Chapter Review 4 - 综合练习

基础练习 / Basic Exercises

基础题

练习1 / Exercise 1

帕斯卡三角形的第16行如下所示：

The 16th row of Pascal's triangle is shown below:

\(1 \quad 15 \quad 105 \quad \ldots \quad \ldots\)

a) 求该行的下两个值。

a) Find the next two values in the row.

b) 因此求 \((1+2x)^{15}\) 展开式中 \(x^3\) 的系数。

b) Hence find the coefficient of \(x^3\) in the expansion of \((1+2x)^{15}\).

答案 / Answer

a) 第16行的完整序列：1, 15, 105, 455, 1365, 3003, 5005, 6435, 6435, 5005, 3003, 1365, 455, 105, 15, 1

下两个值是 455 和 1365

b) \((1+2x)^{15}\) 中 \(x^3\) 的系数是 \(\binom{15}{3} \cdot 2^3 = 455 \times 8 = 3640\)

基础题

练习2 / Exercise 2

已知 \(\binom{45}{17} = \frac{45!}{17!a!}\)，写出 \(a\) 的值。

Given that \(\binom{45}{17} = \frac{45!}{17!a!}\), write down the value of \(a\).

答案 / Answer

\(\binom{45}{17} = \frac{45!}{17!(45-17)!} = \frac{45!}{17!28!}\)

所以 \(a = 28\)

中等题

练习3 / Exercise 3

20个人在学校集市上玩游戏。恰好有 \(n\) 个人获奖的概率建模为：

20 people play a game at a school fair. The probability that exactly \(n\) people win a prize is modelled as:

\(\binom{20}{n}p^n(1-p)^{20-n}\)

其中 \(p\) 是任何一个人获胜的概率。计算以下情况的概率：

where \(p\) is the probability of any one person winning. Calculate the probability of:

a) 当 \(p = \frac{1}{2}\) 时，5个人获胜

b) 当 \(p = 0.7\) 时，没有人获胜

c) 当 \(p = 0.6\) 时，13个人获胜

答案保留3位有效数字。

Give your answers to 3 significant figures.

答案 / Answer

a) \(P(5) = \binom{20}{5}(\frac{1}{2})^5(\frac{1}{2})^{15} = 15504 \times \frac{1}{32} \times \frac{1}{32768} = 0.0148\)

b) \(P(0) = \binom{20}{0}(0.7)^0(0.3)^{20} = 1 \times 1 \times (0.3)^{20} = 3.49 \times 10^{-11}\)

c) \(P(13) = \binom{20}{13}(0.6)^{13}(0.4)^7 = 77520 \times (0.6)^{13} \times (0.4)^7 = 0.165\)

进阶练习 / Advanced Exercises

中等题

练习4 / Exercise 4

当 \((1-\frac{3}{2}x)^p\) 按 \(x\) 的升幂展开时，\(x\) 的系数是 -24。

When \((1-\frac{3}{2}x)^p\) is expanded in ascending powers of \(x\), the coefficient of \(x\) is -24.

a) 求 \(p\) 的值。

a) Find the value of \(p\).

b) 求展开式中 \(x^2\) 的系数。

b) Find the coefficient of \(x^2\) in the expansion.

c) 求展开式中 \(x^3\) 的系数。

c) Find the coefficient of \(x^3\) in the expansion.

答案 / Answer

a) \((1-\frac{3}{2}x)^p = 1 + p(-\frac{3}{2}x) + \ldots\)

\(x\) 的系数是 \(-\frac{3p}{2} = -24\)

所以 \(p = 16\)

b) \(x^2\) 的系数是 \(\binom{16}{2}(-\frac{3}{2})^2 = 120 \times \frac{9}{4} = 270\)

c) \(x^3\) 的系数是 \(\binom{16}{3}(-\frac{3}{2})^3 = 560 \times (-\frac{27}{8}) = -1890\)

中等题

练习5 / Exercise 5

已知：

Given that:

\((2-x)^{13} \equiv A + Bx + Cx^2 + \ldots\)

求整数 \(A\)、\(B\) 和 \(C\) 的值。

find the values of the integers \(A\), \(B\) and \(C\).

答案 / Answer

\((2-x)^{13} = 2^{13} + \binom{13}{1}2^{12}(-x) + \binom{13}{2}2^{11}(-x)^2 + \ldots\)

\(A = 2^{13} = 8192\)

\(B = -\binom{13}{1}2^{12} = -13 \times 4096 = -53248\)

\(C = \binom{13}{2}2^{11} = 78 \times 2048 = 159744\)

中等题

练习6 / Exercise 6

a) 展开 \((1-2x)^{10}\) 按 \(x\) 的升幂排列，直到并包括 \(x^3\) 项，简化展开式中的每个系数。

a) Expand \((1-2x)^{10}\) in ascending powers of \(x\) up to and including the term in \(x^3\), simplifying each coefficient in the expansion.

b) 使用你的展开式求 \((0.98)^{10}\) 的近似值，清楚地说明你为 \(x\) 使用的代换。

b) Use your expansion to find an approximation of \((0.98)^{10}\), stating clearly the substitution which you have used for \(x\).

答案 / Answer

a) \((1-2x)^{10} = 1 + 10(-2x) + 45(-2x)^2 + 120(-2x)^3 + \ldots\)

\(= 1 - 20x + 180x^2 - 960x^3 + \ldots\)

b) 设 \(x = 0.01\)，则 \(2x = 0.02\)

\((0.98)^{10} = (1-0.02)^{10} \approx 1 - 20(0.01) + 180(0.01)^2 - 960(0.01)^3\)

\(= 1 - 0.2 + 0.018 - 0.00096 = 0.817\)

挑战练习 / Challenge Exercises

挑战题

练习7 / Exercise 7

a) 使用二项式级数展开 \((2-3x)^{10}\) 按 \(x\) 的升幂排列，直到并包括 \(x^3\) 项，给出每个系数为整数。

a) Use the binomial series to expand \((2-3x)^{10}\) in ascending powers of \(x\) up to and including the term in \(x^3\), giving each coefficient as an integer.

b) 使用你的级数展开，用 \(x\) 的适当值，获得 1.97^{10} 的估计值，答案保留2位小数。

b) Use your series expansion, with a suitable value for \(x\), to obtain an estimate for \(1.97^{10}\), giving your answer to 2 decimal places.

答案 / Answer

a) \((2-3x)^{10} = 2^{10} + \binom{10}{1}2^9(-3x) + \binom{10}{2}2^8(-3x)^2 + \binom{10}{3}2^7(-3x)^3 + \ldots\)

\(= 1024 - 15360x + 103680x^2 - 414720x^3 + \ldots\)

b) 设 \(x = 0.01\)，则 \(3x = 0.03\)

\((1.97)^{10} = (2-0.03)^{10} \approx 1024 - 15360(0.01) + 103680(0.01)^2 - 414720(0.01)^3\)

\(= 1024 - 153.6 + 10.368 - 0.41472 = 880.35\)

挑战题

练习8 / Exercise 8

a) 展开 \((3+2x)^4\) 按 \(x\) 的升幂排列，给出每个系数为整数。

a) Expand \((3+2x)^4\) in ascending powers of \(x\), giving each coefficient as an integer.

b) 因此或以其他方式，写出 \((3-2x)^4\) 按 \(x\) 的升幂展开式。

b) Hence, or otherwise, write down the expansion of \((3-2x)^4\) in ascending powers of \(x\).

c) 因此通过为 \(x\) 选择合适的值，证明 \((3+2\sqrt{2})^4 + (3-2\sqrt{2})^4\) 是整数并说明其值。

c) Hence by choosing a suitable value for \(x\) show that \((3+2\sqrt{2})^4 + (3-2\sqrt{2})^4\) is an integer and state its value.

答案 / Answer

a) \((3+2x)^4 = 3^4 + 4 \cdot 3^3 \cdot 2x + 6 \cdot 3^2 \cdot (2x)^2 + 4 \cdot 3 \cdot (2x)^3 + (2x)^4\)

\(= 81 + 216x + 216x^2 + 96x^3 + 16x^4\)

b) \((3-2x)^4 = 81 - 216x + 216x^2 - 96x^3 + 16x^4\)

c) 设 \(x = \sqrt{2}\)，则：

\((3+2\sqrt{2})^4 + (3-2\sqrt{2})^4 = (81 + 216\sqrt{2} + 216 \cdot 2 + 96 \cdot 2\sqrt{2} + 16 \cdot 4) + (81 - 216\sqrt{2} + 216 \cdot 2 - 96 \cdot 2\sqrt{2} + 16 \cdot 4)\)

\(= (81 + 216\sqrt{2} + 432 + 192\sqrt{2} + 64) + (81 - 216\sqrt{2} + 432 - 192\sqrt{2} + 64)\)

\(= 577 + 408\sqrt{2} + 577 - 408\sqrt{2} = 1154\)