7.1 递归公式

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  递递递性公式公式

  Have you ever had one of those "advent calendars" around the December holidays? Commonly they have a little piece of chocolate for each day to count down the days until Christmas. If the calendar starts on the 1st, there are 25 candies at the beginning of the month, then 24 remaining on the 2nd, 23 on the 3rd, and so forth.

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  The number of pieces remaining after each day is a series: {25, 24, 23...}. How could you identify how many are left on any specific day of the month? Is there a general formula of calculating the candy remaining on any date?
  ::每天之后的残片数量是一个序列 : { 25, 24, 23... } 。 您如何确定月份任何特定日期还剩下多少? 是否有计算任何日期剩余糖果的一般公式 ?

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  Recursive Formulas
  ::递递递性公式公式

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  A is an ordered list of objects. The simplest way to represent a sequence is by listing some of its terms.
  ::A 是按顺序排列的对象列表。 代表序列的最简单方式是列出其某些术语 。

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  The sequence of odd, positive integers is shown here:
  ::在此显示奇数正数整数的序列 :

  	1, 3, 5, 7 ...

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  In this lesson you will learn to represent a sequence recursively , which means that you need to know the previous term in order to find the next term in the sequence.
  ::在这个教训中,你将学会反复代表一个序列,这意味着你需要知道上一个术语,才能在序列中找到下一个术语。

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  Consider the sequence shown above. What is the next term?
  ::考虑上面显示的顺序。下一个任期是什么?

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  As long as you are familiar with the odd integers (i.e., you can count in 2’s) you can figure out that the next term is 9. If we want to describe this sequence in general, we can do so by stating what the first term is, and then by stating the relationship between successive terms. When we represent a sequence by describing the relationship between its successive terms, we are representing the sequence recursively .
  ::只要您熟悉奇数的整数( 也就是说, 您可以在 2 中计数 ) , 您就可以发现下一学期是 9 。 如果我们想概括地描述这个序列, 我们可以通过说明第一个学期是什么来这样做, 然后通过说明连续学期之间的关系来这样做。 当我们通过描述其连续学期之间的关系来代表一个序列时, 我们反复地代表序列 。

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  The terms in a sequence are often denoted with a variable and a subscript. All of the terms in a given sequence are written with the same variable, and increasing subscripts. So we might list terms in a sequence as a ₁ , a ₂ , a ₃ , a ₄ , a ₅ ...
  ::顺序中的术语通常用变量和下标来表示。 特定顺序中的所有术语都用相同的变量写成, 并增加下标。 所以我们可以按顺序以 as1, a2, a3, a3, a4, a5 列出术语...

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  We can use this notation to represent the example above. This sequence is defined as follows:
  ::我们可以使用这个符号来代表上面的例子。这个序列的定义如下:

  	a 1 = 1
  	a n = a n -1 + 2

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  At first glance this notation may seem confusing. What is important to keep in mind is that the subscript of a term represents its “place in line.” So $\begin{array}{r}{a}_n\end{array}$ just means the n ^th term in the sequence. The term $\begin{array}{r}{a}_{n-1}\end{array}$ just means the term before $\begin{array}{r}{a}_n\end{array}$ . In the sequence of odd numbers above, a ₁ = 1, a ₂ = 3, a ₃ = 5, a ₄ = 7, a ₅ = 9 and so on. If, for example, we wanted to find a ₁₀ , we would need to find the 9 ^th term in the sequence first. To find the 9 ^th term we need to find the 8 ^th term, and so on, back to a term that we know.
  ::乍一看,这个符号可能看起来有点混乱。 重要的是要记住, 一个术语的下标代表着它的“ 行中的位置 ” 。 因此, 一个公正的术语意味着序列中的 nth 。 a - 1 词只是意味着一个词之前的词 。 在以上奇数的顺序中, a1 = 1, a2 = 3, a3 = 5, a4 = 7, a5 = 9 等。 例如,如果我们想要找到 a10, 我们需要在顺序中先找到第 9 个词 。 要找到第 9 个术语, 我们需要找到第 8 个术语, 然后再找到我们已知的术语 。

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked about a general formula for the candy remaining on any given date.
  ::早些时候,有人问过你 糖的通用配方 在任何特定日期留下。

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  The remaining candy in an advent calendar is a standard arithmetic sequence, and can be described as a _n = a _{n - 1} - 1.
  ::进历中剩下的糖果是一种标准的算术序列,可以称为 = an-1。

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  Example 2
  ::例2

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  For the sequence of odd numbers, list a ₆ , a ₇ , a ₈ , a ₉ , and a ₁₀ .
  ::单数序列,列出A6、A7、A8、A9、A9和A10。

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  Each term is two more than the previous term.
  ::每个任期比前一任期多两个任期。

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  a ₆ = a ₅ + 2 = 9 + 2 = 11
  ::a6=a5+2=9+2=11

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  a ₇ = a ₆ + 2 = 11 + 2 = 13
  ::a7 = a6 + 2 = 11 + 2 = 13

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  a ₈ = a ₇ + 2 = 13 + 2 = 15
  ::a8 = a7 + 2 = 13 + 2 = 15

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  a ₉ = a ₈ + 2 = 15 + 2 = 17
  ::a9 = a8 + 2 = 15 + 2 = 17

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  a ₁₀ = a ₉ + 2 = 17 + 2 = 19
  ::a10=a9+2=17+2=19

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  The sequence of odd numbers is linear , and it is referred to as an arithmetic sequence . Every arithmetic sequence has a common difference , or a constant difference between each term. (The common difference is analogous to the slope of a line.) The sequence of odd numbers has a common difference of 2 because for all n , a _n - a _{n - 1} = 2.
  ::奇数的序列是线性的,它被称为算术序列。每个算术序列在每个术语之间有一个共同的差别,或一个不变的差别。 (常见的差别类似于一条线的斜坡。)奇数的序列有一个共同的差别为2,因为对于所有 n, a - a - 1 = 2。)奇数的序列有一个共同的差别,因为对于所有 n, a - 1 = 2 。

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  Finding terms in this sequence is relatively straightforward, as the pattern is familiar. However, this would clearly be tedious if you needed to find the 100 ^th term.
  ::在这个序列中查找术语相对简单,因为模式是熟悉的。 但是,如果需要找到第100个术语,这显然将是无聊的。

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  Example 3
  ::例3

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  Find the 5 ^th term for the sequence:
  ::查找序列的第五个术语 :

  	t 1 = 3
  	t n = 2 t n -1

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  t ₅ = 48
  ::t5=48

  		t 2 = 2 t 1 = 2 × 3 = 6
  t 1 = 3	$\begin{array}{r}\to \end{array}$	t 3 = 2 t 2 = 2 × 6 = 12
  t n = 2 × t n -1		t 4 = 2 t 3 = 2 × 12 = 24
  		t 5 = 2 t 4 = 2 × 24 = 48

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  This example is a geometric sequence . Every geometric sequence has a common ratio , which is 2 in this example, because for all n , $\begin{array}{r}\frac{t_n}{t_{n-1}}=2\end{array}$ . The terms of a geometric sequence follow an exponential pattern.
  ::这是一个几何序列。 每一个几何序列都有共同比率, 在这个示例中为 2, 因为对于所有 n, tntn- 1=2. 几何序列的术语遵循指数模式 。

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  Example 4
  ::例4

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  Find the 4 ^th term for the sequence:
  ::查找序列的第四个术语 :

  	b 1 = 2
  	b n = ( b n -1 ) 2 + 1

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  b ₄ = 677
  ::b4=677

  		b 2 = ( b 1 ) 2 + 1 = 2 2 + 1 = 4 + 1 = 5
  b 1 = 2	$\begin{array}{r}\to \end{array}$	b 3 = ( b 2 ) 2 + 1 = 5 2 + 1 = 25 + 1 = 26
  b n = ( b n -1 ) 2 + 1		b 4 = ( b 3 ) 2 + 1 = 26 2 + 1 = 676 + 1 = 677

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  This sequence is neither arithmetic nor geometric, though its values follow a cubic pattern. As you can see from just a few terms here, the terms in a sequence can grow quickly.
  ::这个序列既不是算术,也不是几何,尽管其值遵循的是立方模式。从这里的几个术语中可以看到,一个序列中的术语可以快速增长。

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  For any of these sequences, as noted above, determining more than a few values by hand can be time consuming. In another lesson, we will introduce , which can be used to define a sequence in a way that makes finding the n ^th term faster.
  ::对于任何这些序列,如上所述,手工确定几个以上的数值会耗费时间。 在另一个教训中,我们将引入,可以用来定义一个序列,以便更快地找到 nth 术语。

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  Example 5
  ::例5

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  Write the next 5 terms of the following sequence:  $\begin{array}{r}{a}_1=-4\text{, }{a}_2=-4\text{,}\end{array}$ and $\begin{array}{r}{a}_n=2a_{n-1}+a_{n-2}\end{array}$ .
  ::写下以下顺序的下五个条件: a14, a24, 和 an=2an-1+an-2。

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  $\begin{array}{r}{a}_2=2(-4)+(-4)=-12\end{array}$
  ::a2=2(-4)+(-4)+___________________________________________________________________________________________________________

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  $\begin{array}{r}{a}_3=2(-12)+(-4)=-28\end{array}$
  ::a3=2(-12)+(-4)_________________________________________________________________________________________________________________________

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  $\begin{array}{r}{a}_4=2(-28)+(-12)=-68\end{array}$
  ::a4=2(-28)+(-12)____________________________________________________________________________________________________________________________________________________________

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  $\begin{array}{r}{a}_5=2(-68)+(-28)=-164\end{array}$
  ::a5=2(-68)+(-28)_164

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  $\begin{array}{r}{a}_6=2(-164)+(-68)=-396\end{array}$
  ::a6=2(-164)+(-68)____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

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  So our answer is: -12, -28, -68, -164, and -396.
  ::我们的答案是: -12,28,68,164,396。

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  Example 6
  ::例6

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  Write a recursive formula that fits the following sequence: 1, 5, 9, 13, 17.
  ::写一个符合以下顺序的递归公式:1、5、9、13、17。

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  In this problem we deduct that each term differs by the same amount. Once, we identify the difference of each term, +4 each time in this case, then we know that the sequence requires that we add the same amount (4) to each term.
  ::在此问题上,我们扣除每个术语的差别相同。 一旦我们确定了每个术语的差别, 此时每次是+4, 那么我们就会知道序列要求我们在每个术语中加上同样的数量(4)。

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  We write that as: $\begin{array}{r}{a}_n=a_{n-1}+4\end{array}$ .
  ::我们写成: an=an-1+4。

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  Example 7
  ::例7

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  Given the following sequence, write a recursive formula, then find the next three numbers in the sequence: 3, -4, -1, -5, -6, -11, -17.
  ::根据以下顺序,写一个递归公式,然后在顺序中找到接下来的三个数字:3,4,4,1,5,6,6,11,17。

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  Getting to the next term is not going to be as easy as the previous example. We need to examine the number sequence more closely to solve this problem.
  ::进入下一个任期并不像上一个例子那么容易。 我们需要更仔细地研究数字序列来解决这个问题。

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  In this sequence, the Fibonacci Series was applied. What this means is that the two previous terms were added together to get the next term in the sequence.
  ::在这一顺序中,应用了Fibonacci系列,这意味着将前两个术语加在一起,以便在顺序中加入下一个术语。

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  It is written as: $\begin{array}{r}{a}_n=a_{n-1}+a_{n-2}\end{array}$ .
  ::其写法为:an=an-1+an-2。

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  Now that we know the formula, we can find the next three numbers in the sequence:
  ::现在我们知道公式了, 我们可以在顺序中找到接下来的三个数字:

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  $\begin{array}{r}{a}_8=(-11)+(-17)=-28\end{array}$
  ::a8=(- 11)+(- 17)__________________________________________________________________________________________________________

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  $\begin{array}{r}{a}_9=(-17)+(-28)=-45\end{array}$
  ::a9=(-17)+(-28)_________________________________________________________________________________________________________________________

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  $\begin{array}{r}{a}_{10}=(-28)+(-45)=-73\end{array}$
  ::a10=(- 28)+(- 45)______________________________________________________________________________________

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  So the next three numbers in the sequence are: -28, -45, and -73.
  ::接下来的3个数字在序列中是: -28, -45,和 -73。

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  Review
  ::回顾

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  1. A sequence in which you know the previous term in order to find the next term is:
     ::为了找到下一个任期,您知道上一个任期的顺序是:
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  2. Why is the sequence of odd numbers linear?
     ::为什么奇数序列是线性的?
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  3. Which type of sequence has a common difference?
     ::哪种序列有共同的区别?
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  4. A sequence that uses the same multiple to get from one term to another is:
     ::从一个术语到另一个术语使用相同倍数的序列是:
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  5. Find the value of a ₆ , given the sequence defined as: a ₁ = 4 a _n = 5a _{n -1}
     ::查找 a6 的值, 根据定义的序列为: a1 = 4 an = 5an-1
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  6. Find the value of a ₅ , given the sequence defined as: a ₁ = 32 a _n = (1/2) a _{n -1}
     ::查找 a5 的值, 其序列定义为: a1 = 32 a = (1/2)an-1

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  Using the given recursive formulas, identify the next 5 terms in the sequences that follow:
  ::使用给定的递归公式,按以下顺序确定接下来的5个术语:

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  7. $\begin{array}{r}{a}_1=-2a_2=1\end{array}$ and $\begin{array}{r}{a}_n=3a_{n-1}-5a_{n-2}\end{array}$
     ::a12a2=1和an=3an-1--5an-2
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  8. $\begin{array}{r}{a}_1=-2\end{array}$ and $\begin{array}{r}{a}_n=3a_{n-1}\end{array}$
     ::a12和 an=3an-1
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  9. $\begin{array}{r}{a}_1=3a_2=-2\end{array}$ and $\begin{array}{r}{a}_n=-5a_{n-1}+a_{n-2}\end{array}$
     ::a1=3a22和 an5an-1+1-2
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  10. $\begin{array}{r}{a}_1=1\end{array}$ and $\begin{array}{r}{a}_n=4a_{n-1}\end{array}$
      ::a1=1和an=4an-1
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  11. $\begin{array}{r}{a}_1=-4a_2=1\end{array}$ and $\begin{array}{r}{a}_n=-a_{n-1}+a_{n-2}\end{array}$
      ::a14a2=1和anan-1+an-2

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  Given the following sequence of numbers find the recursive formula.
  ::根据以下数字序列,可以找到递归公式。

  12. 1, 5, 9, 13, 17
  13. -1, 3, 2, 5, 7, 12, 19
  14. -4, 16, -64, 256, -1024

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  Given the following sequence of numbers find the recursive formula and the next three numbers in the sequence.
  ::根据以下数字序列,在序列中发现递归公式和下三个数字。

  15. 1, -1, 1, -1, 1
  16. -5, -1, -6, -7, -13, -20, -33
  17. 1, - 3, 9, -27, 81
  18. -3, -4, -7, -11, -18, -29, -47
  19. -1, -5, -9, -13, -17
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  20. Write the first five terms of the sequence: $\begin{array}{r}{a}_n=(-1{)}^n\cdot 5a_{n-1}\end{array}$  given  that the first term is $\begin{array}{r}{a}_1=3\end{array}$  
      ::写入序列的前五个条件 : an= (-1) n5an-1, 因为第一个条件为 a1=3
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  21. Given the formula: $\begin{array}{r}{a}_n=4n-1\end{array}$ , is the number 27 a term in the sequence of numbers?
      ::根据公式: an=4n- 1, 数字序列中的27是一个术语吗?
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  22. Given the formula: $\begin{array}{r}{a}_n=4n-1\end{array}$ is the number 97 a term in the sequence of numbers?
      ::根据公式: an=4n-1, 数字序列中的97是一个术语?

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  Review (Answers)
  ::回顾(答复)

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  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 。