2022年秋高中數(shù)學(xué)第六章計(jì)數(shù)原理6.3二項(xiàng)式定理6.3.1二項(xiàng)式定理課后習(xí)題新人教A版選擇性必修第三冊(cè)

歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！6.3二項(xiàng)式定理6.3.1二項(xiàng)式定理A級(jí)必備知識(shí)基礎(chǔ)練1.(x-2y)10的展開(kāi)式中x6y4的系數(shù)是()A.840 B.-840 C.210 D.-2102.在(1-x)5-(1-x)6的展開(kāi)式中,含x3的項(xiàng)的系數(shù)是()A.-5 B.5 C.-10 D.103.使得3x+1xxn(n∈N*)的展開(kāi)式中含有常數(shù)項(xiàng)的最小的n為()A.4 B.5 C.6 D.74.(多選題)對(duì)于二項(xiàng)式1x+x3n(n∈N*),下列判斷正確的有()A.存在n∈N*,展開(kāi)式中有常數(shù)項(xiàng)B.對(duì)任意n∈N*,展開(kāi)式中沒(méi)有常數(shù)項(xiàng)C.對(duì)任意n∈N*,展開(kāi)式中沒(méi)有x的一次項(xiàng)D.存在n∈N*,展開(kāi)式中有x的一次項(xiàng)5.若(x+a)10的展開(kāi)式中,x7的系數(shù)為15,則a=.(用數(shù)字填寫(xiě)答案)

6.若x>0,設(shè)x2+1x5的展開(kāi)式中的第三項(xiàng)為M,第四項(xiàng)為N,則M+N的最小值為7.已知x+23xn的展開(kāi)式中的第9項(xiàng)與第10項(xiàng)二項(xiàng)式系數(shù)相等,求x8.已知在x+2x29.求證:1+2+22+…+25n-1(n∈B級(jí)關(guān)鍵能力提升練10.在(1-x3)(1+x)10的展開(kāi)式中,x5的系數(shù)是()A.-297 B.-252 C.297 D.20711.對(duì)任意實(shí)數(shù)x,有x3=a0+a1(x-2)+a2(x-2)2+a3(x-2)3,則a2的值為()A.3 B.6 C.9 D.2112.(x2+2)1x2-15的展開(kāi)式中的常數(shù)項(xiàng)是()A.-3 B.-2 C.2 D.313.(2022江蘇南京玄武校級(jí)月考)-2Cn1+4Cn2-8Cn3+…+(-2)A.(-1)n-1 B.(-1)nC.3n D.3n-114.(2022山西臨汾模擬)在a+x-1x10的展開(kāi)式中,xA.343 B.423 C.2 15.已知在12x2-1xn的展開(kāi)式中,第9項(xiàng)為常數(shù)項(xiàng),則:(1)n的值為;

(2)含x的整數(shù)次冪的項(xiàng)有個(gè).

16.在(1-x)5+(1-x)6+(1-x)7+(1-x)8的展開(kāi)式中,含x3的項(xiàng)的系數(shù)是.

17.已知(xcosθ+1)5的展開(kāi)式中x2的系數(shù)與x+544的展開(kāi)式中x3的系數(shù)相等,則cosθ=.

18.已知x?124x(1)證明:展開(kāi)式中沒(méi)有常數(shù)項(xiàng);(2)求展開(kāi)式中所有的有理項(xiàng).C級(jí)學(xué)科素養(yǎng)創(chuàng)新練19.已知f(x)=(1+x)m,g(x)=(1+2x)n(m,n∈N*).(1)若m=3,n=4,求f(x)g(x)的展開(kāi)式中含x2的項(xiàng);(2)令h(x)=f(x)+g(x),h(x)的展開(kāi)式中x的項(xiàng)的系數(shù)為12,那么當(dāng)m,n為何值時(shí),含x2的項(xiàng)的系數(shù)取得最小值?6.3.1二項(xiàng)式定理1.A在通項(xiàng)Tk+1=C10k(-2y)kx10-k中,令k=4,即得(x-2y)10的展開(kāi)式中x6y4項(xiàng)的系數(shù)為C104×(-22.D(1-x)5中x3的系數(shù)為-C53=-10,-(1-x)6中x3的系數(shù)為-C63·(-1)3=20,故(1-x)5-(1-x)6的展開(kāi)式中3.B3x+1xxn展開(kāi)式中的第k+1項(xiàng)為Cnk(3x)n-kx-32k=Cnk3n-kxn-52k.若展開(kāi)式中含常數(shù)項(xiàng),4.AD二項(xiàng)式1x+x3n的展開(kāi)式的通項(xiàng)為T(mén)k+1=Cnkx4k-n,由通項(xiàng)可知,當(dāng)n=4k(k∈N*)和n=4k-1(k∈N*)時(shí),展開(kāi)式中分別存在常數(shù)項(xiàng)和x的一次項(xiàng),故選AD5.12二項(xiàng)展開(kāi)式的通項(xiàng)為T(mén)k+1=C10kx10-kak,當(dāng)10-k=7時(shí),k=3,T4=C103a3x7,則C106.522T3=C52·x231x2=54x,T4=C53·x22·1x3=52x,故M+N=5x4+52x7.解∵Cn8=Cn9,∴n=17,T令17-k2?k3∴T10=C179·x4·29·x-3=C179故x的系數(shù)為29C8.解T5=Cn4(x)n-4·24x-8=16Cn4xn-202,T3=Cn由題意知,16Cn44Cn2Tk+1=C10k(x)10-k·2kx-2k=令10-5k2=0,所以展開(kāi)式中的常數(shù)項(xiàng)為C102×229.證明∵1+2+22+…+25n-1=25n-12-1=25n-1=32n-1=(31+1)n-1=Cn0·31n+Cn1·31n-1+…+Cnn-1·31+Cnn-1=31(Cn0·3110.D(1-x3)(1+x)10=(1+x)10-x3(1+x)10,x5的系數(shù)為C105?11.B∵x3=(x-2+2)3=C30(x-2)3+C31(x-2)2·2+C32(x-2)·22+C33·23=8+12(x-2)+6(∴a2=6.12.D1x2-15展開(kāi)式的通項(xiàng)為T(mén)k+1=C5k·1x令10-2k=2或10-2k=0,解得k=4或k=5.故(x2+2)·1(-1)4×C54+2×(-1)5×13.A∵-2Cn1+4Cn2-8Cn3+…+(-2)nCnn=1-2Cn1+4Cn2-8Cn3+…+(-2)nCnn-14.C由多項(xiàng)式的乘法性質(zhì)知每個(gè)括號(hào)里的因式是a,x,-1x,則x8=x9×1x=x8×1×1,共有則對(duì)應(yīng)的x8為C101-1x·x9+C102a2·x8=(-10∵x8的系數(shù)為170,∴-10+45a2=170,則45a2=180,即a2=4,解得a=2.15.(1)10(2)6二項(xiàng)展開(kāi)式的通項(xiàng)為T(mén)k+1=Cnk12x2n-k·-1xk=(-1)k12n-k(1)因?yàn)榈?項(xiàng)為常數(shù)項(xiàng),所以當(dāng)k=8時(shí),2n-52k=解得n=10.(2)要使20-52k為整數(shù),需k為偶數(shù)由于k=0,1,2,3,…,9,10,故符合要求的項(xiàng)有6個(gè),分別為展開(kāi)式的第1,3,5,7,9,11項(xiàng).16.-121展開(kāi)式中含x3的項(xiàng)的系數(shù)為C53(-1)3+C63(-1)3+C73(-1)3+C817.±22(xcosθ+1)5展開(kāi)式中x2的系數(shù)為C53cosx+544展開(kāi)式中x3的系數(shù)為54·C41.由題意可知C53cos2θ=54·C41,18.(1)證明由題意得2Cn1·12即n2-9n+8=0,∴n=8(n=1舍去).∴Tk+1=C8k(x)8-k·-124xk=-12k·若Tk+1是常數(shù)項(xiàng),則16-3即16-3k=0,∵k∈Z,∴等式不可能成立,∴展開(kāi)式中沒(méi)有常數(shù)項(xiàng).(2)解由(1)知,若Tk+1是有理項(xiàng),當(dāng)且僅當(dāng)16-3k4為整數(shù).∵0≤k≤8,k∈Z即展開(kāi)式中有三項(xiàng)有理項(xiàng),分別是T1=x4,T5=358x,T9=1256x-19.解(1)