二項(xiàng)式性質(zhì)試題及答案

單項(xiàng)選擇題（每題2分，共10題）1.二項(xiàng)式\((a+b)^n\)展開式的項(xiàng)數(shù)是（）A.\(n\)項(xiàng)B.\(n+1\)項(xiàng)C.\(n-1\)項(xiàng)D.\(2n\)項(xiàng)2.二項(xiàng)式\((x+2)^5\)展開式中\(zhòng)(x^3\)的系數(shù)是（）A.10B.20C.40D.803.二項(xiàng)式\((1-x)^6\)展開式中第4項(xiàng)的二項(xiàng)式系數(shù)是（）A.\(C_{6}^3\)B.\(C_{6}^4\)C.\(-C_{6}^3\)D.\(-C_{6}^4\)4.若\((1+x)^n\)展開式中第3項(xiàng)與第7項(xiàng)的二項(xiàng)式系數(shù)相等，則\(n\)的值為（）A.8B.9C.10D.115.二項(xiàng)式\((2x-\frac{1}{x})^6\)展開式中的常數(shù)項(xiàng)是（）A.-160B.160C.-20D.206.二項(xiàng)式\((a+b)^n\)展開式中，二項(xiàng)式系數(shù)最大的項(xiàng)在（）A.中間一項(xiàng)B.中間兩項(xiàng)C.當(dāng)\(n\)為偶數(shù)時(shí)在中間一項(xiàng)，\(n\)為奇數(shù)時(shí)在中間兩項(xiàng)D.無(wú)法確定7.二項(xiàng)式\((1+2x)^4\)展開式中\(zhòng)(x^2\)的系數(shù)為（）A.6B.12C.24D.488.已知\((x+\frac{1}{x})^n\)展開式的二項(xiàng)式系數(shù)之和為64，則\(n\)的值為（）A.5B.6C.7D.89.二項(xiàng)式\((3x-1)^5\)展開式中各項(xiàng)系數(shù)之和是（）A.1B.-1C.32D.-3210.二項(xiàng)式\((x-\sqrt{2})^7\)展開式中\(zhòng)(x^5\)的系數(shù)是（）A.14B.-14C.\(14\sqrt{2}\)D.\(-14\sqrt{2}\)多項(xiàng)選擇題（每題2分，共10題）1.對(duì)于二項(xiàng)式\((a+b)^n\)，以下說(shuō)法正確的是（）A.展開式中各項(xiàng)的二項(xiàng)式系數(shù)之和為\(2^n\)B.展開式中第\(r+1\)項(xiàng)的通項(xiàng)公式為\(T_{r+1}=C_{n}^ra^{n-r}b^r\)C.當(dāng)\(n\)為偶數(shù)時(shí)，中間一項(xiàng)的二項(xiàng)式系數(shù)最大D.當(dāng)\(n\)為奇數(shù)時(shí)，中間兩項(xiàng)的二項(xiàng)式系數(shù)相等且最大2.二項(xiàng)式\((x+y)^5\)展開式中（）A.各項(xiàng)系數(shù)之和為\(32\)B.二項(xiàng)式系數(shù)之和為\(32\)C.\(x^3y^2\)的系數(shù)為\(10\)D.奇數(shù)項(xiàng)的二項(xiàng)式系數(shù)之和為\(16\)3.下列關(guān)于二項(xiàng)式\((2x-3)^4\)的說(shuō)法正確的是（）A.展開式的通項(xiàng)為\(T_{r+1}=C_{4}^r(2x)^{4-r}(-3)^r\)B.展開式中\(zhòng)(x^2\)的系數(shù)為\(216\)C.展開式中各項(xiàng)系數(shù)之和為\(1\)D.展開式中所有項(xiàng)的二項(xiàng)式系數(shù)之和為\(16\)4.二項(xiàng)式\((a-\frac{1}{a})^6\)展開式中（）A.常數(shù)項(xiàng)為\(-20\)B.二項(xiàng)式系數(shù)最大的項(xiàng)是第4項(xiàng)C.含\(a^2\)的項(xiàng)的系數(shù)為\(15\)D.所有項(xiàng)的二項(xiàng)式系數(shù)之和為\(64\)5.若\((1+mx)^6=a_0+a_1x+a_2x^2+\cdots+a_6x^6\)，且\(a_1+a_2+\cdots+a_6=63\)，則實(shí)數(shù)\(m\)的值可能為（）A.1B.-1C.2D.-26.二項(xiàng)式\((\sqrt{x}+\frac{1}{x})^n\)展開式中（）A.當(dāng)\(n=6\)時(shí)，展開式中常數(shù)項(xiàng)為\(15\)B.展開式的通項(xiàng)為\(T_{r+1}=C_{n}^r(\sqrt{x})^{n-r}(\frac{1}{x})^r\)C.若展開式中第3項(xiàng)與第5項(xiàng)的二項(xiàng)式系數(shù)相等，則\(n=6\)D.二項(xiàng)式系數(shù)之和為\(2^n\)7.二項(xiàng)式\((x^2+\frac{1}{x})^5\)展開式中（）A.含\(x^7\)的項(xiàng)的系數(shù)為\(5\)B.展開式的通項(xiàng)為\(T_{r+1}=C_{5}^r(x^2)^{5-r}(\frac{1}{x})^r\)C.常數(shù)項(xiàng)為\(10\)D.各項(xiàng)系數(shù)之和為\(32\)8.關(guān)于二項(xiàng)式\((1+x)^n\)展開式，下列說(shuō)法正確的是（）A.當(dāng)\(n=10\)時(shí)，展開式中系數(shù)最大的項(xiàng)是第6項(xiàng)B.展開式中各項(xiàng)系數(shù)之和等于二項(xiàng)式系數(shù)之和C.展開式中第\(r\)項(xiàng)的二項(xiàng)式系數(shù)為\(C_{n}^{r-1}\)D.展開式中\(zhòng)(x^k\)的系數(shù)為\(C_{n}^k\)9.二項(xiàng)式\((3-2x)^7\)展開式中（）A.各項(xiàng)系數(shù)之和為\(1\)B.二項(xiàng)式系數(shù)之和為\(128\)C.\(x^3\)的系數(shù)為\(-2268\)D.奇數(shù)項(xiàng)的二項(xiàng)式系數(shù)之和為\(64\)10.若二項(xiàng)式\((x+\frac{a}{x})^n\)展開式的二項(xiàng)式系數(shù)之和為\(128\)，且展開式中\(zhòng)(x^3\)的系數(shù)為\(21\)，則（）A.\(n=7\)B.\(a=1\)C.展開式中常數(shù)項(xiàng)為\(35\)D.展開式中所有項(xiàng)的系數(shù)之和為\(128\)判斷題（每題2分，共10題）1.二項(xiàng)式\((a+b)^n\)展開式中各項(xiàng)系數(shù)之和為\(2^n\)。（）2.二項(xiàng)式\((x-2)^5\)展開式中第3項(xiàng)的系數(shù)是\(C_{5}^2(-2)^2\)。（）3.二項(xiàng)式\((1+2x)^n\)展開式中各項(xiàng)二項(xiàng)式系數(shù)最大的項(xiàng)一定是中間項(xiàng)。（）4.若\((1+x)^n\)展開式中第\(r\)項(xiàng)的二項(xiàng)式系數(shù)等于第\(s\)項(xiàng)的二項(xiàng)式系數(shù)，則\(r=s\)。（）5.二項(xiàng)式\((a+b)^n\)展開式的通項(xiàng)公式\(T_{r+1}=C_{n}^ra^{n-r}b^r\)中\(zhòng)(r\)從\(0\)到\(n\)。（）6.二項(xiàng)式\((x+\frac{1}{x})^6\)展開式中的常數(shù)項(xiàng)是\(20\)。（）7.二項(xiàng)式\((3x-1)^5\)展開式中各項(xiàng)系數(shù)之和為\(-32\)。（）8.二項(xiàng)式\((a-b)^n\)展開式中奇數(shù)項(xiàng)的二項(xiàng)式系數(shù)之和等于偶數(shù)項(xiàng)的二項(xiàng)式系數(shù)之和。（）9.二項(xiàng)式\((\sqrt{x}+1)^n\)展開式中含\(x^k\)的項(xiàng)的系數(shù)為\(C_{n}^k\)。（）10.二項(xiàng)式\((2x-\frac{1}{x^2})^6\)展開式中\(zhòng)(x^3\)的系數(shù)為\(-192\)。（）簡(jiǎn)答題（每題5分，共4題）1.求二項(xiàng)式\((x+2)^6\)展開式中\(zhòng)(x^4\)的系數(shù)。-答案：根據(jù)通項(xiàng)公式\(T_{r+1}=C_{6}^rx^{6-r}2^r\)，令\(6-r=4\)，得\(r=2\)。則\(x^4\)系數(shù)為\(C_{6}^2\times2^2=\frac{6!}{2!(6-2)!}\times4=15\times4=60\)。2.已知\((1+x)^n\)展開式中第5項(xiàng)與第7項(xiàng)的二項(xiàng)式系數(shù)相等，求\(n\)的值。-答案：由二項(xiàng)式系數(shù)性質(zhì)，\(C_{n}^4=C_{n}^6\)。根據(jù)二項(xiàng)式系數(shù)對(duì)稱性\(C_{n}^m=C_{n}^{n-m}\)，所以\(n=4+6=10\)。3.求二項(xiàng)式\((3x-\frac{1}{x})^4\)展開式中的常數(shù)項(xiàng)。-答案：通項(xiàng)\(T_{r+1}=C_{4}^r(3x)^{4-r}(-\frac{1}{x})^r=C_{4}^r3^{4-r}x^{4-r}(-1)^rx^{-r}=C_{4}^r3^{4-r}(-1)^rx^{4-2r}\)。令\(4-2r=0\)，\(r=2\)。常數(shù)項(xiàng)為\(C_{4}^2\times3^{2}\times(-1)^2=6\times9=54\)。4.若\((2+x)^n\)展開式的二項(xiàng)式系數(shù)之和為\(128\)，求展開式中\(zhòng)(x^3\)的系數(shù)。-答案：因?yàn)槎?xiàng)式系數(shù)之和為\(2^n=128\)，所以\(n=7\)。通項(xiàng)\(T_{r+1}=C_{7}^r2^{7-r}x^r\)，令\(r=3\)，\(x^3\)系數(shù)為\(C_{7}^3\times2^{4}=\frac{7!}{3!(7-3)!}\times16=35\times16=560\)。討論題（每題5分，共4題）1.討論二項(xiàng)式系數(shù)與展開式系數(shù)的區(qū)別與聯(lián)系。-答案：區(qū)別：二項(xiàng)式系數(shù)僅與組合數(shù)\(C_{n}^r\)有關(guān)，展開式系數(shù)還與各項(xiàng)中字母前數(shù)字有關(guān)。聯(lián)系：在\((a+b)^n\)中，當(dāng)\(a=b=1\)時(shí)，展開式系數(shù)之和等于二項(xiàng)式系數(shù)之和\(2^n\)。2.如何利用二項(xiàng)式性質(zhì)求展開式中系數(shù)最大的項(xiàng)？-答案：設(shè)二項(xiàng)式\((a+bx)^n\)，設(shè)第\(r+1\)項(xiàng)系數(shù)最大，列出不等式組\(\begin{cases}C_{n}^ra^{n-r}b^r\geqC_{n}^{r-1}a^{n-r+1}b^{r-1}\\C_{n}^ra^{n-r}b^r\geqC_{n}^{r+1}a^{n-r-1}b^{r+1}\end{cases}\)，求解\(r\)范圍確定項(xiàng)數(shù)。3.舉例說(shuō)明二項(xiàng)式性質(zhì)在實(shí)際計(jì)算中的應(yīng)用。-答案：比如求\((1.02)^5\)近似值，將其寫成\((1+0.02)^5\)，利用二項(xiàng)式展開\(C_{5}^0+C_{5}^1\times0.02+C_{5}^2\times(0.02)^2+\cdots\)，取前幾項(xiàng)計(jì)算得近似值，簡(jiǎn)化復(fù)雜乘法運(yùn)算。4.探討二項(xiàng)式\((a+b)^n\)與\((b+a)^n\)展開式的關(guān)系。-答案：二者展開式本質(zhì)相同，因?yàn)閈((a+b)^n=\sum_{r=0}^{n}C_{n}^ra^{n-r}b^r\)，\((b+a)^n=\sum_{r=0}^{n}C_{n}^rb^{n-r}a^r\)，只是各項(xiàng)中\(zhòng)(a\)、\(b\)順序不同，但二項(xiàng)式系數(shù)和展開式系數(shù)都相同。答案單項(xiàng)選擇題1.B2.C3.A4.A5.A6.C7.C8.B9.B10.A多項(xiàng)選擇題1.ABCD2.ABCD3.ACD4.BD5.AC6.ABCD7