Course: 12.7 使用二元理论论

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使用二进制定理
Solve the puzzle: I am the fifth term in the expansion of $\begin{align*}(2x + 1)^7\end{align*}$ . What am I?
::解开谜题:我是扩展(2x+1)的第五任期。

Using the Binomial Theorem
::使用二进制定理

Using Pascal’s Triangle and the patterns within it are only one way to expand binomials. The Binomial Theorem can also be used to expand binomials and is sometimes more efficient, particularly for higher degree binomials. The Binomial Theorem is given by:
::使用帕斯卡尔的三角形及其内部的图案只是扩大二元论的一种方法。比诺米亚定理也可以用来扩大二元论,有时会更有效,特别是对于更高层次的二元论。

$\begin{aligned} (a + b)^{n} = (\binom{n}{0}) a^{n} b^{0} + (\binom{n}{1}) a^{n - 1} b^{1} + (\binom{n}{2}) a^{n - 2} b^{2} + \dots + (\binom{n}{n - 1}) a^{1} b^{n - 1} + (\binom{n}{n}) a^{0} b^{n} \end{aligned}$
$\begin{align*}(a+b)^n=\dbinom{n}{0} a^n b^0 + \dbinom{n}{1} a^{n-1} b^1 + \dbinom{n}{2} a^{n-2} b^2 + \ldots + \dbinom{n}{n-1} a^1 b^{n-1} + \dbinom{n}{n} a^0 b^n \end{align*}$
:a+b)n=(n0)anb0+(n1)an-1b1+(n2)an-2b2+...+(nn-1)a1bn-1+(nn)a0bn

It can be seen in this rule that the powers of $\begin{align*}a\end{align*}$ and $\begin{align*}b\end{align*}$ decrease and increase, respectively, as we have observed. Recall that the notation $\begin{align*}\dbinom{n}{r}\end{align*}$ refers to the calculation of the number of combinations of $\begin{align*}r\end{align*}$ elements selected from a set of $\begin{align*}n\end{align*}$ elements and that $\begin{align*}\dbinom{n}{r}={_nC}_r = \frac{n!}{r!(n-r)!}\end{align*}$ .
::从这一规则可以看出,正如我们所观察到的,a和b的功率分别下降和增加。回顾注(nr)是指计算从一组 n 元素中选定的r 元素的组合数,以及(nr)=nCr=n!r!!

As it turns out, $\begin{align*}\dbinom{n}{0}, \dbinom{n}{1}, \dbinom{n}{2}, \dbinom{n}{3} \ldots \dbinom{n}{n-1}, \dbinom{n}{n},\end{align*}$ are the elements in the $\begin{align*}(n+1)^{st}\end{align*}$ row of Pascal’s Triangle. If we let $\begin{align*}n=5\end{align*}$ , then we can find the coefficients as follows:
::结果,(n0) (n1) (n2) (n3).(nn)-1(nn),是帕斯卡尔三角形(n+1)行中的元素。如果我们使用n=5,那么我们可以找到下列系数:

$\begin{aligned} (\binom{5}{0}) = \frac{5!}{0! 5!} = 1; (\binom{5}{1}) = \frac{5!}{1! 4!} = 5; (\binom{5}{2}) = \frac{5!}{2! 3!} = 10; (\binom{5}{3}) = \frac{5!}{3! 2!} = 10; (\binom{5}{4}) = \frac{5!}{4! 1!} = 5; (\binom{5}{5}) = \frac{5!}{5! 0!} = 1 \end{aligned}$
$\begin{align*}\dbinom{5}{0}=\frac{5!}{0!5!}=1; \ \dbinom{5}{1}=\frac{5!}{1!4!}=5; \ \dbinom{5}{2}=\frac{5!}{2!3!}=10; \ \dbinom{5}{3}=\frac{5!}{3!2!}=10; \ \dbinom{5}{4}=\frac{5!}{4!1!}=5; \ \dbinom{5}{5}=\frac{5!}{5!0!}=1\end{align*}$

These are the elements of the $\begin{align*}6^{th}\end{align*}$ row of Pascal’s Triangle: $\begin{align*}1 \ \ 5 \ \ 10 \ \ 10 \ \ 5 \ \ 1\end{align*}$
::这些是帕斯卡尔三角区第六排: 1 5 10 10 5 1

The Binomial Theorem allows us to determine the coefficients of the terms in the expansion without having to extend the triangle to the appropriate row.
::Binomial定理使我们可以确定扩展条件的系数,而不必将三角形扩展至合适的行。

Let's use the Binomial Theorem to expand $\begin{align*}(x+2y)^6\end{align*}$ .
::让我们使用二进制理论来扩展 (x+2y) 6。

First, in this problem, $\begin{align*}a=x\end{align*}$ , $\begin{align*}b=2y\end{align*}$ and $\begin{align*}n=6\end{align*}$ . Now we can substitute into the rule.
::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。

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

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

What if we just want to find a single term in the expansion? We can use the following rule to represent the $\begin{align*}(r+1)^{st}\end{align*}$ term in the expansion: $\begin{align*}\dbinom{n}{r}a^{n-r}b^r\end{align*}$ . The rule is for the $\begin{align*}(r+1)^{st}\end{align*}$ term because if we want the $\begin{align*}1^{st}\end{align*}$ term, then $\begin{align*}r=0\end{align*}$ (refer back to the Binomial Theorem expansion rule). The value of $\begin{align*}r\end{align*}$ in the expansion is always one less than the term number.
::如果我们只想在扩展中找到一个单一的术语呢? 我们可以使用以下的规则来代表扩展中的第一个术语r+1) : (nr) an-rbr。规则是针对(r+1) 的术语, 因为如果我们想要第一个术语, 那么r=0 (再参照Binomial理论扩展规则 ) 。扩展中的 r 值总是比术语数少一个。

Now, let's find the $\begin{align*}4^{th}\end{align*}$ term in the expansion of $\begin{align*}(3x-5)^8\end{align*}$ .
::现在,让我们在扩大(3x - 5)8时找到第四学期。

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

$\begin{aligned} (\binom{8}{3}) (3 x)^{8 - 3} (- 5)^{3} = (56) (243 x^{5}) (- 125) = - 1, 701, 000 x^{5} \end{aligned}$
$\begin{align*}\dbinom{8}{3}(3x)^{8-3}(-5)^3=(56)(243x^5)(-125)=-1,701,000x^5\end{align*}$
:83)(3x)8-3(-5)3=(56)(243x5)(-125)1,701,000x5)

Finally, let's find the coefficient of the term containing $\begin{align*}y^5\end{align*}$ in the expansion of $\begin{align*}(4+y)^9\end{align*}$ .
::最后,让我们在扩大(4+y)9时找到包含 y5 的术语系数。

This time, think about the rule, $\begin{align*}\dbinom{n}{r}a^{n-r}b^r\end{align*}$ , and that we know that $\begin{align*}b^r=y^5\end{align*}$ and thus $\begin{align*}r=5\end{align*}$ . We also know that $\begin{align*}n=9\end{align*}$ and $\begin{align*}a=4\end{align*}$ . Now we can fill in the rest of the rule:
::这一次,想想规则,(nr)an-rbr,我们知道br=y5,因此r=5,我们也知道n=9和a=4。现在我们可以填写规则的其余部分:

$\begin{aligned} (\binom{9}{5}) (4)^{9 - 5} y^{5} = (126) (256) y^{5} = 32, 256 y^{5} \end{aligned}$
$\begin{align*}\dbinom{9}{5}(4)^{9-5} y^5=(126)(256)y^5=32,256 y^5\end{align*}$
:95(4)(4)9-5-5y5=(126)(256)-55=32,256y5)

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the 5th term when you expand $\begin{align*}(2x + 1)^7\end{align*}$ .
::早些时候,在扩展(2x+1)7时,有人要求你找到第五学期。

Since we want the $\begin{align*}5^{th}\end{align*}$ term, $\begin{align*}r=4\end{align*}$ . Now we can set up the formula with $\begin{align*}a=2x\end{align*}$ , $\begin{align*}b=1\end{align*}$ , $\begin{align*}n=7\end{align*}$ and $\begin{align*}r=4\end{align*}$ and evaluate:
::既然我们想要第五学期, r=4。现在我们可以用 a=2x, b=1, n=7和 r=4来设置公式, 并评估 :

$\begin{aligned} (\binom{7}{4}) (2 x)^{7 - 4} (1)^{4} = (35) (8 x^{3}) (1) = 280 x^{3} \end{aligned}$
$\begin{align*}\dbinom{7}{4}(2x)^{7-4}(1)^4=(35)(8x^3)(1)=280x^3\end{align*}$
:74)(2x)7-4(1)4=(35)(8x3)(1)=280x3

Example 2
::例2

Use the Binomial Theorem to show that $\begin{align*}(a+b)^2=a^2+2ab+b^2\end{align*}$ .
::使用 Binomial 定理来显示 (a+b) 2= a2+2ab+b2。

$\begin{aligned} (a + b)^{2} & = (\binom{2}{0}) a^{2} b^{0} + (\binom{2}{1}) a^{1} b^{1} + (\binom{2}{2}) a^{0} b^{2} \\ = (1) a^{2} (1) + (2) a b + (1) (1) b^{2} \\ = a^{2} + 2 a b + b^{2} \end{aligned}$
$\begin{align*}(a+b)^2 &=\dbinom{2}{0}a^2b^0+\dbinom{2}{1}a^1b^1+\dbinom{2}{2}a^0b^2 \\ &=(1)a^2(1)+(2)ab+(1)(1)b^2 \\ &=a^2+2ab+b^2 \end{align*}$
:a+b)2=(20)a2b0+(21)a1b1+(22)a1b1+(22)a0b2=(1)a2(1)+(2)ab+(1)(1)(1)b2=a2+2abb2+b2

Example 3
::例3

Find the coefficient of the $\begin{align*}5^{th}\end{align*}$ term in the expansion of $\begin{align*}(1-3x)^{10}\end{align*}$ .
::在扩展(1-3x)10时查找第5学期的系数。

$\begin{align*}r=4\end{align*}$ in the $\begin{align*}5^{th}\end{align*}$ term so, $\begin{align*}\dbinom{10}{4}(1)^{10-4}(-3x)^4=210(1)(81x^4)=17,010x^4\end{align*}$ . Since only the coefficient is required, we can drop the variable for the final answer: 17,010.
::r=4,第5学期为(104)(1)10-4(-3x)4=210(1)(81x4)=17,010x4,因为只需要系数,我们可以将变量下调为17,010。

Example 4
::例4

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

The constant term occurs when the power of $\begin{align*}x\end{align*}$ is zero. Let $\begin{align*}r\end{align*}$ remain unknown for the time being: $\begin{align*}\dbinom{4}{r}(4x^3)^{4-r}\left(\frac{1}{x}\right)^r\end{align*}$ . Now isolate the variables to determine when the power of $\begin{align*}x\end{align*}$ will be zero as shown:
::当 x 的功率为零时, 恒定值就会发生。 r 暂时不为人知 : (4r)( 4x3)4-r(1x)r。现在将变量分离出来, 以确定 x 的功率何时为零 :

We can set the variable portion of the expanded term rule equal to $\begin{align*}x^0\end{align*}$ .
::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, $\begin{align*}x\end{align*}$ , on both sides and can drop the bases to set the exponents equal to each other and solve for $\begin{align*}r\end{align*}$ .
::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, 并且可以丢下基座, 使推手对等, 并解决 r 。

$\begin{aligned} (x^{3})^{4 - r} (x^{- 1})^{r} & = x^{0} \\ x^{12 - 3 r} \cdot x^{- r} & = x^{0} \\ x^{12 - 3 r - r} & = x^{0} \\ x^{12 - 4 r} & = x^{0} \\ 12 - 4 r & = 0 \\ 12 & = 4 r \\ r & = 3 \end{aligned}$
$\begin{align*}(x^3)^{4-r} (x^{-1})^r &=x^0 \\ x^{12-3r} \cdot x^{-r} &=x^0 \\ x^{12-3r-r} &=x^0 \\ x^{12-4r} &=x^0 \\ 12-4r &=0 \\ 12 &=4r \\ r &=3 \end{align*}$
:x3)4-r(x-1)4-r(x-1)=x0x12-3r__x-r=x0x12-3r__x-r=x0x12-3r=x0x12-4r=x012-4r=x012-4r=x012-4r=012-4r=012=4rr=3)

Now, plug the value of $\begin{align*}r\end{align*}$ into the rule to get the constant term in the expansion.
::现在,在规则中插入 r 值, 以便在扩展时获得常数值。

$\begin{aligned} (\binom{4}{3}) (4 x^{3})^{4 - 3} {(\frac{1}{x})}^{3} = 4 (4 x^{3}) {(\frac{1}{x})}^{3} = 16. \end{aligned}$
$\begin{align*}\dbinom{4}{3}(4x^3)^{4-3}\left(\frac{1}{x}\right)^3=4(4x^3)\left(\frac{1}{x}\right)^3=16.\end{align*}$
:43)(4x3)4-3(1x)3=4(4x3)(1x)3=16)。

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。

$\begin{align*}(x-a)^7\end{align*}$
:x-a)7

$\begin{align*}(2a+3)^4\end{align*}$
:2a+3)4

Find the $\begin{align*}n^{th}\end{align*}$ term in the expansions of the following binomials.
::在以下二进制的扩展中查找 nth 术语。

$\begin{align*}(7x-2)^5; \ n = 4\end{align*}$
:7x-2);5;n=4

$\begin{align*}\left(6x+\frac{1}{2} \right)^7; \ n = 3\end{align*}$
:6x+12);7;n=3

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

$\begin{align*}\left(\frac{2}{3}x+9y \right)^6; \ n = 4\end{align*}$
:23x+9y) 6; n=4

Find the term with $\begin{align*}\ x^5 \end{align*}$ in the expansion of $\begin{align*}(3x-2)^7\end{align*}$ .
::在(3x-2)7的扩展中查找 x5 的术语。

Find the term with $\begin{align*}\ y^6 \end{align*}$ in the expansion of $\begin{align*}(5-y)^8\end{align*}$ .
::在(5-y)8的扩展中用 y6 查找该词。

Find the term with $\begin{align*}\ a^3 \end{align*}$ in the expansion of $\begin{align*}(2a-b)^{10}\end{align*}$ .
::在扩大(2a-b)10后以 a3 查找该词。

Find the term with $\begin{align*}\ x^4 \end{align*}$ in the expansion of $\begin{align*}(8-3x)^5\end{align*}$ .
::在扩展( 8- 3x) 5 时用 x4 查找该词。

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

Find the constant term in the expansion of $\begin{align*}\left(\frac{5}{x^3}-x \right)^8\end{align*}$ .
::在扩展的( 5x3- x) 8 中查找常数值。

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

使用二进制定理

Using the Binomial Theorem ::使用二进制定理

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)

Using the Binomial Theorem
::使用二进制定理

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Review
::回顾

Review (Answers)
::回顾(答复)