Section: 二进制理论论 | 14.3二元论理论

Section outline

Introduction
::导言

The Binomial Theorem is used to generate IP addresses that are assigned to different computers, make economic predictions, and determine the cost and time needed to complete projects. The Binomial Theorem is a formula you can use to expand a binomial such as $\begin{aligned} (2 x - 3)^{5} \end{aligned}$ without having to compute the repeated distribution. What is the expanded version of $\begin{aligned} (2 x - 3)^{5} \end{aligned}$ ?
::Binomial 理论用于生成分配给不同计算机的IP地址,作出经济预测,并确定完成项目所需的成本和时间。 Binomial 理论是一种公式,可以用来扩大二进制,例如(2x-3)5,而不必计算重复的分布。扩展版(2x-3)是什么?

Binomial Theorem
::二进制理论论

The Binomial Theorem states
::比约莫定理论国

$\begin{aligned} (a + b)^{n} = \sum_{i = 0}^{n} (\binom{n}{i}) a^{i} b^{n - i} . \end{aligned}$
:a+b)ni=0n(ni)aibn-i。

Writing out a few terms of the summation symbol helps you to understand how this theorem works:
::写出总和符号的几个术语, 帮助您了解这个定理是如何运作的:

$\begin{aligned} (a + b)^{n} = (\binom{n}{0}) a^{n} + (\binom{n}{1}) a^{n - 1} b^{1} + (\binom{n}{2}) a^{n - 2} b^{2} + \dots + (\binom{n}{n}) b^{n} \end{aligned}$

:a+b)n=(n0)an+(n1)an-1b1+(n2)an-2b2(nn)bn

Going from one term to the next in the expansion, you should notice that the exponents of $\begin{aligned} a \end{aligned}$ decrease while the exponents of $\begin{aligned} b \end{aligned}$ increase. You should also notice that the coefficients of each term are combinations. Recall that $\begin{aligned} (\binom{n}{0}) \end{aligned}$ is the number of ways to choose $\begin{aligned} 0 \end{aligned}$ objects from a set of $\begin{aligned} n \end{aligned}$ objects.
::从一个任期到扩展的下一个任期, 您应该注意到, b 增加的指数是减少的指数。您也应该注意到, 每个术语的系数是组合的。回顾 (n0) 是从一组点选择 0 对象的方法数。

Binomial Theorem
::二进制理论论

The Binomial Theorem is a theorem that states how to expand binomials that are raised to a power using combinations. The Binomial Theorem is $\begin{aligned} (a + b)^{n} = \sum_{i = 0}^{n} (\binom{n}{i}) a^{i} b^{n - i} \end{aligned}$ .
::Binomial 定理是一个理论理论,它说明了如何将二进制发展成使用组合的力量。 Binomial定理是(a+b)ni=0n(ni)aibn-i。

The following video demonstrates how to apply the Binomial Theorem:
::以下影片展示了如何应用Binomial定理:

Pascal's Triangle
::帕斯卡尔三角

Another way to think about the coefficients in the Binomial Theorem is that they are the numbers from Pascal's Triangle:
::思考Binomial定理中系数的另一种方式是,

This triangle was named for mathematician Blaise Pascal (shown below).
::此三角形命名为数学家Blaise Pascal(如下表所示)。

Look at the expansions of $\begin{aligned} (a + b)^{n} \end{aligned}$ below, and notice how the coefficients of the terms are the numbers in Pascal's Triangle:
::看看下面(a+b)n的扩展,

Be extremely careful when working with binomials of the form $\begin{aligned} (a - b)^{n} \end{aligned}$ . You need to remember to capture the negative with the 2nd term as you write out the expansion: $\begin{aligned} (a - b)^{n} = (a + (- b))^{n} \end{aligned}$ .
::在处理表(a-b)n的二进制时,要非常小心。在写扩展内容a-b)n=(a+(-b))n时,要记住用第二个学期来记录负数。

The following video explains binomial expansion using Pascal's Triangle.
::以下影片用Pascal的三角形解释二进制扩张。

Play, Learn, and Explore with Pascal's Triangle:
::与帕斯卡三角一起玩、学习、探索:

Examples
::实例

Example 1
::例1

Expand the binomial below using the Binomial Theorem.
::使用 Binomial 定理展开下面的二进制。

$\begin{aligned} (m - n)^{6} \end{aligned}$

:m-n)6

Solution:
::解决方案 :

$\begin{aligned} (m - n)^{6} & = (\binom{6}{0}) m^{6} + (\binom{6}{1}) m^{5} (- n)^{1} + (\binom{6}{2}) m^{4} (- n)^{2} + (\binom{6}{3}) m^{3} (- n)^{3} \\ + (\binom{6}{4}) m^{2} (- n)^{4} + (\binom{6}{5}) m^{1} (- n)^{5} + (\binom{6}{6}) (- n)^{6} \\ = 1 m^{6} - 6 m^{5} n + 15 m^{4} n^{2} - 20 m^{3} n^{3} + 15 m^{2} n^{4} - 6 m^{1} n^{5} + 1 n^{6} \end{aligned}$

:m-n)6=(60)m6+(61)5m51+(62)m42+(62)42+(63)33+(64)m24+(65)m15+(66)6=1m6-6m5n+6m5n+15m4n2-20m3n3+15m2n3+15m2n4-6m1n5+1n6)

Example 2
::例2

What is the coefficient of the term $\begin{aligned} x^{7} y^{9} \end{aligned}$ in the expansion of the binomial $\begin{aligned} (x + y)^{16} \end{aligned}$ ?
::二进制( X+y) 16 扩展中的 X7y9 术语系数是多少?

Solution:
::解决方案 :

The Binomial Theorem allows you to calculate just the coefficient you need.
::Binomial 定理允许您只计算您需要的系数。

$\begin{aligned} (\binom{16}{9}) = \frac{16!}{9! 7!} = \frac{16 \cdot 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10}{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 11, 440 \end{aligned}$

Note that symmetry occurs in combinations. That is, 16 choose 9 and 16 choose 7 will give the same result.
::请注意对称在组合中发生。也就是说, 16选择 9 和 16选择 7 将产生相同的结果。

Example 3
::例3

What is the coefficient of $\begin{aligned} x^{6} \end{aligned}$ in the expansion of $\begin{aligned} (4 - 3 x)^{7} \end{aligned}$ ?
::在扩大(4-3x)7时x6系数是多少?

Solution:
::解决方案 :

For this problem, you should calculate the whole term, since the 3 and the 4 in $\begin{aligned} (3 - 4 x) \end{aligned}$ will impact the coefficient of $\begin{aligned} x^{6} \end{aligned}$ as well. $\begin{aligned} (\binom{7}{6}) 4^{1} (- 3 x)^{6} = 7 \cdot 4 \cdot 729 x^{6} = 20, 412 x^{6} \end{aligned}$ . The coefficient is 20,412.
::对于这个问题,你应该计算整个术语,因为3-4x中的3和4也将影响x6的系数。 (7641(-3x)6=74729x6=20,412x6. 系数为20,412。)

Example 4
::例4

Recall the problem from the Introduction: What is the expanded version of $\begin{aligned} (2 x - 3)^{5} \end{aligned}$ ?
::回顾导言中的问题:扩大版(2x-3)是什么?

Solution:
::解决方案 :

The expanded version of $\begin{aligned} (2 x - 3)^{5} \end{aligned}$ is
::扩大版(2x-3)5

$\begin{aligned} (2 x - 3)^{5} & = (\binom{5}{0}) (2 x)^{5} + (\binom{5}{1}) (2 x)^{4} (- 3)^{1} + (\binom{5}{2}) (2 x)^{3} (- 3)^{2} \\ + (\binom{5}{3}) (2 x)^{2} (- 3)^{3} + (\binom{5}{4}) (2 x)^{1} (- 3)^{4} + (\binom{5}{5}) (- 3)^{6} \\ = (2 x)^{5} + 5 (2 x)^{4} (- 3)^{1} + 10 (2 x)^{3} (- 3)^{2} \\ + 10 (2 x)^{2} (- 3)^{3} + 5 (2 x)^{1} (- 3)^{4} + (- 3)^{5} \\ = 32 x^{5} - 240 x^{4} + 720 x^{3} - 1080 x^{2} + 810 x - 243. \end{aligned}$

:2x3)5=(50)2(2x)5+(51)2(2x)4(3)1+(52)2(2x)3(3)2+(53)2(2x)(3)2(2x)(3)3+(54)2(2x)(1)(3)4+(5)4(3)6=(2x)5+5(2x)(3)4(3)1+(2x)(3)2x)(3)2+10(2x)(3)2(2x)(3)3)(3)3+5(2x)(2x)1(3)4)4+(3)5=32x5-240x4(4)720x3-1080x2+810x-243)。

Example 5
::例5

What is the coefficient of $\begin{aligned} x^{3} \end{aligned}$ in the expansion of $\begin{aligned} (x - 4)^{5} \end{aligned}$ ?
::在(x-4)5的扩展中x3系数是多少?

Solution:
::解决方案 :

$\begin{aligned} (\binom{5}{2}) \cdot 1^{3} (- 4)^{2} = 160 \end{aligned}$

Example 6
::例6

Compute the following summation:
::计算下列总和:

$\begin{aligned} \sum_{i = 0}^{4} (\binom{4}{i}) \end{aligned}$

::i=04(4i)

Solution:
::解决方案 :

This problem is asking for $\begin{aligned} (\binom{4}{0}) + (\binom{4}{1}) + \dots + (\binom{4}{4}) \end{aligned}$ , which are the sum of all the coefficients of $\begin{aligned} (a + b)^{4} \end{aligned}$ .
$\begin{aligned} 1 + 4 + 6 + 4 + 1 = 16 \end{aligned}$
Note that the sum of the coefficients of the expansion of $\begin{aligned} (a + b)^{4} \end{aligned}$ is the same as what happens when $\begin{aligned} a = b = 1 \end{aligned}$ . So, computing directly,
$\begin{aligned} (1 + 1)^{4} = 2^{4} = 16. \end{aligned}$
Example 7
::这个问题正在要求(40)+(41)(44),这是(a+b)4.1+4+6+4+4+4+1=16的所有系数的总和。注a+b)4的扩展系数总和与a=b=1时发生的情况相同。因此,直接计算(1+1)4=24=16。

Collapse the following binomial expansion using the Binomial Theorem:
$\begin{aligned} 32 x^{5} - 80 x^{4} + 80 x^{3} - 40 x^{2} + 10 x - 1 \end{aligned}$

::使用二进制定理折叠以下二进制扩展 : 32x5- 80x4+80x3- 40x2+10x- 1

Solution:
::解决方案 :

Since the last term is -1 and the power on the 1st term is a 5, you can conclude that the 2nd half of the binomial is $\begin{aligned} (? - 1)^{5} \end{aligned}$ . The 1st term is positive and $\begin{aligned} (2 x)^{5} = 32 x^{5} \end{aligned}$ , so the 1st term in the binomial must be $\begin{aligned} 2 x \end{aligned}$ . The binomial is $\begin{aligned} (2 x - 1)^{5} \end{aligned}$ .
::由于最后一个学期是-1,第一个学期的权力是5,你可以得出结论,二进制的第二半是(?)-1,5。第一个学期是正的,第二个学期是2x5=325,因此二进制的第一个学期必须是2x。二进制是(2x-1),5。

Summary
::摘要

The Binomial Theorem (or binomial expansion ) describes the algebraic expansion of powers of a binomial.
::二元论(或二元论扩展)描述了二元论力量的代数扩张。

The coefficients in the binomial expansion appear as the entries of Pascal's Triangle. In Pascal's Triangle, each entry is the sum of the two above it.
::二进制扩展中的系数显示为帕斯卡尔三角的条目。在帕斯卡尔三角中,每个条目是上面两个条目的总和。

The coefficients can also be generated using combinations. The numbers in the combination are associated with a particular row and column in Pascal's Triangle. For example, 5 combinations of 3 would be associated with the 5th row and 3rd column of the Triangle.
::也可以使用组合生成系数。组合中的数字与Pascal三角的某一行和列相关。例如, 5 组合3 与三角的第 5 行和 3 列相关。

The Binomial Theorem is $\begin{aligned} (a + b)^{n} = \sum_{i = 0}^{n} (\binom{n}{i}) a^{i} b^{n - i} \end{aligned}$ .
::Binomial定理是(a+b)ni=0n(ni)aibn-i。

Review
::回顾

Expand each of the following binomials using the Binomial Theorem:
::使用 Binomial 定理展开下列二进制 :

1. $\begin{aligned} (x - y)^{4} \end{aligned}$
::1. (x-y)4

2. $\begin{aligned} (x - 3 y)^{5} \end{aligned}$
::2. (x-3y)5

3. $\begin{aligned} (2 x + 4 y)^{7} \end{aligned}$
::3. (2x+4y)7

4. What is the coefficient of $\begin{aligned} x^{4} \end{aligned}$ in $\begin{aligned} (x - 2)^{7} \end{aligned}$ ?
::4. (x-2)7中的x4系数是多少?

5. What is the coefficient of $\begin{aligned} x^{3} y^{5} \end{aligned}$ in $\begin{aligned} (x + y)^{8} \end{aligned}$ ?
::5. (x+y)8中的x3y5系数是多少?

6. What is the coefficient of $\begin{aligned} x^{5} \end{aligned}$ in $\begin{aligned} (2 x - 5)^{6} \end{aligned}$ ?
::6. 乘以x5的系数是多少(2x-5)6 ?

7. What is the coefficient of $\begin{aligned} y^{2} \end{aligned}$ in $\begin{aligned} (4 y - 5)^{4} \end{aligned}$ ?
::7. y2在(4Y-5)中的系数是多少? 4

8. What is the coefficient of $\begin{aligned} x^{2} y^{6} \end{aligned}$ in $\begin{aligned} (2 x + y)^{8} \end{aligned}$ ?
::8. 在(2x+y)8中x2y6系数是多少?

9. What is the coefficient of $\begin{aligned} x^{3} y^{4} \end{aligned}$ in $\begin{aligned} (5 x + 2 y)^{7} \end{aligned}$ ?
::9. 乘以x3y4(5x+2y)7的系数是多少?

Compute the following summations:
::计算下列总和:

10. $\begin{aligned} \sum_{i = 0}^{9} (\binom{9}{i}) \end{aligned}$
::10. i=09(9i)

11. $\begin{aligned} \sum_{i = 0}^{12} (\binom{12}{i}) \end{aligned}$
::11. i=012(12i)

12. $\begin{aligned} \sum_{i = 0}^{8} (\binom{8}{i}) \end{aligned}$
::12. i=08(8)i

Collapse the following polynomials using the Binomial Theorem:
::使用 Binomial 定理折叠下列多符号 :

13. $\begin{aligned} 243 x^{5} - 405 x^{4} + 270 x^{3} - 90 x^{2} = 15 x - 1 \end{aligned}$
::13. 243x5-405x4+270x3-90x2=15x-1

14. $\begin{aligned} x^{7} - 7 x^{6} y + 21 x^{5} y^{2} - 35 x^{4} y^{3} + 35 x^{3} y^{4} - 21 x^{2} y^{5} + 7 x y^{6} - y^{7} \end{aligned}$
::14. x7-7x6y+21x5y2-35x4y3+35x4y3+35x3y4-21x2y5+7xy6-y7

15. $\begin{aligned} 128 x^{7} - 448 x^{6} y + 672 x^{5} y^{2} - 560 x^{4} y^{3} + 280 x^{3} y^{4} - 84 x^{2} y^{5} + 14 x y^{6} - y^{7} \end{aligned}$
::15. 128x7-448x6y+672x5y2-560x4y3+280x3y4-84x2y5+14xy6-y7

Review (Answers)
::回顾(答复)

Please see the Appendix.
::请参看附录。

Section outline

Introduction ::导言

Binomial Theorem ::二进制理论论

Binomial Theorem ::二进制理论论

Pascal's Triangle ::帕斯卡尔三角

Examples ::实例

( 5 2 ) ⋅ 1 3 ( − 4 ) 2 = 160

Summary ::摘要

Review ::回顾

Review (Answers) ::回顾(答复)

Introduction
::导言

Binomial Theorem
::二进制理论论

Binomial Theorem
::二进制理论论

Pascal's Triangle
::帕斯卡尔三角

Examples
::实例

$\begin{aligned} (\binom{5}{2}) \cdot 1^{3} (- 4)^{2} = 160 \end{aligned}$

Summary
::摘要

Review
::回顾

Review (Answers)
::回顾(答复)