新教材2024-2025學(xué)年高中數(shù)學(xué)第五章計(jì)數(shù)原理4二項(xiàng)式定理4.1二項(xiàng)式定理的推導(dǎo)分層作業(yè)北師大版選擇性必修第一冊(cè)

第五章§4二項(xiàng)式定理4.1二項(xiàng)式定理的推導(dǎo)A級(jí)必備學(xué)問(wèn)基礎(chǔ)練1.(2-x)5綻開(kāi)式中的第三項(xiàng)為()A.10x2 B.40x3 C.-40x3 D.80x22.[2024山東濰坊?？茧A段練習(xí)]x-2x6綻開(kāi)式中的常數(shù)項(xiàng)是(A.-160 B.-140 C.160 D.1403.(多選題)[2024福建漳州期中]在二項(xiàng)式3x2-2x5的綻開(kāi)式中,有()A.含x的項(xiàng) B.含1xC.含x4的項(xiàng) D.含1x4.(1-x)20的二項(xiàng)綻開(kāi)式中,x的系數(shù)與x9的系數(shù)之差為()A.190 B.380 C.-190 D.05.在x-12x5的綻開(kāi)式中,6.(x-y)(x+y)8的綻開(kāi)式中x2y7的系數(shù)為(用數(shù)字作答).

7.[2024廣東韶關(guān)高二統(tǒng)考期末]x-1x26的綻開(kāi)式中常數(shù)項(xiàng)為(用數(shù)字作答).

8.已知ax2+bx6的二項(xiàng)綻開(kāi)式中x3項(xiàng)的系數(shù)為20,求a2+b2的最小值.B級(jí)關(guān)鍵實(shí)力提升練9.已知(2-x)10=a0+a1x+a2x2+…+a10x10,則a8等于()A.180 B.-180 C.45 D.-4510.-2Cn1+4Cn2-8Cn3+…+(-2)A.(-1)n-1 B.(-1)nC.3n D.3n-111.(x-2y)6的綻開(kāi)式中,x4y2的系數(shù)為()A.15 B.-15 C.60 D.-6012.(3x+2y+z)5綻開(kāi)式中xy3z項(xiàng)的系數(shù)為()A.120 B.240 C.360 D.48013.(多選題)若(2-x)6=a0+a1x+a2x2+a3x3+a4x4+a5x5+a6x6,則()A.a0=64B.a0+a1+a2+a3+a4+a5+a6=1C.a0-a1+a2-a3+a4-a5+a6=36D.a3是a0,a1,a2,a3,a4,a5,a6中的最大值14.若(1+x)10=∑i=010ai(1-x)i,則a915.二項(xiàng)式x+2x27的綻開(kāi)式中,16.設(shè)a≠0,n是大于1的自然數(shù),1+xan的綻開(kāi)式為a0+a1x+a2x2+…+anxn.若點(diǎn)Ai(i,ai)的位置如圖所示,則17.[2024江西撫州高二校聯(lián)考階段練習(xí)]在二項(xiàng)式x-2xn的綻開(kāi)式中,.①全部偶數(shù)項(xiàng)的二項(xiàng)式系數(shù)之和為256;②前三項(xiàng)的二項(xiàng)式系數(shù)之和等于46.試在上面兩個(gè)條件中選擇一個(gè)補(bǔ)充在橫線(xiàn)上,并解答下列問(wèn)題:(1)求x-2xn(2)求(1-2x)n綻開(kāi)式中系數(shù)肯定值最大的項(xiàng).C級(jí)學(xué)科素養(yǎng)創(chuàng)新練18.已知正實(shí)數(shù)m,若x10=a0+a1(m-x)+a2(m-x)2+…+a10(m-x)10,其中a8=180,則m的值為.

參考答案§4二項(xiàng)式定理4.1二項(xiàng)式定理的推導(dǎo)1.D由題設(shè),綻開(kāi)式通項(xiàng)為T(mén)r+1=C5r25-r(-x)r=(-1)r25-rC5第三項(xiàng)有r=2,則T3=(-1)223C52x2=80x故選D.2.A綻開(kāi)式通項(xiàng)為T(mén)r+1=C6r·(x)6-r·-2xr=(-2)r·C6令3-r=0,所以r=3,所以常數(shù)項(xiàng)為(-2)3·C633.ABC二項(xiàng)式3x2-2x5的綻開(kāi)式的通項(xiàng)為T(mén)k+1=C5k·35-k·(-2)k·x10-3k,當(dāng)10-3k=1時(shí),k=3,知A正確;當(dāng)10-3k=-2時(shí),k=4,知B正確;當(dāng)10-3k=4時(shí),k=2,知C正確;當(dāng)10-3k=-4時(shí),k=143,知D錯(cuò)誤故選ABC.4.D5.6.-20x2y7=x·(xy7),其系數(shù)為C8x2y7=y·(x2y6),其系數(shù)為-C8∴x2y7的系數(shù)為C87-C86=7.15x-1x26綻開(kāi)式的通項(xiàng)為T(mén)r+1=C6rx6-r(-1)rx-2r=(-1)rC6rx6-3r,則有T3=(-1)28.解ax2+bx6的二項(xiàng)綻開(kāi)式的通項(xiàng)為T(mén)r+1=C6ra6-rbrx12-3r(r=0,1,2,…,6)令12-3r=3,則r=3,于是C63a3b3=20,可得ab=1,所以a2+b2≥2ab=2(當(dāng)且僅當(dāng)a=b=1時(shí)等號(hào)成立),即a2+b29.A10.A-2Cn1+4Cn2-8Cn3+…=1-2Cn1+4Cn2-8Cn3+…+(=(1-2)n-1=(-1)n-1.11.C12.D(3x+2y+z)5表示5個(gè)因式3x+2y+z的乘積,故它的綻開(kāi)式中含xy3z的項(xiàng)是由其中一個(gè)因式取3x,三個(gè)因式取2y,剩下的一個(gè)因式取z得到的,故xy3z的系數(shù)為C51×3×C43×213.ABC∵(2-x)6=a0+a1x+a2x2+a3x3+a4x4+a5x5+a6x6,∴令x=0,有26=a0=64,故選項(xiàng)A正確;令x=1,有(2-1)6=1=a0+a1+a2+a3+a4+a5+a6,故選項(xiàng)B正確;令x=-1,有(2+1)6=a0-a1+a2-a3+a4-a5+a6=36,故選項(xiàng)C正確;又a3=C63·23(-1)3<a故選ABC.14.-20因?yàn)?1+x)10=[2-(1-x)]10=∑i=010ai(1-x所以a9=C109×2×(-1)915.8416.317.解(1)x-2xn的二項(xiàng)綻開(kāi)式的通項(xiàng)為T(mén)k+1=Cnk(x)n-k·-2xk=(選①,全部偶數(shù)項(xiàng)的二項(xiàng)式系數(shù)之和為2n-1=256,可得n=9.選②,前三項(xiàng)的二項(xiàng)式系數(shù)之和為Cn0+Cn1+Cn由上知,綻開(kāi)式的通項(xiàng)為T(mén)k+1=(-2)kC9常數(shù)項(xiàng)即當(dāng)9-3k2=0時(shí),k=3,∴常數(shù)項(xiàng)為T(mén)4=(-2)3(2)由(1)得n=9,(1-2x)9的二項(xiàng)綻開(kāi)式的通項(xiàng)為T(mén)r+1=C9r(-2x)r=(-2)rC9故第(r+1)項(xiàng)的系數(shù)的肯定值為2rC由題設(shè),令2rC9r≥∴r=6,即第7項(xiàng)系數(shù)的肯定值最大,且