高考數(shù)學(xué)一輪復(fù)習(xí) 考點(diǎn)規(guī)范練45 二項(xiàng)式定理（含解析）新人教A版-新人教A版高三數(shù)學(xué)試題

考點(diǎn)規(guī)范練45二項(xiàng)式定理一、基礎(chǔ)鞏固1.x(1+x)6的展開(kāi)式中x3的系數(shù)為()A.30 B.20 C.15 D.102.設(shè)n為正整數(shù),x-1xx2A.16 B.10 C.4 D.23.(4x-2-x)6(x∈R)展開(kāi)式中的常數(shù)項(xiàng)是()A.-20 B.-15C.15 D.204.若(1+3)4=a+b3(a,b為有理數(shù)),則a+b等于()A.36 B.46C.34 D.445.已知數(shù)列{an}為等差數(shù)列,且滿足a1+a5=90.若(1-x)m展開(kāi)式中x2項(xiàng)的系數(shù)等于數(shù)列{an}的第三項(xiàng),則m的值為()A.6 B.8 C.9 D.106.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()A.74 B.121C.-74 D.-1217.使3x+1xxn(n∈NA.4 B.5C.6 D.78.已知(x+1)10=a1+a2x+a3x2+…+a11x10.若數(shù)列a1,a2,a3,…,ak(1≤k≤11,k∈Z)是一個(gè)單調(diào)遞增數(shù)列,則k的最大值是()A.6 B.7 C.8 D.59.二項(xiàng)式2x-1x10.若ax2+1x5的展開(kāi)式中x5的系數(shù)是11.設(shè)(x-2)5=a0+a1(x+1)+a2(x+1)2+…+a5(x+1)5,則a1+a2+…+a5=.

12.已知(1+ax)(1+x)5的展開(kāi)式中x3的系數(shù)為5,則實(shí)數(shù)a=.

二、能力提升13.若x+axA.-40 B.-20C.20 D.4014.若(1+x+x2)6=a0+a1x+a2x2+…+a12x12,則a2+a4+…+a12=()A.256 B.364C.296 D.51315.(x+y)(2x-y)5的展開(kāi)式中x3y3的系數(shù)為()A.-80 B.-40 C.40 D.8016.已知多項(xiàng)式(x+1)3(x+2)2=x5+a1x4+a2x3+a3x2+a4x+a5,則a4=,a5=.

17.若x9=a0+a1(x-1)+a2(x-1)2+…+a9(x-1)9,則a1+a3三、高考預(yù)測(cè)18.已知二項(xiàng)式x2+1xn

考點(diǎn)規(guī)范練45二項(xiàng)式定理1.C解析因?yàn)?1+x)6的展開(kāi)式的第(k+1)項(xiàng)為Tk+1=C6kxk,所以x(1+x)6的展開(kāi)式中x3的項(xiàng)為C62x3=152.B解析因?yàn)閤-1xx2n展開(kāi)式的通項(xiàng)公式為Tk+1=C2nkx2n-k-1xxk=C3.C解析設(shè)展開(kāi)式中的常數(shù)項(xiàng)是第(k+1)項(xiàng),則Tk+1=C6k·(4x)6-k·(-2-x)k=C6k·(-1)k·212x-2kx·2-kx=C6k·(-1)k·令12x-3kx=0,解得k=4,故常數(shù)項(xiàng)為T5=C64·(-1)4=4.D解析(1+3)4=1+C41·3+C42(3)2+C43(3)3+故a+b=44.5.D解析由題意,a3=a1+a52=902=45,(1-x)m展開(kāi)式中x2項(xiàng)的系數(shù)為6.D解析展開(kāi)式中含x3項(xiàng)的系數(shù)為C53(-1)3+C63(-1)3+C73(-1)3+C87.B解析Tr+1=Cnr(3x)n-r1xxr當(dāng)Tr+1是常數(shù)項(xiàng)時(shí),有n-52r=0,故選B8.A解析由二項(xiàng)式定理知an=C10n-1(n=1,2,3,…,11).又(x+1)10展開(kāi)式中二項(xiàng)式系數(shù)最大項(xiàng)是第6項(xiàng),故a6=C9.-160解析二項(xiàng)式2x-1x6的通項(xiàng)為Tr+1=C6r(2x)6-令3-r=0,則r=3.故(-1)3×C63×23=-20×8=-10.-2解析因?yàn)門r+1=C5r(ax2)5-r1xr=C5ra5-r·x10-5r2,所以由10-5r2=11.211解析將(x-2)5=a0+a1(x+1)+a2(x+1)2+…+a5(x+1)5化為[(x+1)-3]5=a0+a1(x+1)+a2(x+1)2+…+a5(x+1)5,令x+1=0,得a0=-35,令x+1=1,得a0+a1+a2+a3+a4+a5=-25,則a1+a2+a3+a4+a5=-25+35=211.12.-12解析∵(1+x)5=1+C51x+C52x2+C53x3+C∴(1+ax)(1+x)5的展開(kāi)式中x3的系數(shù)為a·C52即10a+10=5,解得a=-1213.D解析在x+ax得(1+a)(2-1)5=2,即a=1.原式=x·2x故常數(shù)項(xiàng)為x·C53(2x)2-1x3+1x·C52(214.B解析令x=1,則a0+a1+a2+…+a12=36,①令x=-1,則a0-a1+a2-…+a12=1,②由①+②,可得a0+a2+a4+…+a12=36令x=0,則a0=1,故a2+a4+…+a12=36+12-115.C解析(2x-y)5的展開(kāi)式的通項(xiàng)公式Tr+1=C5r(2x)5-r(-y)當(dāng)r=3時(shí),x(2x-y)5的展開(kāi)式中x3y3的系數(shù)為C53×22×(-1)3當(dāng)r=2時(shí),y(2x-y)5的展開(kāi)式中x3y3的系數(shù)為C52×23×(-1)2=故展開(kāi)式中x3y3的系數(shù)為80-40=40.16.164解析由二項(xiàng)式展開(kāi)式可得通項(xiàng)公式為C3rx3-r·C2mx2-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=13×217.649解析令x=2,得29=a0+a1+a2+…+a8+a9令x=0,得0=a0-a1+a2-…+a8-a9,所以a1+a3+a5+a7+a9=a0+a2+a4+a6+a