Course: 11.3 使用和编写序列规则

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序列使用和写入n期规则

You buy new furniture at zero percent interest on a monthly installment plan. The total of your furniture is $4800. The following 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?
::您在每月分期付款计划中以零利率购买新家具。您的家具总数为 4800 美元。以下显示您在每月初仍欠家具的余额。您如何为序列写一条通则 ?

4800, 4600, 4400, 4200,...

nth Term
::nth 任期

Recursive rules can help us generate multiple sequential terms in a sequence but are not helpful in determining a particular single term . Consider the sequence: $\begin{aligned} 3, 5, 7, \dots, a_{n} \end{aligned}$ . The recursive rule for this sequence is $\begin{aligned} a_{n} = a_{n - 1} + 2 \end{aligned}$ . What if we want to find the $\begin{aligned} 100^{t h} \end{aligned}$ term? The recursive rule only allows us to find a term in the sequence if we know the previous term. An $\begin{aligned} n^{t h} \end{aligned}$ term or general rule, however, will allow us to find the $\begin{aligned} 100^{t h} \end{aligned}$ term by replacing $\begin{aligned} n \end{aligned}$ in the formula with 100.
::递归规则可以帮助我们按顺序生成多个顺序术语, 但对于确定一个特定术语没有帮助。考虑顺序 : 3, 5, 7,..., a。此序列的递归规则是 an=an-1+2 。如果我们想要找到第100个术语, 递归规则只能让我们在序列中找到一个术语, 如果我们知道前一个术语。但是, nthterm 或 general 规则可以让我们找到第100个术语, 将公式中的 n 替换为 100 。

Let's solve the following problems.
::让我们解决以下的问题。

Write the first 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: $\begin{aligned} a_{n} = n^{2} - 1 \end{aligned}$ .
::前三个术语、第15个术语和顺序的第40个术语用一般规则写成:an=n2-1。

We can find each of these terms by replacing $\begin{aligned} n \end{aligned}$ with the appropriate term number:
::以适当的用词编号取代n,可以找到这些术语中的每一个:

\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 = 1599 \end{aligned}

::a1=(1)2-1=0a2=(2)-2-1=3a3=(3)2-1=8a15=(15)2-1=224a40=(40)2-1=1599

These terms can also be found using a graphing calculator. First press $\begin{aligned} 2^{n d} \end{aligned}$ STAT (to get to the List menu). Arrow over to OPS, select option 5: seq( and type in (expression, variable, begin, end). For this particular problem, the calculator yields the following:
::也可以使用图形计算器找到这些术语。第一次按 2 STAT (以获取列表菜单 ) 。向 OPS 箭头, 选择选项 5 : 后( 并输入类型 ( 表达、变量、开始、结束) 。对于这个特定的问题, 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 first three terms
::后(x2-1,x,1,3,3,3,0 3,8}前三个任期的后(x2- 1,x,1,3,3)

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

$\begin{aligned} s e q (x^{2} - 1, x, 40, 40) = {1599} \end{aligned}$ for the $\begin{aligned} 40^{t h} \end{aligned}$ term
::第40学期的后(x2-1,x,40,40,40,40)1599}

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

The previous problem 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.
::前一个问题说明了一般规则如何直接将术语数与术语值相匹配。另一种说法是,一般规则以n的函数表示 nth 术语。让我们将上述顺序中的术语放在一个带有术语数的表格中,以帮助识别规则。

Looking at the terms and term numbers together helps us to 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

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,...

Let’s make the table again to begin to analyze the relationship between the term number and the term value.
::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. To start, write each term as a product of the term number and a second factor. Then it can be observed that the second factor is always one less that the term number and the general rule can be written as $\begin{aligned} a_{n} = n (n - 1) \end{aligned}$
::在此情况下, 模式不那么明显。首先, 将每个术语写成术语数和第二个系数的产物。然后可以看到, 第二个系数总是少一个, 术语数和一般规则可以写成 an=n( n- 1) 。

Examples
::实例

Example 1
::例1

Earlier, you were asked how would you write a general rule for the sequence 4800, 4600, 4400, 4200, ...
::之前有人问过你你如何为序列4800,4600,4400,4400,4200,...

Let’s put the terms in the sequence in a table with their term numbers to help identify the rule.
::让我们把术语在顺序中的顺序放在一个带有术语数字的表格中,以帮助确定规则。

$\begin{aligned} n \end{aligned}$	1	2	3	4
$\begin{aligned} a \end{aligned}$	4800	4600	4400	4200

Looking at the terms and term numbers together helps us to see that each term is the result of subtracting 200 times one less than the term from the first term. The general rule is $\begin{aligned} a_{n} = a_{n - 1} - 200 (n - 1) \end{aligned}$ .
::将术语和术语数字放在一起看有助于我们看到,每个术语都是从第一个术语中减去200倍于第一个术语的二百倍的结果。一般规则是 an=1-1-200(n-1)。

Example 2
::例2

Given the general rule: $\begin{aligned} a_{n} = 3 n - 13 \end{aligned}$ , write the first five terms, $\begin{aligned} 25^{t h} \end{aligned}$ term and the $\begin{aligned} 200^{t h} \end{aligned}$ term of the sequence.
::根据一般规则: an=3n-13, 写下序列的前五个任期、第25个任期和第200个任期。

Plug in the term numbers as shown:
::插入显示的字数 :

\begin{aligned} a_{1} & = 3 (1) - 13 = - 10 \\ a_{2} & = 3 (2) - 13 = - 7 \\ a_{3} & = 3 (3) - 13 = - 4 \\ a_{4} & = 3 (4) - 13 = - 1 \\ a_{5} & = 3 (5) - 13 = 2 \\ a_{25} & = 3 (25) - 13 = 62 \\ a_{200} & = 3 (200) - 13 = 587 \end{aligned}

::a1=3(1)-1310a2=3(2)-137a3=3(3)-134a4=3(4)-131a5=3(5)-13=2a25=3(25)-13=62a200=3(200)-13=587

Example 3
::例3

Write the general rule for the sequence: $\begin{aligned} 4, 5, 6, 7, \dots \end{aligned}$
::4,5,6,7,...

Put the values in a table with the term numbers and see if there is a way to write the term as a function of the term number.
::将数值放入含有术语数字的表格,并查看是否有一种方法将术语作为术语数字的函数写入。

$\begin{aligned} n \end{aligned}$	1	2	3	4
$\begin{aligned} a_{n} \end{aligned}$	4	5	6	7
$\begin{aligned} n \pm (?) \end{aligned}$	$\begin{aligned} (1) + 3 \end{aligned}$	$\begin{aligned} (2) + 3 \end{aligned}$	$\begin{aligned} (3) + 3 \end{aligned}$	$\begin{aligned} (4) + 3 \end{aligned}$

Each term appears to be the result of adding three to the term number. Thus, the general rule is $\begin{aligned} a_{n} = n + 3 \end{aligned}$
::每个术语似乎都是在术语数中增加三个词的结果。因此,一般规则是 an=n+3。

Example 4
::例4

Write the general rule and find the $\begin{aligned} 35^{t h} \end{aligned}$ term of the sequence: $\begin{aligned} - 1, 0, 3, 8, 15, \dots \end{aligned}$
::写上一般规则并找到序列的第35个术语 1,0,3,8,15...

$\begin{aligned} n \end{aligned}$	1	2	3	4	5
$\begin{aligned} a_{n} \end{aligned}$	-1	10	3	8	15
$\begin{aligned} n (?) \end{aligned}$	$\begin{aligned} (1) (- 1) \end{aligned}$	$\begin{aligned} (2) (0) \end{aligned}$	$\begin{aligned} (3) (1) \end{aligned}$	$\begin{aligned} (4) (2) \end{aligned}$	$\begin{aligned} (5) (3) \end{aligned}$

Each term appears to be the result of multiplying the term number by two less than the term number. Thus, the general rule is $\begin{aligned} a_{n} = n (n - 2) \end{aligned}$ .
::每个术语似乎都是将术语数乘以比术语数少两倍的结果。因此,一般规则是 an=n(n-2)。

Review
::回顾

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。

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

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

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

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

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

Use your calculator to generate the first 5 terms in each sequence. Use MATH > FRAC , on your calculator to convert decimals to fractions.
::使用您的计算器生成每个序列的前5个条件。在计算器上使用 MATH > FRAC,将小数点转换为分数。

$\begin{aligned} 4 n - 3 \end{aligned}$
::4n-3

$\begin{aligned} - \frac{1}{2} n + 5 \end{aligned}$
::- 12n+5

$\begin{aligned} {(\frac{2}{3})}^{n} + 1 \end{aligned}$
:23n+1)

$\begin{aligned} 2 n (n - 1) \end{aligned}$
::2n(n- 1)

$\begin{aligned} \frac{n (n + 1) (2 n + 1)}{6} \end{aligned}$
::n(n+1)(2n+1)6

Write the $\begin{aligned} n^{t h} \end{aligned}$ term rule for the following sequences.
::为以下序列写入 nth 术语规则。

$\begin{aligned} 3, 5, 7, 9, \dots \end{aligned}$
$\begin{aligned} 1, 7, 25, 79, \dots \end{aligned}$
$\begin{aligned} 6, 14, 24, 36, \dots \end{aligned}$
$\begin{aligned} 6, 5, 4, 3, \dots \end{aligned}$
$\begin{aligned} 2, 5, 9, 14, \dots \end{aligned}$

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

序列使用和写入n期规则

nth Term ::nth 任期

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)