二項(xiàng)式定理與楊輝三角第1課時(shí)　二項(xiàng)式定理+練習(xí) 高二上學(xué)期數(shù)學(xué)人教B版（2019）選擇性必修第二冊(cè)

3.3二項(xiàng)式定理與楊輝三角第1課時(shí)二項(xiàng)式定理一、選擇題1.(x-2y)10的展開(kāi)式共有 ()A.10項(xiàng) B.11項(xiàng)C.12項(xiàng) D.9項(xiàng)2.[2023·福建泉州高二期末]1x-x10的展開(kāi)式中含x2的項(xiàng)的系數(shù)為A.-45 B.-10C.10 D.453.已知(1-2x)n的展開(kāi)式中含x2的項(xiàng)的系數(shù)為40,則n為 ()A.5 B.6 C.7 D.84.[2023·河北唐山開(kāi)灤一中高二月考](a-2b-3c)4的展開(kāi)式中含abc2的項(xiàng)的系數(shù)為 ()A.208 B.-216C.217 D.-2185.[2023·北京廣渠門(mén)中學(xué)高二月考]若x-2xn的展開(kāi)式中的第4項(xiàng)和第5項(xiàng)的二項(xiàng)式系數(shù)相等,則展開(kāi)式中含x的項(xiàng)的系數(shù)為A.280 B.-280C.560 D.-5606.若(1+ax)(1+x)5的展開(kāi)式中,含x2的項(xiàng)與含x3的項(xiàng)的系數(shù)之和為-10,則a= ()A.-2 B.-1 C.10 D.17.[2023·江西九江高二期末]已知(x-1)4+2x5=a0+a1(x+1)+a2(x+1)2+a3(x+1)3+a4(x+1)4+a5(x+1)5,則a3= ()A.-2 B.2 C.4 D.128.(多選題)記2x+1xn的展開(kāi)式中第m項(xiàng)的系數(shù)為bm(m,n∈A.當(dāng)n=6時(shí),b3=240B.當(dāng)n=6時(shí),展開(kāi)式中的常數(shù)項(xiàng)是160C.若b3=2b4,則n=5D.若展開(kāi)式中含常數(shù)項(xiàng),則n的最小值是49.(多選題)在3x+1xn(n∈N*)的展開(kāi)式中,有理項(xiàng)恰有兩項(xiàng),則nA.8 B.12 C.13 D.15二、填空題10.[2023·江蘇南通高二期末]1x3·x+11.若m≠0,且(x2-x+m)6=a1+a1x+a2x2+a3x3+…+a12x12,則m的值為.

12.[2023·江蘇徐州高二期末]若x+1+axx+2三、解答題13.在x2(1)含x3的項(xiàng)的系數(shù);(2)常數(shù)項(xiàng).14.在2x+(1)求n的值;(2)求展開(kāi)式中有理項(xiàng)的系數(shù)之和(用數(shù)字作答).3.3二項(xiàng)式定理與楊輝三角第1課時(shí)二項(xiàng)式定理1.B[解析](x-2y)10的展開(kāi)式共有11項(xiàng).2.D[解析]1x-x10的展開(kāi)式的通項(xiàng)為T(mén)k+1=C10k1x10-k(-x)k=(-1)kC10kx32k-10,令3.A[解析](1-2x)n的展開(kāi)式的通項(xiàng)為T(mén)k+1=Cnk(-2x)k,展開(kāi)式中含x2的項(xiàng)的系數(shù)為Cn2(-2)2=2n(n-1),所以2n(n-1)=40,可得n=4.B[解析](a-2b-3c)4的展開(kāi)式中,含abc2的項(xiàng)為C41·a1·C31·(-2b)·C22·(-3c)2=-5.B[解析]∵x-2xn的展開(kāi)式中的第4項(xiàng)和第5項(xiàng)的二項(xiàng)式系數(shù)相等,∴Cn3=Cn4,可得n=7,∴x-2x7的展開(kāi)式的通項(xiàng)為T(mén)k+1=C7kx7-k-2xk=(-2)kC7kx7-2k,令76.A[解析](1+ax)(1+x)5的展開(kāi)式中含x2的項(xiàng)為1×C52x2+ax×C51x=(10+5a)x2,展開(kāi)式中含x3的項(xiàng)為1×C53x3+ax×C52x2=(10+10a)x3,所以10+5a+10+7.D[解析]令t=x+1,得x=t-1,則原式可化為(t-2)4+2(t-1)5=a0+a1t+a2t2+a3t3+a4t4+a5t5.(t-2)4的展開(kāi)式的通項(xiàng)為T(mén)k+1=C4kt4-k(-2)k=C4k(-2)kt4-k,令4-k=3,得k=1,則T2=C41(-2)1t3=-8t3.(t-1)5的展開(kāi)式的通項(xiàng)為T(mén)'r+1=C5rt5-r(-1)r=C5r(-1)rt5-r,令5-r=3,得r=2,則T'3=C52(-1)2t3=10t3.綜上,a8.ABC[解析]當(dāng)n=6時(shí),2x+1x6的展開(kāi)式中第3項(xiàng)的系數(shù)b3=C62×24=240,故A正確;2x+1x6的展開(kāi)式的通項(xiàng)為T(mén)k+1=C6k(2x)6-k1xk=C6k·26-kx6-2k,令6-2k=0,得k=3,故其展開(kāi)式中的常數(shù)項(xiàng)為C63×23=160,故B正確;若b3=2b4,則Cn9.AC[解析]3x+1xn(n∈N*)的展開(kāi)式的通項(xiàng)為T(mén)k+1=Cnk(3x)n-k1xk=Cnk·x2n-5k6(0≤k≤n).對(duì)于A,展開(kāi)式的通項(xiàng)為T(mén)k+1=C8k·x16-5k6,由16-5k6∈Z,可得k=2或k=8,此時(shí)展開(kāi)式中有兩個(gè)有理項(xiàng),故A正確;對(duì)于B,展開(kāi)式的通項(xiàng)為T(mén)k+1=C12k·x24-5k6,由24-5k6∈Z,可得k=10.60[解析]∵x+2x6的展開(kāi)式的通項(xiàng)為T(mén)k+1=C6kx6-k2xk=C6k2kx6-32k,∴當(dāng)k=2時(shí),T3=C611.-6[解析]由題意得(x2-x+m)6的展開(kāi)式中的常數(shù)項(xiàng)與含x的項(xiàng)的系數(shù)相等,則m6=C61(-1)m5,解得m=-6或m=12.9[解析]在x+1+axx+2x5中,令x=1,得(1+1+a)(1+2)5=243,解得a=-1.x+2x5的展開(kāi)式的通項(xiàng)為T(mén)k+1=C5kx5-k(2x-1)k=C5k2kx5-2k.當(dāng)5-2k=3時(shí),k=1,則T2=C51×2x3=10x3;當(dāng)5-2k=5時(shí),k=0,則T1=C50x5=x5;當(dāng)5-2k=4時(shí),k=13.解:(1)x2+1x6的展開(kāi)式的通項(xiàng)為T(mén)k+1=C6k(x2)6-k1xk=C6kx所以展開(kāi)式中含x3的項(xiàng)的系數(shù)為C63=(2)令12-3r=0,得r=4,所以展開(kāi)式中的常數(shù)項(xiàng)為C64=14.解:(1)2x+13xn的展開(kāi)式的通項(xiàng)為T(mén)k+1=Cnk(2x)n-k