Course: 13.2 序列 | The Way To Learn

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序列

The look-and-say follows the pattern 1, 11, 21, 1211, 111221, .... ¹ How do we recognize what this pattern is? We discuss sequences and how to find the next element in a sequence in this section.
::外观和观察遵循模式1、11、21、1211、111221、111221、......1。我们如何认识这种模式?我们讨论顺序以及如何在本节的顺序中找到下一个元素。

lesson content

Sequences
::序列

A sequence is a numbered collection of terms, usually with a pattern. It can have an infinite number of terms, in which case it is called an infinite sequence , or it can have a finite number of terms, in which case it is called a finite sequence .
::一个序列是一个有编号的术语集合, 通常使用一个模式。它可以有无限的多个术语, 在这样的情况下, 它被称为一个无限的序列, 或者它可以有一定的术语数量, 在这样的情况下, 它被称为一个限定的序列。

When looking at a sequence of numbers, you will find there are many possibilities for what the pattern can be.
::当查看数字序列时,你会发现,对于该图案的图案,有很多可能性。

There could be a common difference: the same value is added or subtracted to progress from each term to the next.	$\begin{aligned} 5, 8, 11, 14, \dots \end{aligned}$ (add $\begin{aligned} 3 \end{aligned}$ )
There could be a common ratio: the same factor is multiplied to progress from one term to the next.	$\begin{aligned} 9, 3, 1 \end{aligned}$ , $\begin{aligned} \frac{1}{3}, \dots \end{aligned}$ $\begin{aligned} ( \end{aligned}$ multiply by $\begin{aligned} \frac{1}{3}) \end{aligned}$
If the terms are fractions, perhaps there is one pattern in the numerator and a different pattern in the denominator.	$\begin{aligned} \frac{1}{9}, \frac{3}{8}, \frac{5}{7}, \frac{7}{6}, \dots \end{aligned}$ (numerator ( $\begin{aligned} + 2 \end{aligned}$ ), denominator ( $\begin{aligned} - 1 \end{aligned}$ ))
If the terms are growing rapidly, perhaps the difference between the term values is increasing by some constant factor.	$\begin{aligned} 2, 5, 9, 14, \dots \end{aligned}$ (add $\begin{aligned} 3 \end{aligned}$ , add $\begin{aligned} 4 \end{aligned}$ , add $\begin{aligned} 5 \end{aligned}$ , ...)
The terms may represent a particular type of number, such as prime numbers, perfect squares, perfect cubes, etc.	$\begin{aligned} 2, 3, 5, 7, \dots \end{aligned}$ (prime numbers)
Each term may be the result of performing an operation on prior terms.	$\begin{aligned} 2, 5, 7, 12, 19, \dots \end{aligned}$ (add the previous two terms)
The value may be connected to the term number.	$\begin{aligned} 0, 2, 6, 12, \dots \end{aligned}$ In this example, $\begin{aligned} ((0 \times 1) = 0, (1 \times 2) = 2, (2 \times 3) = 6, (3 \times 4) = 12, \dots) \end{aligned}$
The terms may alternate in sign. A sequence like this is called an alternating sequence .	$\begin{aligned} - 1, 1, - 1, 1, - 1, 1, \dots \end{aligned}$ (raise $\begin{aligned} - 1 \end{aligned}$ to the term number)

This is not a comprehensive list of all possible patterns that may be present in a sequence, but it demonstrates the many ways in which a sequence can be formed.
::这并不是一个按顺序排列的所有可能模式的综合清单,而是表明可以以多种方式形成一个序列。

Notation
::标注

The n ^th term of a sequence is denoted as

\begin{aligned} a_{n} \end{aligned}

. Note

\begin{aligned} n \end{aligned}

is a counting number.

by CK-12 reviews the definitions above.
::CK-12审查了上述定义。

Let's consider some examples in which we have to recognize the patterns to determine the next term in a sequence.
::让我们考虑一些例子, 我们必须承认模式来决定下一个词的顺序。

Example 1
::例1

Find the next two terms in the sequence: $\begin{aligned} 160, 80, 40, 20, \underline{}, \underline{} \end{aligned}$
::在顺序中查找下两个条件:160,80,40,20,_,_

Solution: Each term is the result of multiplying the previous term by $\begin{aligned} \frac{1}{2} \end{aligned}$ . Therefore, the next terms are:
::解决办法:每个术语是前一术语乘以12的结果。因此,下一个术语是:

\begin{aligned} \frac{1}{2} (20) = 10 & \frac{1}{2} (10) = 5 \end{aligned}

Example 2
::例2

Find the next two terms in the sequence: $\begin{aligned} - 5, - 1, 3, 7, \underline{}, \underline{} \end{aligned}$
::在顺序中找到下两个条件: 5 -1,3,7,_,_

Solution: Each term is the previous term plus 4. Therefore, the next two terms are 11 and 15.
::解决办法:每个任期为前一任期加4。因此,下两个任期为11和15。

Example 3
::例3

Find the next two terms in the sequence: $\begin{aligned} 1, 4, 9, 16, \underline{}, \underline{} \end{aligned}$
::在顺序中查找下两个条件:1,4,9,16,_,_

Solution: This sequence is the set of perfect squares or the term number squared. Therefore, the $\begin{aligned} 5^{t h} \end{aligned}$ and $\begin{aligned} 6^{t h} \end{aligned}$ terms will be $\begin{aligned} 5^{2} = 25 \end{aligned}$ and $\begin{aligned} 6^{2} = 36 \end{aligned}$ .
::解决方案: 此序列是完整的方形或术语编号的平方形。因此, 第5和第6个术语将是 52=25 和 62=36 。

Example 4
::例4

Find the next two terms in t he look-and-say sequence: 1, 11, 21, 1211, 111221, ....
::发现接下来两个术语在一对一的顺序中: 1,11,21,1211,111221,111221,...

Solution: This sequence follows the pattern that you look at the previous number and then say how many there are of each number. For example, the first number is 1. Then the next number there is one 1: 11. Then the next number there is two 1's: 21. Then, there is one 2 and one 1: 1211. Next, there is one 1, one 2, and two 1's: 111221.
::解决方案 : 此序列遵循的是您查看上一个数字的模式, 然后表示每个数字中有多少。例如, 第一个数字是 1 。然后下一个数字是 1 1: 11. 然后下一个数字是 2 1 : 21 : 21 : 然后是 1 2 和 1 1: 1211 。接下来是 1 1 、 1 2 和 2 1 : 11 221 。

The next number in this sequence is three 1's, two 2's, and one 1: 312211.
::这个序列的下一个号码是 3 1, 2 2, 1 1 1, 312 211.

Then there is one 3, one 1, two 2's, and two 1's: 13112221.
::然后是1,3,1,1,1,2,2,2,2,1,1,13112221。

by Nayan Mallick discusses how mathematician John Conway struggled with this sequence and how to form a look-and-say sequence.
::讨论数学家约翰·康威如何与这个序列搏斗, 以及如何形成一个一对一对一的序列。

Example 5
::例5

Find the next two terms in the sequence: $\begin{aligned} 0, 3, 7, 12, 18, \underline{}, \underline{} \end{aligned}$
::在顺序中查找下两个条件: 0,3,7,12,18,_,_

Solution: The difference between the 1st two terms $\begin{aligned} (3 - 0) \end{aligned}$ is 3, the difference between 2nd and 3rd terms $\begin{aligned} (7 - 3) \end{aligned}$ is 4, the difference between the 3rd and 4th terms $\begin{aligned} (12 - 7) \end{aligned}$ is 5, and the difference between the 4th and 5th terms $\begin{aligned} (18 - 12) \end{aligned}$ is 6. Each time we add one more to get the next term. The next difference will be 7, so it is $\begin{aligned} 18 + 7 = 25 \end{aligned}$ for the 6th term. To get the 7th term, we add 8, so $\begin{aligned} 25 + 8 = 33 \end{aligned}$ .
::解答:第一和第二任期(3-0)的差别是3,第二和第三任期(7-3)的差别是4,第三和第四任期(12-7)的差别是5,第四和第五个任期(18-12)的差别是6,每次我们再增加一个任期以获得下一个任期,下一个差额是7,因此第六任期是18+7=25。为了获得第七任期,我们增加8, 因此25+8=33。

We can also write out the terms of a sequence if we know the rule for the sequence.
::我们也可以写出序列的条款如果我们知道序列的规则。

Example 6
::例6

Write the 1st three terms, the $\begin{aligned} 15^{t h} \end{aligned}$ term, and the $\begin{aligned} 40^{t h} \end{aligned}$ term of the sequence with the general rule: $\begin{aligned} a_{n} = n^{2} - 1 \end{aligned}$ .
::写上第一三学期、第十五学期和第40学期的顺序,通则为:an=n2-1。

Solution: We can find each of these terms by replacing $\begin{aligned} n \end{aligned}$ with the appropriate term number:
::解决方案:我们可以用适当的术语编号代替n,找到其中的每一个术语:

\begin{aligned} a_{1} & = (1)^{2} - 1 = 0 \\ a_{2} & = (2)^{2} - 1 = 3 \\ a_{3} & = (3)^{2} - 1 = 8 \\ a_{15} & = (15)^{2} - 1 = 224 \\ a_{40} & = (40)^{2} - 1 = 1, 599 \end{aligned}

::a1=(1)2-1=0a2=(2)-2-1=3a3=(3)2-1=8a15=(15)2-1=224a40=(40)2-1=1,599

How to Find the Terms of a Sequence Using a TI-83/84
::如何确定使用TI-83/84的序列的条件

1. First press 2 ^nd STAT (to get to the List menu).
::1. 第2版STAT报(获得《清单》菜单)。

2. Arrow over to OPS, select option 5: seq( and type in (expression, variable, begin, end). Note you have to use the X button as the variable.
::2. 向 OPS 箭头,选择选项 5 : 后(和输入类型( 表达、变量、开始、结束) ) 。请注意, 您必须使用 X 按钮作为变量。

For this particular problem, the calculator yields the following:
::对于这一特殊问题,计算器产生以下结果:

$\begin{aligned} s e q (x^{2} - 1, x, 1, 3) = {0 3 8} \end{aligned}$ for the 1st three terms
:x2- 1,x,1,3,3,1,3,0,3,8}第1个任期的后继(x2- 1,x,1,3,3) =0,3,8}

$\begin{aligned} s e q (x^{2} - 1, x, 15, 15) = {224} \end{aligned}$ for the $\begin{aligned} 15^{t h} \end{aligned}$ term
::第15学期的后(x2-1,x,15,15,15,15)224}

$\begin{aligned} s e q (x^{2} - 1, x, 40, 40) = {1, 599} \end{aligned}$ for the $\begin{aligned} 40^{t h} \end{aligned}$ term
::第40学期的后(x2-1,x,40,40,40,40) 1,599}

Example 7
::例7

Write a general rule for the sequence: $\begin{aligned} 5, 10, 15, 20, \dots \end{aligned}$
::5,10,15,20,...

Solution: The previous example illustrates how a general rule maps a term number directly to the term value. Another way to say this is that the general rule expresses the $\begin{aligned} n^{t h} \end{aligned}$ term as a function of $\begin{aligned} n \end{aligned}$ . Let's put the terms in the above sequence in a table with their term numbers to help identify the rule.
::解答: 上一个示例说明了一条通则如何将术语数直接映射到术语值。另一种说法是, 通则将 n 表示为 n 的函数。让我们将上述顺序中的术语用术语数放在一个表格中, 以帮助识别规则。

Looking at the terms and term numbers together helps us see that each term is the result of multiplying the term number by 5. The general rule is $\begin{aligned} a_{n} = 5 n \end{aligned}$
::将术语和术语数字放在一起看有助于我们看到,每个术语都是将术语数乘以5的结果。一般规则是 an=5n。

$\begin{aligned} n \end{aligned}$	1	2	3	4
$\begin{aligned} a \end{aligned}$	5	10	15	20

Example 8
::例8

Find the $\begin{aligned} n^{t h} \end{aligned}$ term rule for the sequence: $\begin{aligned} 0, 2, 6, 12, \dots \end{aligned}$
::为序列查找 nth 术语规则: 0,2,6,12,...

Solution: Let's make the table again to begin to analyze the relationship between the term number and the term value.
::解答:让我们再次使表格开始分析术语数和术语值之间的关系。

$\begin{aligned} n \end{aligned}$	1	2	3	4
$\begin{aligned} a_{n} \end{aligned}$	0	2	6	12
$\begin{aligned} n (?) \end{aligned}$	$\begin{aligned} (1) (0) \end{aligned}$	$\begin{aligned} (2) (1) \end{aligned}$	$\begin{aligned} (3) (2) \end{aligned}$	$\begin{aligned} (4) (3) \end{aligned}$

This time the pattern is not so obvious, but if we write each term as a product of the term number and a 2nd factor, then it can be observed that the 2nd factor is always one less that the term number, and the general rule can be written as $\begin{aligned} a_{n} = n (n - 1) . \end{aligned}$
::这一次模式并不那么明显,但如果我们把每个术语作为术语数和第二个系数的产物来写,那么可以发现,第二个系数总是少一个,即术语数和一般规则可以写成 an=n(n-1) 。

WARNING
::警告

The 1st few terms in a sequence do not necessarily lead to only one rule.
::按顺序排列的最初几个术语不一定只产生一条规则。

For example, for the sequence $\begin{aligned} \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \dots \end{aligned}$ we can say the formula is $\begin{aligned} a_{n} = \frac{1}{2^{n}} \end{aligned}$ . If that is the formula, then $\begin{aligned} a_{4} = \frac{1}{16} \end{aligned}$ .
::例如,对于序列 12,14,18,... 我们可以说公式是 an=12n。如果这是公式,那么a4=116。

However, the formula $\begin{aligned} a_{n} = \frac{1}{n^{2} - n + 2} \end{aligned}$ also works. In this case, $\begin{aligned} a_{4} = \frac{1}{14} \end{aligned}$ .
::然而,公式a=1n2-n+2也有效,在这种情况下,a4=114。

by CK-12 shows how to write an expression for the nth term of a sequence.
::的 CK-12 显示如何写入序列 nth 术语的表达式。

Recursive Sequences
::递递递序列

A recursive sequence is a sequence in which a term depends on the previous terms. The variable $\begin{aligned} a_{n} \end{aligned}$ represents the $\begin{aligned} n^{t h} \end{aligned}$ term, and $\begin{aligned} a_{n - 1} \end{aligned}$ represents the term preceding $\begin{aligned} a_{n} \end{aligned}$ .
::递归序列是一个术语取决于先前术语的序列。变量a 代表 nth 术语,而 an-1 代表前术语。

Example 9
::例9

Write a recursive rule for the sequence: $\begin{aligned} 3, 9, 27, 81, \dots \end{aligned}$
::序列的递归规则: 3,9,27,81,...

Solution: In this sequence, each term is multiplied by 3 to get the next term. We can write a recursive rule: $\begin{aligned} a_{1} = 3, a_{n} = 3 a_{n - 1} \end{aligned}$
::解决方案 : 在此序列中, 每个术语乘以 3 以获得下一个术语。我们可以写入一个递归规则 : a1= 3, an= 3an-1 。

Example 10
::例10

Write a recursive rule for the sequence: $\begin{aligned} 1, 1, 2, 3, 5, 8, \dots \end{aligned}$
::序列的递归规则: 1,1、2,3,5,8,...

Solution: This is a special sequence called the Fibonacci sequence. In this sequence, each term is the sum of the previous two terms. We can write the recursive rule for this sequence as follows: $\begin{aligned} a_{1} = 1, a_{2} = 1, a_{n} = a_{n - 2} + a_{n - 1} . \end{aligned}$
::解答 : 这是一个叫做 Fibonacci 序列的特殊序列。在这个序列中, 每个词是前两个词的总和。我们可以写出此序列的递归规则如下: a1=1, a2=1,an=an-2+an-1 。

Example 11
::例例11

Write a recursive rule for the sequence: $\begin{aligned} 1, 2, 4, 7, \dots \end{aligned}$
::序列的递归规则: 1,2,4,7,...

Solution: This one is a little trickier to express. Try looking at each term as shown below:
::解答: 此词比较容易表达。尝试查看以下每个词 :

\begin{aligned} a_{1} & = 1 \\ a_{2} & = a_{1} + 1 \\ a_{3} & = a_{2} + 2 \\ a_{4} & = a_{3} + 3 \\ ⋮ \\ a_{n} & = a_{n - 1} + (n - 1) \end{aligned}

::a1=1a2=a1+1a3=a2+2a4=a3+3 *an=an-1+(n-1)

Feature: Just a Fluke
::特点:只是一瞬间

by Deirdre Mundy
::由Deirdre Mundy 编辑

In the meadow, the sheep graze calmly. They do not realize that the grass they are eating is alive with parasites. They are surrounded by flukes, parasitic flatworms that will invade their livers and turn each sheep into a parasite-producing factory. People who eat plants from the meadow or drink water from the stream are at risk, too. Is there a way to predict how fast this parasite will spread throughout the population?
::在草原上,羊群平静地放牧,他们不知道自己吃的草是用寄生虫生存的,他们周围都是木头、寄生虫,它们会侵入他们的肝脏,把每只羊变成产生寄生虫的工厂。吃草地上的植物或从溪流中喝水的人也处于危险之中。有办法预测这种寄生虫会迅速蔓延到整个人口吗?

lesson content

Time Enough for Worms
::时间够虫子了

Biologists use a pair of recursive functions known as the Nicholson-Bailey equations to predict how quickly parasites will infect a population of hosts. The equations $\begin{aligned} N_{t + 1} = L N_{t} e_{t}^{- a p} \end{aligned}$ and $\begin{aligned} P_{t + 1} = N_{t} [1 - e_{t}^{- a p}] \end{aligned}$ use the populations of hosts and parasites at time $\begin{aligned} t \end{aligned}$ to predict their populations at time $\begin{aligned} t + 1 \end{aligned}$ . In these equations, $\begin{aligned} N \end{aligned}$ stands for the number of hosts, $\begin{aligned} P \end{aligned}$ for the number of parasites, $\begin{aligned} a \end{aligned}$ for the sample area, and $\begin{aligned} L \end{aligned}$ for the rate at which the hosts reproduce. Eventually, this system reaches equilibrium, where the number of infected hosts remains stable. For liver flukes, equilibrium is often reached when every available host in the area carries the infection. This is because the flatworms have an unusual life cycle that allows them to spread very quickly.
::生物学家使用被称为 Nicholson-Bailey 方程式的一对递归函数来预测寄生虫如何迅速感染宿主人口。等式Nt+1=LNtet-ap和Pt+1=Nt[1-et-ap]使用宿主和寄生虫的人口来预测其人口,t+1。在这些方程式中, N代表宿主数量, P代表寄生虫数量,一个样本区域,L代表宿主繁殖速度。最终, 这个系统达到了平衡, 受感染宿主的数量保持稳定。对于肝脏流感, 当该地区所有可用的宿主携带感染时, 通常会达到平衡。这是因为扁虫的生命周期异常, 使得它们能够迅速扩散。

An infected host, like a sheep or person, passes liver fluke eggs in his stools. The eggs can remain dormant for a long time. When the eggs hit water, they hatch, and the liver flukes enter a free-living stage. In this stage, they infect snails or fish. Later, they leave their 1st hosts and attach themselves to plants growing in or near a stream. When a mammal eats the plants, the flukes travel through the walls of its intestines and into the liver. There, the larval flukes grow to adulthood and begin to lay eggs.
::受感染的宿主像绵羊或人一样,在自己的凳子里传过肝结蛋。卵可以长期休眠。当卵子打水时,它们孵化,而肝结将进入一个自由生存的阶段。在这个阶段,它们感染蜗牛或鱼。后来,它们离开第一个宿主,在溪水中生长或附近生长的植物上附着。哺乳动物吃植物时,它们会穿过肠壁,进入肝脏。在那里,幼虫长到成年,开始产卵。

In some parts of the world, liver flukes infect huge numbers of people. Their attacks on the liver can cause serious pain, digestive issues, liver failure, and even cancer. While a diagnosed infection can be treated with drugs, public health experts hope to reduce the infection rate through education. If people in areas with high liver fluke populations change their food preparation practices, they can avoid infection.
::在世界某些地方,肝脏的爆发会感染大量人口,肝脏的侵袭会造成严重疼痛、消化问题、肝衰竭甚至癌症。尽管诊断的感染可以用药物治疗,但公共卫生专家希望通过教育降低感染率。如果肝脏紧缺人口高的地区的人改变他们的食品准备做法,他们可以避免感染。

by Biresonance Therapy shows the removal of a liver fluke from a patient's liver.
::肝脏断裂从病人的肝脏取出

Summary
::摘要

A sequence is a list of numbers or terms that follow a particular pattern.
::序列是按特定模式排列的数字或术语列表。

This pattern can take on many forms.
::这种模式可采取多种形式。

A recursive sequence is a sequence in which the terms depend on the values of the previous terms.
::递归序列是指术语取决于先前术语值的顺序。

Review
::回顾

Find the next three terms in each sequence.
::在每个序列中找出接下来的三个术语。

1. $\begin{aligned} 15, 21, 27, 33, \underline{}, \underline{}, \underline{} \end{aligned}$

2. $\begin{aligned} - 4, 12, - 36, 108, \underline{}, \underline{}, \underline{} \end{aligned}$

3. $\begin{aligned} 51, 47, 43, 39, \underline{}, \underline{}, \underline{} \end{aligned}$

4. $\begin{aligned} 100, 10, 1, 0.1, \underline{}, \underline{}, \underline{} \end{aligned}$

5. $\begin{aligned} 1, 2, 4, 8, \underline{}, \underline{}, \underline{} \end{aligned}$

6. $\begin{aligned} \frac{7}{2}, \frac{5}{3}, \frac{3}{4}, \frac{1}{5}, \underline{}, \underline{}, \underline{} \end{aligned}$

Find the missing terms in the sequences.
::在序列中找到缺失的术语。

7. $\begin{aligned} 1, 4, \underline{}, 16, 25, \underline{} \end{aligned}$

8. $\begin{aligned} \frac{2}{3}, \frac{3}{4}, \underline{}, \frac{5}{6}, \underline{} \end{aligned}$

9. $\begin{aligned} 0, 2, \underline{}, 9, 14, \underline{} \end{aligned}$

10. $\begin{aligned} 1, \underline{}, 27, 64, 125, \underline{} \end{aligned}$

11. $\begin{aligned} 5, \underline{}, 11, 17, 28, \underline{}, 73 \end{aligned}$

12. $\begin{aligned} 3, 8, \underline{}, 24, \underline{}, 48 \end{aligned}$

13. $\begin{aligned} 1, 1, 2, \underline{}, 5, \underline{}, 13 \end{aligned}$

Describe the pattern and write a general rule for the following sequences:
::描述图案并写出下列序列的一般规则:

14. $\begin{aligned} \frac{1}{4}, - \frac{1}{2}, 1, - 2 \dots \end{aligned}$

15. $\begin{aligned} 5, 11, 17, 23, \dots \end{aligned}$

16. $\begin{aligned} 33, 28, 23, 18, \dots \end{aligned}$

17. $\begin{aligned} 1, 4, 16, 64, \dots \end{aligned}$

18. $\begin{aligned} 21, 30, 39, 48, \dots \end{aligned}$

19. $\begin{aligned} 1, 5, 10, 16, 23, \dots \end{aligned}$

20. $\begin{aligned} \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5} \dots \end{aligned}$

21. $\begin{aligned} 4, 5, 9, 14, 23, \dots \end{aligned}$

Use the $\begin{aligned} n^{t h} \end{aligned}$ term rule to generate the indicated terms in each sequence.
::使用 nth 术语规则在每个序列中生成指定的术语。

22. $\begin{aligned} 2 n + 7 \end{aligned}$ , terms $\begin{aligned} 1 - 5 \end{aligned}$ and the $\begin{aligned} 10^{t h} \end{aligned}$ term.
::22. 2n+7,第1-5和第10期。

23. $\begin{aligned} - 5 n - 1 \end{aligned}$ , terms $\begin{aligned} 1 - 3 \end{aligned}$ and the $\begin{aligned} 50^{t h} \end{aligned}$ term.
::23.5n-1, 任期1-3和第50届。

24. $\begin{aligned} 2^{n} - 1 \end{aligned}$ , terms $\begin{aligned} 1 - 3 \end{aligned}$ and the $\begin{aligned} 10^{t h} \end{aligned}$ term.
::24. 2n-1,第1-3和第10期。

25. $\begin{aligned} {(\frac{1}{2})}^{n} \end{aligned}$ , terms $\begin{aligned} 1 - 3 \end{aligned}$ and the $\begin{aligned} 8^{t h} \end{aligned}$ term.
::25. (12)n, 术语1-3和第八条。

26. $\begin{aligned} \frac{n (n + 1)}{2} \end{aligned}$ , terms $\begin{aligned} 1 - 4 \end{aligned}$ and the $\begin{aligned} 20^{t h} \end{aligned}$ term.
::26. n(n+112),术语1-4和第20个任期。

Explore More
::探索更多

1. In 2013, the days a full moon appeared was in the sequence below (with Jan. 1 being Day 1). ² Write a recursive formula for the sequence.
::1. 2013年,满月出现的日子按以下顺序排列(1月1日为第1天)。 2 编写序列的递归公式。

9, 38, 67, 96, ...

2. You buy new furniture at 0% interest on a monthly installment plan. The total of your furniture is $4,800. The sequence below shows the balance you still owe on the furniture at the beginning of each month. How would you write a general rule for the sequence?
::2. 在每月分期付款计划上以零息购买新家具,共计4 800美元,以下顺序显示每个月初家具上仍欠的余额。您将如何为顺序写出一条通则?

4,800, 4,600, 4,400, 4,200,...

3. Write the 1st five terms of a sequence for a culture of bacteria, 15 milligrams, that doubles every day. How would you write the general rule for this sequence?
::3. 为细菌文化的顺序写第五个条件,即每天加倍的15毫克,你如何写出这一顺序的一般规则?

4. Find the formula for the following sequence: $\begin{aligned} \frac{1}{3}, \frac{2}{3}, \frac{7}{9}, \frac{5}{6} \underline{}, \underline{} \end{aligned}$
::4. 为下列序列确定公式:13,23,79,56_,_

Answers for Review and Explore More Problems
::回顾和探讨更多问题的答复

Please see the Appendix.
::请参看附录。

References
::参考参考资料

1. "Look-and-say sequence," last modified March 10, 2017,
::1. 2017年3月10日最后一次修改的 "一对一序列"

2. "Full Moon Calendar,"
::2. 满月历程,

Section outline

General

序列

Sequences ::序列

Notation ::标注

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Example 6 ::例6

How to Find the Terms of a Sequence Using a TI-83/84 ::如何确定使用TI-83/84的序列的条件

Example 7 ::例7

Example 8 ::例8

WARNING ::警告

Recursive Sequences ::递递递序列

Example 9 ::例9

Example 10 ::例10

Example 11 ::例例11

Feature: Just a Fluke ::特点:只是一瞬间

Time Enough for Worms ::时间够虫子了

Summary ::摘要

Review ::回顾

Explore More ::探索更多

Answers for Review and Explore More Problems ::回顾和探讨更多问题的答复

References ::参考参考资料