Course: 12.8 二元论理论 | The Way To Learn

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二进制理论论

The Binomial Theorem tells you how 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 理论告诉您如何扩大二进制( 2x-3) 5 , 而不必计算重复的分布。扩展版本 ( 2x-3) 5 是什么 ?

Introduction to the Binomial Theorem
::Binomial定理介绍

The Binomial Theorem states:
::Binomial定理指出:

$\begin{aligned} (a + b)^{n} = \sum_{i = 0}^{n} (\binom{n}{i}) a^{i} b^{n - i} \end{aligned}$
: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 objects from a set of $\begin{aligned} n \end{aligned}$ objects.
::从一个词到扩展的下一个词,您应该注意到,在 b 增加的引言中,减幅的指数是减少的指数。您也应该注意到, 每个词的系数是组合的。回顾 (n0) 是从一组 n 对象中选择对象的方法数。

Take the following binomial:
::采取以下二进制 :

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

It can be expanded using the Binomial Theorem:
::使用Binomial定理可以扩大:

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

m-n)6=(60)m6+(61)5m5

1+(62)m4

2+(62)4

2+(63)3

3+(64)m2

4+(65)m1

5+(66)

6=1m6-6m5n+6m5n+15m4n2-20m3n3+15m2n3+15m2n4-6m1n5+1n6)

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 second 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(a-b)n=(a+(-b))n时,需要记住用第二学期来记录负数。n。

Another way to think about the coefficients in the Binomial Theorem is that they are the numbers from Pascal’s Triangle. 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.
::思考Binomial理论中系数的另一种方式是它们来自帕斯卡尔三角。看看下面(a+b)的扩展,并注意这些术语的系数是如何在帕斯卡尔三角中的数字的。

Examples
::实例

Example 1
::例1

Earlier, you were asked to expand $\begin{aligned} (2 x - 3)^{5} \end{aligned}$ . The expanded version of $\begin{aligned} (2 x - 3)^{5} \end{aligned}$ is:
::早些时候,你被要求扩大(2x-3)5. 扩大版(2x-3)5 如下:

\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)(2x)5+(50)(2x)5+(51)(2x)4(3)1+(52)(2x)3(3)3(3)2+(53)(2x)(3)(2x)(2x)(2)(3)(3)3+(54)(2x)(1)(3)4(3)(3)(3)6=(2x)5(5)(2x)(4)(3)(3)1(2x)(2x)(3)5+10(2x)(2x)(2x)(3)3)(3)3+5(2x)(2x)(3)3)(3)(3)3(3)3(3)(3)3(3)(3)+(3)(3)(3)5(3)(3)(3)(3)(3x)(3)(2x)(2x)(2x)(2x)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(2x)(2x)(2x)(3)(3)(2x)(3)(3)(3)(2x)(3)(3)(3)(3)(3)(3)(3)(2x)(2x)(3)(3)(3)(3)(3)(3)(3)(3)(3)(2x)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3))))))))))+5)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(2x)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(2x)(3)(3

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?

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。

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

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?

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.该系数为20,412。)

Example 4
::例4

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

$\begin{aligned} \sum_{i = 0}^{4} (\binom{4}{i}) \end{aligned}$
::i=04(4i)

This 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 $\begin{aligned} (a + b)^{4} \end{aligned}$ .
::这是要求(40)+(41)(44),这是(a+b)4的所有系数之和。

$\begin{aligned} 1 + 4 + 6 + 4 + 1 = 16 \end{aligned}$

Example 5
::例5

Collapse the following polynomial using the Binomial Theorem.
::使用 Binomial 理论折叠以下的多义理论。

$\begin{aligned} 32 x^{5} - 80 x^{4} + 80 x^{3} - 40 x^{2} + 10 x - 1 \end{aligned}$
::32x5-80x4+80x3-40x2+10x-1

Since the last term is -1 and the power on the first term is a 5 you can conclude that the second half of the binomial is $\begin{aligned} (? - 1)^{5} \end{aligned}$ . The first term is positive and $\begin{aligned} (2 x)^{5} = 32 x^{5} \end{aligned}$ , so the first term in the binomial must be $\begin{aligned} 2 x \end{aligned}$ . The binomial is $\begin{aligned} (2 x - 1)^{5} \end{aligned}$ .
::由于上一个学期是-1,而第一个学期的权力是5,你可以得出结论,下半学期是(?)-1。第一个学期是正数,第二个学期是2x5=32x5,因此二学期的第一个学期必须是2x。二学期是(2x-1-5)。

Summary

The Binomial Theorem allows you to expand a binomial without computing the repeated distribution.
- The theorem states: $\begin{aligned} (a + b)^{n} = \sum_{i = 0}^{n} (\binom{n}{i}) a^{n - i} b^{i} . \end{aligned}$
  ::Theorem指出a+b)ni=0n(ni)an-ibi。
::Binomial 理论允许您在不计算重复分布的情况下扩展二进制。理论指出 a+b) ni=0n( ni) an- ibi 。

In the expansion, exponents of 'a' decrease while exponents of 'b' increase, and coefficients of each term are combinations.
::在扩展中,“a”减少的指数和“b”增加的指数,以及每个术语的系数是组合。

Be careful with binomials of the form $\begin{aligned} (a - b)^{n}, \end{aligned}$ as you need to capture the negative with the second term in the expansion.
::请注意表(a-b)n的二进制,因为您需要在扩展的第二个任期内捕捉负数。

The coefficients in the Binomial Theorem are the numbers from Pascal's Triangle.
::Binomial定理中的系数是帕斯卡尔三角的数字。

Review
::回顾

Expand each of the following binomials using the Binomial Theorem.
::使用二进制定理, 展开以下的二进制。

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 理论折叠以下多边名词。

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)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Section outline

General

二进制理论论

Introduction to the Binomial Theorem ::Binomial定理介绍

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Review ::回顾

Review (Answers) ::回顾(答复)