(x-2x)6的展開(kāi)式中含x3項(xiàng)的系數(shù)為A,二項(xiàng)式系數(shù)為B,則AB=()A.1B.2C.3D.46.若(x6+1xx)

2、n的展開(kāi)式中含有常數(shù)項(xiàng),則n的最小值等于()A.3B.4C.5D.67.(x2+3y-y2)7展開(kāi)式中x12y2的系數(shù)為()A.7B.-7C.42D.-428.(多選)(2020江蘇宿城宿遷中學(xué)期中)對(duì)于(a+b)n(nN*)展開(kāi)式的二項(xiàng)式系數(shù)下列結(jié)論正確的是()A.Cnm=Cnn-mB.Cnm+Cnm-1=Cn+1mC.當(dāng)n為偶數(shù)時(shí),Cn0+Cn1+Cn2+Cnn=2nD.Cn1+Cn2+Cnn=2n9.(多選)(2020山東淄博五中高三月考)在(2x-1)7的展開(kāi)式的各項(xiàng)中,二項(xiàng)式系數(shù)最大的項(xiàng)是()A.第2項(xiàng)B.第3項(xiàng)C.第4項(xiàng)D.第5項(xiàng)10.(x-1)7(x+1)3的展開(kāi)式中x的系數(shù)是.

3、綜合提升組11.已知nN*,設(shè)5x2-1xn的展開(kāi)式的各項(xiàng)系數(shù)之和為M,二項(xiàng)式系數(shù)之和為N,若M-N=992,則展開(kāi)式中x的系數(shù)為()A.-250B.250C.-500D.50012.(2x2-x+1)8的展開(kāi)式中x5的系數(shù)是()A.1 288B.1 280C.-1 288D.-1 28013.若a0 x2 016+a1x2 015(1-x)+a2x2 014(1-x)2+a2 016(1-x)2 016=1,則a0+a1+a2+a2 016的值為()A.1B.0C.22 016D.22 01514.在2x+3y-49的展開(kāi)式中,不含x的各項(xiàng)系數(shù)之和為. 15.已知在(3x2+3x2)n的展開(kāi)式中各項(xiàng)系數(shù)的和比它的二項(xiàng)式系數(shù)的和大992,則展開(kāi)式中系數(shù)最大的項(xiàng)為.創(chuàng)新應(yīng)用組16.已知實(shí)數(shù)x(-1,1),nN*,記X=(1-x)