课程： 13.10 二元论理论

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选择节二元论论

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二元论论
In the Gilbert and Sullivan comic opera, The Pirates of Penzance , Major General Stanley sings about how educated he is, including this part about math:
::在吉尔伯特和苏利文的漫画《彭赞斯海盗》中,斯坦利少将歌唱他受教育程度,包括数学部分:

"I'm very well acquainted, too, with matters mathematical,
::"我也非常熟悉数学上的问题

I understand equations, both the simple and quadratical,
::我理解方程式, 简单和二次方程式,

About binomial theorem I'm teeming with a lot o' news,
::关于二元论的理论... ...我充满了很多新闻,

With many cheerful facts about the square of the hypotenuse." ¹
::有许多关于下层平方的令人愉快的事实。"1

In this book, we have covered many of these same topics, except for the Binomial Theorem. We study this below.
::在这本书中,除了Binomial定理之外,我们讨论了许多同样的题目,下文将对此进行研究。

by English National Opera shows a performance of the Major General's Song.
::英国国家歌剧院展示了少将之歌的表演。

T he Binomial Theorem
::二元论论

The Binomial Theorem can be used to raise binomials to powers.
::Binomial定理可用来将二元论提升至权力。

The Binomial Theorem
::二元论论

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

It can be seen in this rule that the powers of $\begin{aligned} a \end{aligned}$ and $\begin{aligned} b \end{aligned}$ decrease and increase, respectively. The notation $\begin{aligned} (\binom{n}{r}) \end{aligned}$ refers to the calculation of the number of combinations of $\begin{aligned} r \end{aligned}$ elements selected from a set of $\begin{aligned} n \end{aligned}$ elements, and that $\begin{aligned} (\binom{n}{r}) = {_{n} C}_{r} = \frac{n!}{r! (n - r)!} \end{aligned}$ . We can also use the rows from Pascal's Triangle to find the coefficients.
::从这一规则可以看出, a 和 b 的功率分别下降和增加。符号( nr) 是指从一组 n 元素中选择的 r 元素的组合数的计算, (nr) = nCr=n!r!! ! 我们也可以使用 Pascal 三角形的行来找到系数。

Example 1
::例1

Use the Binomial Theorem to expand $\begin{aligned} (x + 2 y)^{6} \end{aligned}$ .
::使用 Binomial 定理扩展 (x+2y) 6。

Solution: First, in this problem, $\begin{aligned} a = x \end{aligned}$ , $\begin{aligned} b = 2 y, \end{aligned}$ and $\begin{aligned} n = 6 \end{aligned}$ . Now we can substitute into the rule.
::解决办法:首先,在这个问题上,a=x,b=2y和n=6。现在我们可以替代规则。

$\begin{aligned} (x + 2 y)^{6} & = (\binom{6}{0}) x^{6} (2 y)^{0} + (\binom{6}{1}) x^{5} (2 y)^{1} + (\binom{6}{2}) x^{4} (2 y)^{2} + (\binom{6}{3}) x^{3} (2 y)^{3} + \\ (\binom{6}{4}) x^{2} (2 y)^{4} + (\binom{6}{5}) = x^{1} (2 y)^{5} + (\binom{6}{6}) x^{0} (2 y)^{6} \end{aligned}$

:x+2y)6=(60x6(2y)0+(61x5(2y)1+(62x4(2y)2+(62x4(2y)2+(63x3(2y)3+(64x2(2y)3+(64x2(2y)4+(65)=x1(2y)5+(66x0(2y)6)6

Now we can simplify:
::现在我们可以简化:

$\begin{aligned} = (1) x^{6} (1) + (6) x^{5} (2 y) + (15) x^{4} (4 y^{2}) + (20) x^{3} (8 y^{3}) + \\ (15) x^{2} (16 y^{4}) + (6) x (32 y^{5}) + (1) (1) (64 y^{6}) \\ = x^{6} + 12 x^{5} y + 30 x^{4} y^{2} + 160 x^{3} y^{3} + 240 x^{2} y^{4} + 192 x y^{5} + 64 y^{6} \end{aligned}$

::=(1)x6(1)+(6)x5x5(2y)+(15x4)(4y2)+(20x3(8y3)+(15x2(16y4)+(6)x(32y5)+(1)(1)(64y6)+(1)(1)(64y6)=x6+12x5y+30x4y2+160x3y3+240x2y4+192xy5+64y6

by Mathispower4u shows how to use the Binomial Theorem with combinations.
::Mathispower4u 展示了如何用组合来使用二进制定理。

Example 2
::例2

E xpand $\begin{aligned} (a - b)^{6} \end{aligned}$ .
::扩展(a-b) 6.

Solution: The degree of this expansion is 6, so the powers of $\begin{aligned} a \end{aligned}$ will begin with 6 and decrease by 1 for each term until the powers reach 0, while the powers of $\begin{aligned} - b \end{aligned}$ will begin with 0 and increase by one each term until 6. We can write the variables in the expansion (leaving space for the coefficients) as follows:
::解决办法:扩大的程度是6,因此,一个的权力从6开始,在每个任期中减少1,直到权力达到0,而-b的权力从0开始,每个任期增加1,直到6。我们可以将扩大中的变量(系数的留置空间)写如下:

$\begin{aligned} \underline{} a^{6} + \underline{} a^{5} (- b) + \underline{} a^{4} (- b)^{2} + \underline{} a^{3} (- b)^{3} + \underline{} a^{2} (- b)^{4} + \underline{} a (- b)^{5} + \underline{} (- b)^{6} \end{aligned}$

::=====================================-b)=======================b)=================

Let's use Pascal's Triangle for the coefficients. Here we have a $\begin{aligned} 6^{t h} \end{aligned}$ -degree binomial, so the coefficients will be found in the $\begin{aligned} 6^{t h} \end{aligned}$ row of Pascal's Triangle. Now we can fill in the blanks with the correct coefficients.
::让我们用帕斯卡尔三角形来计算系数。我们这里有一个六度二进制的系数, 因此系数将出现在帕斯卡尔三角形的第六行。现在我们可以用正确的系数填充空白。

$\begin{aligned} a^{6} - 6 a^{5} b + 15 a^{4} b^{2} - 20 a^{3} b^{3} + 15 a^{2} b^{4} - 6 a b^{5} + b^{6} \end{aligned}$

::a6-6a5b+15a4b2-20a3b3+15a2b4-6ab5+b6

by Mathispower4u shows how to use the Binomial Theorem with the coefficients found in Pascal's Triangle.
::Mathispower4u 展示了如何使用Binomial定理, 使用帕斯卡尔三角发现的系数。

If we just want to find a single term in the expansion, we can use the following rule to represent the $\begin{aligned} r^{t h} \end{aligned}$ term in the expansion:
::如果我们只想在扩大中找到一个单一的术语,我们可以使用以下规则在扩大中代表第1个术语:

The rth term in the Expansion
::扩大任期中的第1个任期

The rth term in the expansion of $\begin{aligned} (a + b)^{n} \end{aligned}$ is
:a+b)n 扩展的第rth 术语是

$\begin{aligned} (\binom{n}{r - 1}) a^{n - (r - 1)} b^{r - 1} . \end{aligned}$

:nr-1)an-(r-1)br-1。

Example 3
::例3

Find the $\begin{aligned} 4^{t h} \end{aligned}$ term in the expansion of $\begin{aligned} (3 x - 5)^{8} \end{aligned}$ .
::将第四学期作为(3x-5)8的扩展部分。

Solution: Since we want the $\begin{aligned} 4^{t h} \end{aligned}$ term, $\begin{aligned} r = 4 \end{aligned}$ . Now we can set up the formula with $\begin{aligned} a = 3 x \end{aligned}$ , $\begin{aligned} b = - 5 \end{aligned}$ , $\begin{aligned} n = 8, \end{aligned}$ and $\begin{aligned} r = 4, \end{aligned}$ and evaluate:
::解答:既然我们想要第四个学期, r=4。现在我们可以用 a=3x, b=5, n=8, r=4来设置公式, 并评估:

$\begin{aligned} (\binom{8}{4 - 1}) (3 x)^{8 - (4 - 1)} (- 5)^{4 - 1} = (56) (243 x^{5}) (- 125)^{3} = - 1, 701, 000 x^{5} \end{aligned}$

:84-1)(3x)8-(4-1)(5)4-1=(56)(243x5)(-125)3=-1,701,000x5)

Example 4
::例4

F ind the 5th term when you expand $\begin{aligned} (2 x + 1)^{7} . \end{aligned}$
::当扩展(2x+1) 时查找第五学期。

Solution: Since we want the $\begin{aligned} 5^{t h} \end{aligned}$ term, $\begin{aligned} r = 5 \end{aligned}$ . Now we can set up the formula with $\begin{aligned} a = 2 x \end{aligned}$ , $\begin{aligned} b = 1 \end{aligned}$ , $\begin{aligned} n = 7, \end{aligned}$ and $\begin{aligned} r = 5, \end{aligned}$ and evaluate:
::解答:既然我们想要第五学期, r=5。现在我们可以用 a=2x, b=1, n=7, r=5来设置公式, 并评估:

$\begin{aligned} (\binom{7}{5 - 1}) (2 x)^{7 - (5 - 1)} (1)^{5 - 1} = (35) (8 x^{3}) (1)^{4} = 280 x^{3} \end{aligned}$

:75-1)(2x)7-(5-1)(1)(1)(5)-1=(35)(8)x3)(1)(4)=280x3

Example 5
::例5

Find the constant term in the expansion of $\begin{aligned} {(4 x^{3} + \frac{1}{x})}^{4} \end{aligned}$ .
::在( 4x3+1x) 4 的扩展中查找常数值。

Solution: The constant term occurs when the power of $\begin{aligned} x \end{aligned}$ is 0. Let $\begin{aligned} r \end{aligned}$ remain unknown for the time being: $\begin{aligned} (\binom{4}{r - 1}) (4 x^{3})^{4 - (r - 1)} {(\frac{1}{x})}^{r - 1} \end{aligned}$ . Now, isolate the variables to determine when the power of $\begin{aligned} x \end{aligned}$ will be 0 as shown.
::解析度: 当 x 的功率为 0 时, 恒定值会发生。让 r 暂时不为人知 : (4r-1)( 4x3)4- (r-1)( 1x)r-1。现在, 分离变量以确定 x 的功率何时为 0 。

We can set the variable portion of the expanded term rule equal to $\begin{aligned} x^{0} \end{aligned}$ .
::我们可以设定扩展术语规则的可变部分等于 x0 。

Then, simplify using the rules of exponents on the left-hand side of the equation until we have like bases, $\begin{aligned} x \end{aligned}$ , on both sides, and can drop the bases to set the exponents equal to each other and solve for $\begin{aligned} r \end{aligned}$ .
::然后,简化使用方程式左侧的推手规则, 直到我们有相似的基座, x, 在两侧, 并且可以放下基座, 使推手对等, 并解决r。

$\begin{aligned} (x^{3})^{4 - (r - 1)} (x^{- 1})^{r - 1} & = x^{0} \\ x^{15 - 3 r} \cdot x^{- r + 1} & = x^{0} \\ x^{15 - 3 r - r + 1} & = x^{0} \\ x^{16 - 4 r} & = x^{0} \\ 16 - 4 r & = 0 \\ 16 & = 4 r \\ r & = 4 \end{aligned}$

:x3)4-(r-1)(x-1r-1-1=x0x15-3rxxx-r+1=x0x15-3r-r+1=x0x15-3r-r+1=x0x16-4r=x016-4r=016=4rr=4)

Now, plug the value of $\begin{aligned} r \end{aligned}$ into the rule to get the constant term in the expansion.
$\begin{aligned} (\binom{4}{3}) (4 x^{3})^{4 - 3} {(\frac{1}{x})}^{3} = 4 (4 x^{3}) {(\frac{1}{x})}^{3} = 16 \end{aligned}$

::现在, 在规则中插入 r 的值, 以获得扩展的常值。 (43) (4x3) 4- 3( 1x) 3= 4( 4x3) (1x) 3=16

Summary
::摘要

The Binomial Theorem is $\begin{aligned} (a + b)^{n} = (\binom{n}{0}) a^{n} b^{0} + (\binom{n}{1}) a^{n - 1} b^{1} + (\binom{n}{2}) a^{n - 2} b^{2} + \dots + (\binom{n}{n - 1}) a^{1} b^{n - 1} + (\binom{n}{n}) a^{0} b^{n} . \end{aligned}$
::Binomial定理是 (a+b)n=(n0)anb0+(n1)an-1b1+(n2)an-2b2+...+(nn-1)a1bn-1+(nn)a0bn。

The rth term in an expansion is $\begin{aligned} (\binom{n}{r - 1}) a^{n - (r - 1)} b^{r - 1} \end{aligned}$ .
::扩展中的“Rth”一词是(nr-1)an-(r-1)br-1。

Review
::回顾

Expand the following binomials using the Binomial Theorem:
::使用二进制定理展开以下二进制:

1. $\begin{aligned} (x - a)^{7} \end{aligned}$
::1. (x-a)7

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

Find the $\begin{aligned} n^{t h} \end{aligned}$ term in the expansions of the following binomials:
::在以下二进制扩展中查找 nth 术语 :

3. $\begin{aligned} (7 x - 2)^{5}; n = 4 \end{aligned}$
::3. (7x-2)5;n=4

4. $\begin{aligned} {(6 x + \frac{1}{2})}^{7}; n = 3 \end{aligned}$
::4. (6x+12)7;n=3

5. $\begin{aligned} (5 - a)^{9}; n = 7 \end{aligned}$
::5. (5-a)9;n=7

6. $\begin{aligned} {(\frac{2}{3} x + 9 y)}^{6}; n = 4 \end{aligned}$
::6. (23x+9y)6; n=4

7. Find the term with $\begin{aligned} x^{5} \end{aligned}$ in the expansion of $\begin{aligned} (3 x - 2)^{7} \end{aligned}$ .
::7. 在(3x-2)7的扩展中找到X5这一术语。

8. Find the term with $\begin{aligned} y^{6} \end{aligned}$ in the expansion of $\begin{aligned} (5 - y)^{8} \end{aligned}$ .
::8. 在扩大(5-y)8时,将Y6改为Y6。

9. Find the term with $\begin{aligned} a^{3} \end{aligned}$ in the expansion of $\begin{aligned} (2 a - b)^{10} \end{aligned}$ .
::9. 在扩大(2a-b)10后,将 " a3 " 改为 " a3 " 。

10. Find the term with $\begin{aligned} x^{4} \end{aligned}$ in the expansion of $\begin{aligned} (8 - 3 x)^{5} \end{aligned}$ .
::10. 在扩大(8-3x)5时用x4来查找该术语。

11. Find the constant term in the expansion of $\begin{aligned} {(x^{2} + \frac{3}{x})}^{6} \end{aligned}$ .
::11. 在(x2+3x) 6 的扩展中查找常数。

12. Find the constant term in the expansion of $\begin{aligned} {(\frac{5}{x^{3}} - x)}^{8} \end{aligned}$ .
::12. 在(5x3-x)8的扩展中查找常数。

Answers for Review Problems
::回顾问题的答复

Please see the Appendix.
::请参看附录。

References
::参考参考资料

1. "Major-General's Song," last edited May 8, 2017,
::1. 2017年5月8日编辑的《主将之歌》,

章节大纲

常规

二元论论

T he Binomial Theorem ::二元论论

The Binomial Theorem ::二元论论

Example 1 ::例1

Example 2 ::例2

The rth term in the Expansion ::扩大任期中的第1个任期

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Summary ::摘要

Review ::回顾

Answers for Review Problems ::回顾问题的答复

References ::参考参考资料

T he Binomial Theorem
::二元论论

The Binomial Theorem
::二元论论

Example 1
::例1

Example 2
::例2

The rth term in the Expansion
::扩大任期中的第1个任期

Example 3
::例3

Example 4
::例4

Example 5
::例5

Summary
::摘要

Review
::回顾

Answers for Review Problems
::回顾问题的答复

References
::参考参考资料