Course: 13.6 找出有限自算序列的总和

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查找有限自算序列的总和
A theater's seating is arranged so that each row has two more seats than the one in front of it. The 1st row has five seats, and there are 30 rows of seats in the theater. How many total seats are in the theater?
::剧院的座位安排是让每排的座位比前面的座位多两个。第一排有五个座位,剧院有30排座位。剧院共有多少座位?

The number of seats in each row is an arithmetic . We discuss how to find the sum of an arithmetic sequence in this section.
::每个行的座位数是算术。我们讨论如何在本节中找到算术序列的总和。

Sum of a Finite Arithmetic Sequence
::有限自算序列总和

In the previous section, we looked at how to sum the terms of a sequence. In this section, we will explore an algebraic method unique to arithmetic series. Let's look at an example to illustrate this and develop a formula to find the sum of a finite arithmetic sequence.
::在上一节中,我们研究了如何对一个序列的条件进行总结。在本节中,我们将探讨算术序列所特有的代数法。让我们看看一个示例来说明这一点,并制定一个公式来找到一个有限的算术序列的总和。

Example 1
::例1

Find the sum of the arithmetic sequence $\begin{aligned} 1 + 3 + 5 + 7 + 9 + 11 + \dots + 35 + 37 + 39. \end{aligned}$
::查找算术序列 1+3+5+7+9+11+...+35+37+39的总和。

Solution: Now, while we could just add up all the terms to get the sum, if we had to sum a large number of terms, this would be quite time-consuming. However, notice that if we write out all the numbers twice, in ascending and descending order, the sum of each pair of numbers is the same:
::解答:现在,当我们可以把所有条件加在一起来得到总和时,如果我们不得不加一大批条件,这将非常费时。然而,请注意,如果我们把所有数字都写成两次,按升降顺序排列,每对数字的总和是相同的:

$\begin{aligned} 1 3 5 7 9 11 \dots 35 37 39 \\ 39 37 35 33 31 29 \dots 5 3 1 \\ ⋮ \\ 40 40 40 40 40 40 \dots 40 40 40 \end{aligned}$

Notice that the sum of the corresponding terms in reverse order is always equal to 40, which is the sum of the 1st and last terms in the sequence.
::注意逆序相应条件的总和始终等于40,即顺序中第一个和最后一个条件的总和。

Now, we know the 1st and last terms, but how many terms are there? We need to find $\begin{aligned} n \end{aligned}$ to find the sum. T he $\begin{aligned} n^{t h} \end{aligned}$ term is 39.
::现在,我们知道第一个和最后一个条件,但有多少条件?我们需要找到n,才能找到总和。N是39。

$\begin{aligned} a_{n} & = a_{1} + d (n - 1) \\ 39 & = 1 + 2 (n - 1) \\ 38 & = 2 (n - 1) \\ 19 & = n - 1 \\ 20 & = n \end{aligned}$
Now the sum is $\begin{aligned} \frac{20 (1 + 39)}{2} = 400 \end{aligned}$ . We divide by 2 because we listed the terms in the sequence twice.
::a=a1+d(n-1)39=1+2(n-1)38=2(n-1)19=n-120=n 现在的总数是20(1+39)2=400。我们除以2是因为我们在序列中两次列出了术语。

Let's generalize this process to find a rule for finding the sum of an arithmetic sequence with a finite number of terms. First we start with the $\begin{aligned} n^{t h} \end{aligned}$ term rule $\begin{aligned} a_{n} = a_{1} + (n - 1) d \end{aligned}$ . We need to find the sum of $\begin{aligned} n \end{aligned}$ terms so we will use the index , $\begin{aligned} i \end{aligned}$ , in a summation as shown below:
::让我们普遍化这个过程, 以找到一个规则来找到计算序列的总和, 并有一定数量的条件。首先, 我们从 nth 术语规则 a=a1+( n- 1) d 开始。我们需要找到 n 条件的总和, 这样我们就可以使用索引, i, 以如下的总和 :

$\begin{aligned} \sum_{i = 1}^{n} [a_{1} + (i - 1) d] \end{aligned}$

::i=1n[a1+(i-1)d]

Keep in mind that $\begin{aligned} a_{1} \end{aligned}$ and $\begin{aligned} d \end{aligned}$ are constants in this expression, so we can separate this into two separate summations:
::记住 a1 和 d 是这个表达式中的常数, 所以我们可以将它分为两个不同的相加:

$\begin{aligned} \sum_{i = 1}^{n} a_{1} + \sum_{i = 1}^{n} (i - 1) d \end{aligned}$

::i=1na1i=1n(i-1)d

Expanding the 1st summation, $\begin{aligned} \sum_{i = 1}^{n} a_{1} = a_{1} + a_{1} + a_{1} + \dots + a_{1}, \end{aligned}$ notice that $\begin{aligned} a_{1} \end{aligned}$ is added to itself $\begin{aligned} n \end{aligned}$ times. We can simplify this expression to $\begin{aligned} a_{1} n \end{aligned}$ .
::正在展开第一个和数, “i=1na1=a1+a1+a1+...+a1+...”+a1, 注意将 a1 添加到它本身中 n 次。我们可以将这个表达式简化为 a1n 。

In the 2nd summation, $\begin{aligned} d \end{aligned}$ can be factored out of the summation, and the difference inside can be split up to get $\begin{aligned} d [\sum_{i = 1}^{n} i - \sum_{i = 1}^{n} 1] \end{aligned}$ . Using rules from the section Series and Summation Notation, $\begin{aligned} \sum_{i = 1}^{n} i = \frac{1}{2} n (n + 1) \end{aligned}$ and $\begin{aligned} \sum_{i = 1}^{n} 1 = n \end{aligned}$ .
::在第二个总和中,d可以从总和中计出,而内部的差数可以分开,以获得d[i=1nii=1n1]。使用区域序列和总和符号的规则, i=1ni=12n(n+1) 和i=1n1=n。

Putting it all together, we can write an expression without any summation symbols and simplify.
::综合起来,我们可以写一个表达式, 没有任何总和符号和简化。

$\begin{aligned} \sum_{i = 1}^{n} [a_{1} + (i - 1) d] = & a_{1} n + d [\frac{1}{2} n (n + 1) - n] \\ = a_{1} n + \frac{1}{2} d n (n + 1) - d n & Distribute d \\ = \frac{1}{2} n [2 a_{1} + d (n + 1) - 2 d] & Factor out \frac{1}{2} n \\ = \frac{1}{2} n [2 a_{1} + d n + d - 2 d] \\ = \frac{1}{2} n [2 a_{1} + d n - d] \\ = \frac{1}{2} n [2 a_{1} + d (n - 1)] & Use this if n^{t h} term is not known \\ = \frac{1}{2} n [a_{1} + (a_{1} + d (n - 1))] \\ = \frac{1}{2} n (a_{1} + a_{n}) \end{aligned}$

::*i=1n[a1+(i-1-1)d]=a1n+d[12n(n+1)-n]=a1n+12dn(n+1)-dn分布 d=12n[2a1+d(n+1)-2d]乘以12n=12n[2a1+dn+dn+d-2d]=12n[2a1+dn+dn]=12n[2a1+d(n-1)]使用此值,如果 n第 n期未知=12n[a1+(a1+d)-n-1)]=12n(a1+an)

Sum of a Finite Arithmetic Sequence
::有限自算序列总和

The sum, $\begin{aligned} S_{n} \end{aligned}$ , of the 1st n terms of an arithmetic sequence is
::Sn,算术序列第1n项中的总和是

$\begin{aligned} S_{n} = \frac{1}{2} n (a_{1} + a_{n}), \end{aligned}$
where $\begin{aligned} n \end{aligned}$ is the number of terms, $\begin{aligned} a_{1} \end{aligned}$ is the 1st term, and $\begin{aligned} a_{n} \end{aligned}$ is the nth term.
::Sn=12n(a1+an),其中n是术语数,a1是第一个术语,N1是第一个术语。

We can also use the formula
::我们也可以使用公式

$\begin{aligned} S_{n} = \frac{1}{2} n [2 a_{1} + d (n - 1)] \end{aligned}$

::Sn=12n[2a1+d(n-1)]

if we don't know the nth term, but we know the common difference, d .
::如果我们不知道第n个学期但我们知道共同的区别 d

Example 2
::例2

Find the sum of the sequence $\begin{aligned} 87 + 79 + 71 + 63 + \dots + - 105 \end{aligned}$ .
::查找87+79+71+63+...+105序列的总和。

Solution: $\begin{aligned} d = - 8 \end{aligned}$ , so
$\begin{aligned} - 105 & = 87 + (- 8) (n - 1) \\ - 192 & = - 8 n + 8 \\ - 200 & = - 8 n \\ n & = 25 \end{aligned}$
and then use the rule to find the sum is $\begin{aligned} \frac{1}{2} (25) (87 - 105) = - 225 \end{aligned}$ .
::解答:d=8, so-105=87+(-8(n-1)-192=8n+8-200=8nn=25),然后使用规则确定金额为12(25)(87-105)=225。

Example 3
::例3

Find the sum of the 1st 40 terms in the arithmetic sequence $\begin{aligned} 35 + 31 + 27 + 23 + \dots \end{aligned}$
::在算术序列 35+31+27+23+中查找第40个条件的总和...

Solution: For this particular sequence, we know the 1st term and the common difference, so let's use the rule that does not require the $\begin{aligned} n^{t h} \end{aligned}$ term: $\begin{aligned} \frac{1}{2} n [2 a_{1} + d (n - 1)] \end{aligned}$ , where $\begin{aligned} n = 40, d = - 4, \end{aligned}$ and $\begin{aligned} a_{1} = 35 \end{aligned}$ .
::解答:对于这个特定的序列,我们知道第一个术语和共同的区别,所以让我们使用不需要第一个术语的规则:12n[2a1+d(n-1)],其中n=40,d=4和a1=35。

$\begin{aligned} \frac{1}{2} (40) [2 (35) + (- 4) (40 - 1)] = 20 [70 - 156] = - 1720 \end{aligned}$

We could also find the $\begin{aligned} n^{t h} \end{aligned}$ term and use the rule $\begin{aligned} \frac{1}{2} n (a_{1} + a_{n}) \end{aligned}$ , where $\begin{aligned} a_{n} = a_{1} + d (n - 1) \end{aligned}$ .
::我们还可以找到 nth 术语并使用规则12n(a1+an), 即 an=a1+d(n-1) 。

$\begin{aligned} a_{40} & = 35 + (- 4) (40 - 1) = 35 - 156 = - 121 \\ S_{40} & = \frac{1}{2} (40) (35 - 121) = 20 (- 86) = - 1720 \end{aligned}$

::a40=35+(4)(40-1)=35-1=156=-121S40=12(40)(35-121)=20(86)=1720

by Mathispower4u shows how to find the formula for an arithmetic series and provides examples.
::by Mathispower4u 显示如何找到算术序列的公式并提供示例。

Example 4
::例4

A theater's seating is arranged so that each row has two more seats than the one in front of it. The 1st row has five seats, and there are 30 rows of seats in the theater. How many total seats are in the theater?
::剧院的座位安排是让每排的座位比前面的座位多两个。第一排有五个座位,剧院有30排座位。剧院共有多少座位?

Solution: For this particular series, we know the 1st term and the common difference, so let's use the rule that does not require the $\begin{aligned} n^{t h} \end{aligned}$ term: $\begin{aligned} \frac{1}{2} n [2 a_{1} + d (n - 1)] \end{aligned}$ , where $\begin{aligned} n = 30, d = 2, \end{aligned}$ and $\begin{aligned} a_{1} = 5 \end{aligned}$ .
::解决方案:对于这个特定系列,我们知道第一个术语和共同的区别,所以让我们使用不需要第一个术语的规则:12n[2a1+d(n-1)],其中n=30,d=2,和a1=5。

$\begin{aligned} \frac{1}{2} (30) [2 (5) + (2) (30 - 1)] = 15 [10 + 58] = 1, 020 \end{aligned}$

Therefore, there are a total of 1,020 seats in the theater.
::因此,剧院共有1,020个席位。

Example 5
::例5

Given that in an arithmetic sequence $\begin{aligned} a_{21} = 165 \end{aligned}$ and $\begin{aligned} a_{35} = 277 \end{aligned}$ , find the sum of terms 21 to 35.
::在算术序列a21=165和a35=277中,找到术语21至35的总和。

Solution: This time we have the "1st" and "last" terms of the sequence , but not the number of terms or the common difference. Since our sequence starts with the $\begin{aligned} 21^{s t} \end{aligned}$ term and ends with the $\begin{aligned} 35^{t h} \end{aligned}$ term, there are 15 terms in this series . Now we can use the rule to find the sum as shown.
::解答: 这次我们有序列的“ 1 ” 和“ 最后一个 ” 术语, 但不包括术语数或共同差异。由于我们的序列以21个术语开始, 以35个术语结束, 这一系列中有15个术语。现在我们可以使用规则来找到所显示的金额。

$\begin{aligned} \frac{1}{2} (15) (165 + 277) = 3, 315 \end{aligned}$

Example 6
::例6

Find the value of the arithmetic series $\begin{aligned} \sum_{i = 1}^{8} (12 - 3 i) . \end{aligned}$
::查找计算序列“i”=18(12-3i)的值。

Solution: From the summation notation, we know that we need to sum 8 terms. We can use the expression $\begin{aligned} 12 - 3 i \end{aligned}$ to find the 1st and last terms, and then use the rule to find the sum.
::解答:从总和符号中,我们知道我们需要总计8个术语。我们可以使用12-3i表达式来找到第1个也是最后的术语,然后使用规则来找到总和。

First term: $\begin{aligned} 12 - 3 (1) = 9 \end{aligned}$
::第一学期:12-3(1)=9

Last term: $\begin{aligned} 12 - 3 (8) = - 12 \end{aligned}$
::最后一学期:12-3(8)=12

$\begin{aligned} \sum_{i = 1}^{8} (12 - 3 i) = \frac{1}{2} (8) (9 - 12) = 4 (- 3) = - 12 \end{aligned}$
.
::i=18(12-3i)=12(8)(8)(9)-(12)=4(3)=12。

Example 7
::例7

Find $\begin{aligned} \sum_{i = 10}^{50} (3 i - 90) \end{aligned}$ .
::查找i=1050(3i-90).

Solution: The $\begin{aligned} 10^{t h} \end{aligned}$ term is $\begin{aligned} 3 (10) - 90 = - 60 \end{aligned}$ , $\begin{aligned} 50^{t h} \end{aligned}$ term is $\begin{aligned} 3 (50) - 90 = 60 \end{aligned}$ and $\begin{aligned} n = 50 - 10 + 1 = 41 \end{aligned}$ (add 1 to include the $\begin{aligned} 10^{t h} \end{aligned}$ term). The value of the series is $\begin{aligned} \frac{1}{2} (41) (- 60 + 60) = 0 \end{aligned}$ .
::解答:第10个学期是3(10)-9060,第50个学期是3(50)-90=60和n=50-10+1=41(增加1个学期包括第10个学期),系列值为12(41)(-60+60)=0。

Note here we are redefining the 10 ^th term as the 1st term, and the 50 ^th term as the 41 ^st term, so we can use the formula. To use the formula, the index of summation must begin at 1.
::请注意,我们重新定义第10个术语为第1个术语,第50个术语为第41个术语,这样我们就可以使用公式。要使用公式,总和指数必须从1开始。

Summary
::摘要

The sum of an arithmetic sequence with a finite number of terms is $\begin{aligned} S_{n} = \frac{1}{2} n (a_{1} + a_{n}), \end{aligned}$ where $\begin{aligned} n \end{aligned}$ is the number of terms, $\begin{aligned} a_{1} \end{aligned}$ is the 1st term, and $\begin{aligned} a_{n} \end{aligned}$ is the nth term.
::Sn=12n(a1+an)是一个算术序列,有一定数目的术语,其总和为Sn=12n(a1+an),其中n为术语数,a1为第一个术语,而a1为第n期。

Review
::回顾

Find the sums of the following arithmetic sequence:
::查找下列算术序列的数值:

1. $\begin{aligned} - 6 + - 1 + 4 + \dots + 119 \end{aligned}$

2. $\begin{aligned} 72 + 60 + 48 + \dots + - 84 \end{aligned}$

3. $\begin{aligned} 3 + 5 + 7 + \dots + 99 \end{aligned}$

4. $\begin{aligned} 25 + 21 + 17 + \dots + - 23 \end{aligned}$

5. Find the sum of the 1st 25 terms of the series $\begin{aligned} 215 + 200 + 185 + \dots \end{aligned}$
::5. 查找系列215+200+185+第125个条件的总和...

6. Find the sum of the 1st 14 terms in the series $\begin{aligned} 3 + 12 + 21 + \dots \end{aligned}$
::6. 查找系列3+12+21+中第114个条件的总和...

7. Find the sum of the 1st 32 terms in the series $\begin{aligned} - 70 + - 65 + - 60 + \dots \end{aligned}$
::7. 查找序列 - 70+- 65+- 60+中第32个条件的总和...

8. Find the sum of the 1st 200 terms in $\begin{aligned} - 50 + - 49 + - 48 + \dots \end{aligned}$
::8. 在 - 50+49+48+中查找第1200个条件的总和...

Evaluate the following series:
::评估以下系列:

9. $\begin{aligned} \sum_{i = 4}^{10} (5 i - 22) \end{aligned}$
::9. i=410(5i-22)

10. $\begin{aligned} \sum_{i = 2}^{25} (- 3 i + 37) \end{aligned}$
::10. i=225(-3i+37)

11. $\begin{aligned} \sum_{i = 11}^{48} (i - 20) \end{aligned}$
::11. i=1148(i-20)

12. $\begin{aligned} \sum_{i = 5}^{40} (50 - 2 i) \end{aligned}$
::12. i=540(50-2i)

Find the sum of the series bounded by the terms given. Include these terms in the sum.
::查找受给定条件约束的序列总和。在总和中包括这些条件。

13. $\begin{aligned} a_{7} = 39 \end{aligned}$ and $\begin{aligned} a_{23} = 103 \end{aligned}$
::13.7=39和A23=103

14. $\begin{aligned} a_{8} = 1 \end{aligned}$ and $\begin{aligned} a_{30} = - 43 \end{aligned}$
::a 14.8=1和a30=-43

15. $\begin{aligned} a_{4} = - 15 \end{aligned}$ and $\begin{aligned} a_{17} = 24 \end{aligned}$
::15.4=-15和a17=24

Explore More
::探索更多

1. How many cans are needed to make a triangular arrangement of cans if the bottom row has 35 cans and each successive row has one fewer can than the row below it?
::1. 如果底排有35个罐头,每连行的罐头比下排少1个罐头,需要多少罐头才能作出罐头三角安排?

2. Thomas gets a weekly allowance. The 1st week it is $1, the 2nd week it is $2, the 3rd week it is $3, and so on. If Thomas puts all of his allowance in the bank, how much will he have at the end of one year?
::2. Thomas每周领取津贴,第一周为1美元,第2周为2美元,第3周为3美元,等等,如果Thomas将其所有津贴都存入银行,他一年结束时将有多少?

3. An electronics store sells $900 the 1st month it is in operation. The managers set a goal of increasing sales by $600 each month for 11 months. Assuming this goal is met, find the total sales for the 1st year the store is in operation.
::3. 一家电子商店在营业的1个月里售出900美元,管理人员设定了11个月每月销售额增加600美元的目标,假设这一目标得以实现,则发现该商店第一年的总销售额。

4. A speech organization is holding a competition that awards the top 8 speakers with cash prizes. First place receives a cash prize of $3,200, 2nd place receives $2,800, 3rd place receives $2,400, and so on. What is the total amount of prize money?
::4. 一个演讲组织举办竞赛,向前8名发言者颁发现金奖,第一位获得3 200美元的现金奖,第2位获得2 800美元,第3位获得2 400美元等等。

5. A boss issues year-end bonuses to the top 5 employees. First place receives a bonus of $3,500, 2nd place receives $3,000, 3rd place receives $2,500, and so on. What is the total amount of bonus money?
::5. 老板向前5名雇员发放年终奖金,第1位领取3 500美元的奖金,第2位领取3 000美元,第3位领取2 500美元等等。

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。

PLIX
::PLIX

Try these interactives to reinforce the concepts explored in this section:
::尝试这些互动来强化本节所探讨的概念 :

Section outline

General

查找有限自算序列的总和

Sum of a Finite Arithmetic Sequence ::有限自算序列总和

Example 1 ::例1

Sum of a Finite Arithmetic Sequence ::有限自算序列总和

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Example 6 ::例6

Example 7 ::例7

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

PLIX ::PLIX

Sum of a Finite Arithmetic Sequence
::有限自算序列总和

Example 1
::例1

Sum of a Finite Arithmetic Sequence
::有限自算序列总和

Example 2
::例2

Example 3
::例3

Example 4
::例4

Example 5
::例5

Example 6
::例6

Example 7
::例7

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

PLIX
::PLIX