福建專用2019高考數(shù)學(xué)一輪復(fù)習(xí)課時(shí)規(guī)范練57二項(xiàng)式定理理新人教A版.doc

課時(shí)規(guī)范練57二項(xiàng)式定理一、基礎(chǔ)鞏固組1.(2017湖南邵陽模擬)在(1+3x)n的展開式中,若x5與x6的系數(shù)相等,則x4的二項(xiàng)式系數(shù)為()A.21B.35C.45D.282.若Cn1+3Cn2+32Cn3+3n-2Cnn-1+3n-1=85,則n=()A.6B.5C.4D.33.設(shè)n為正整數(shù),x-1xx2n展開式中存在常數(shù)項(xiàng),則n的一個(gè)可能取值為()A.16B.10C.4D.24.(2017河南鄭州一中質(zhì)檢一,理7)若a= 0sin xdx,則二項(xiàng)式ax-1x6展開式的常數(shù)項(xiàng)是()A.160B.20C.-20D.-1605.(x2-3)1x2+15的展開式中的常數(shù)項(xiàng)是()A.-2B.2C.-3D.36.若(1+3)4=a+b3(a,b為有理數(shù)),則a+b等于()A.36B.46C.34D.447.(x2+3y-y2)7展開式中x12y2的系數(shù)為()A.7B.-7C.42D.-428.(2017甘肅會寧月考)1-90C101+902C102-903C103+(-1)k90kC10k+9010C1010除以88的余數(shù)是()A.-1B.1C.-87D.87導(dǎo)學(xué)號215005889.(2017浙江,13)已知多項(xiàng)式(x+1)3(x+2)2=x5+a1x4+a2x3+a3x2+a4x+a5,則a4=,a5=.10.(2017遼寧沈陽三模,理14)(1+2x)3(1-x)4展開式中x2的系數(shù)為.二、綜合提升組11.(2017遼寧鞍山一模,理3)若(x2+m)x-2x6的展開式中x4的系數(shù)為30,則m的值為()A.-52B.52C.-152D.15212.(2017江西宜春二模,理8)若x3+1xn的展開式中含有常數(shù)項(xiàng),且n的最小值為a,則-aa a2-x2dx=()A.0B.6863C.492D.49導(dǎo)學(xué)號2150058913.在(x+2y)7的展開式中,系數(shù)最大的項(xiàng)是()A.68y7B.112x3y4C.672x2y5D.1 344x2y514.x+ax2x-1x5的展開式中各項(xiàng)系數(shù)的和為2,則該展開式中的常數(shù)項(xiàng)為()A.-40B.-20C.20D.4015.(2017河南重點(diǎn)校聯(lián)考)在2x+3y-49的展開式中,不含x的各項(xiàng)系數(shù)之和為.三、創(chuàng)新應(yīng)用組16.若多項(xiàng)式x3+x10=a0+a1(x+1)+a9(x+1)9+a10(x+1)10,則a9=()A.9B.10C.-9D.-10導(dǎo)學(xué)號2150059017.在(1+x)6(1+y)4的展開式中,記xmyn項(xiàng)的系數(shù)為f(m,n),則f(3,0)+f(2,1)+f(1,2)+f(0,3)=.導(dǎo)學(xué)號21500591課時(shí)規(guī)范練57二項(xiàng)式定理1.BTr+1=Cnr(3x)r=3rCnrxr,由已知得35Cn5=36Cn6,即Cn5=3Cn6,n=7.因此,x4的二項(xiàng)式系數(shù)為C74=35,故選B.2.CCn1+3Cn2+3n-2Cnn-1+3n-1=13(1+3)n-1=85,解得n=4.3.Bx-1xx2n展開式的通項(xiàng)公式為Tk+1=C2nkx2n-k-1xxk=C2nk(-1)kx4n-5k2,令4n-5k2=0,得k=4n5,n可取10.4.Da= 0sin xdx=-cos x|0=2,ax-1x6=2x-1x6的展開式的通項(xiàng)為Tr+1=(-1)r26-rC6rx3-r.令3-r=0,得r=3.故展開式的常數(shù)項(xiàng)是-8C63=-160,故選D.5.B(x2-3)1x2+15=(x2-3)(C50x-10+C51x-8+C52x-6+C53x-4+C54x-2+C55),展開式的常數(shù)項(xiàng)是x2C54x-2-3C55=2.6.D(1+3)4=1+C413+C42(3)2+C43(3)3+(3)4=28+163,由題設(shè)可得a=28,b=16,故a+b=44.7.B將(x2+3y-y2)7看作7個(gè)因式相乘,要得到x12y2項(xiàng),需要7個(gè)因式中有6個(gè)因式取x2,1個(gè)因式取-y2,故x12y2的系數(shù)為C76(-1)=-7.8.B1-90C101+902C102-903C103+(-1)k90kC10k+9010C1010=(1-90)10=8910=(88+1)10=8810+C101889+C10988+1.前10項(xiàng)均能被88整除,余數(shù)是1.9.164由二項(xiàng)式展開式可得通項(xiàng)公式為C3rx3-rC2mx2-m2m,分別取r=3,m=1和r=2,m=2可得a4=4+12=16,令x=0可得a5=1322=4.10.-6展開式中x2項(xiàng)為C3013(2x)0C4212(-x)2+C3112(2x)1C4113(-x)1+C3211(2x)2C4014(-x)0,所求系數(shù)為C30C42+C312C41(-1)+C3222C40=6-24+12=-6.11.Bx-2x6的展開式的通項(xiàng)公式為Tr+1=C6rx6-r-2xr=(-2)rC6rx6-2r,令6-2r=2,得r=2,所以x2項(xiàng)的系數(shù)為(-2)2C62=60,令6-2r=4,得r=1,所以x4項(xiàng)的系數(shù)為(-2)1C61=-12,所以(x2+m)x-2x6的展開式中x4的系數(shù)為60-12m=30,解得m=52,故選B.12.C由題意知展開式的通項(xiàng)公式為Tr+1=Cnr(x3)n-r1xr=Cnrx3n-72r,因?yàn)檎归_式中含有常數(shù)項(xiàng),所以3n-72r=0有整數(shù)解,所以n的最小值為7.故定積分-77 72-x2dx=492.13.C設(shè)第r+1項(xiàng)的系數(shù)最大,則有C7r2rC7r-12r-1,C7r2rC7r+12r+1,即7!r!(7-r)!2r7!(r-1)!(7-r+1)!2r-1,7!r!(7-r)!2r7!(r+1)!(7-r-1)!2r+1,解得r163,r133.rZ,r=5,系數(shù)最大的項(xiàng)為T6=C75x225y5=672x2y5.故選C.14.D在x+ax2x-1x5中,令x=1,得(1+a)(2-1)5=2,即a=1.原式=x2x-1x5+1x2x-1x5,故常數(shù)項(xiàng)為xC53(2x)2-1x3+1xC52(2x)3-1x2=-40+80=40.15.-12x+3y-49的展開式中不含x的項(xiàng)為C99(2x)03y-49=3y-49,令y=1,得各項(xiàng)系數(shù)之和為(3-4)9=-1.16.Dx3+x10=x3+(x+1)-110,題中a9只是(x+1)-110的展開式中(x+1)9的系數(shù),故a9=C101(-1)1=-10.17.120(1+x)6展開式的通項(xiàng)公式為Tr+1=C6r