Course: 11.7 找出有限自算系列的总和

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查找一个精精度亚学系列的和
A theater's seating is arranged such that each row has two more seats than the one in front of it. The first row has five seats and there are 30 rows of seats in the theater. How many total seats are in the theater?
::第一排有五个座位,剧院有30排座位。剧院共有多少座位?

Sum of a Finite Arithmetic Series
::有限亚理学系列总和

The method of using the calculator to evaluate the sum of a series can be used to find the sum of an arithmetic series as well. However, in this concept we will explore an algebraic method unique to arithmetic series. As we discussed earlier in the unit a series is simply the sum of a so an arithmetic series is a sum of an arithmetic sequence. Let’s look at a problem to illustrate this and develop a formula to find the sum of a finite arithmetic series.
::使用计算器来计算一个序列的总和的方法也可以用来查找一个算术序列的总和。但是,在这个概念中,我们将探索一个算术序列独有的代数法。正如我们早些时候在单位中所讨论的那样,一个序列仅仅是一个算术序列的总和,它是一个算术序列的总和。让我们研究一个问题来说明这一点,并开发一个公式来寻找一个算术序列的有限总和。

Let's find the sum of the arithmetic series: $\begin{aligned} 1 + 3 + 5 + 7 + 9 + 11 + \dots + 35 + 37 + 39 \end{aligned}$ .
::让我们找到计算序列的总和:1+3+5+7+9+11+...+35+37+39。

Now, while we could just add up all of the terms to get the sum, if we had to sum a large number of terms that would be very time consuming. A famous German mathematician, Johann Carl Friedrich Gauss, used the method described here to determine the sum of the first 100 integers in grade school. First, we can write out all the numbers twice, in ascending and descending order, and observe that the sum of each pair of numbers is the same:
::现在,尽管我们可以把所有条件加在一起来得到总和,如果我们不得不加起来大量非常耗时的术语。一位著名的德国数学家约翰·卡尔·弗里德里希·高斯用这里描述的方法来决定小学头100个整数的总和。首先,我们可以用上下顺序写出所有数字两次,并观察每对数字的总和是一样的:

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

Notice that the sum of the corresponding terms in reverse order is always equal to 40, which is the sum of the first 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。

What Gauss realized was that this sum can be multiplied by the number of terms and then divided by two (since we are actually summing the series twice here) to get the sum of the terms in the original sequence. For the problem he was given in school, finding the sum of the first 100 integers, he was able to just use the first term , $\begin{aligned} a_{1} = 1 \end{aligned}$ , the last term, $\begin{aligned} a_{n} = 100 \end{aligned}$ , and the total number of terms, $\begin{aligned} n = 100 \end{aligned}$ , in the following formula:
::高斯意识到的是,这个总和可以乘以条款数目,然后除以两个(因为我们实际上在这里两次对序列进行缩放)以获得最初顺序中的条款总和。对于他在学校中遇到的问题,他找到了第一个100整数的总和,他能够在以下公式中使用第一个术语,a1=1,最后一个术语,a=100,以及术语总数,n=100:

$\begin{aligned} \frac{n (a_{1} + a_{n})}{2} = \frac{100 (1 + 100)}{2} = 5050 \end{aligned}$

::n(a1+an)2=100(1+1002=50)

In our problem we know the first and last terms, but how many terms are there? We need to find $\begin{aligned} n \end{aligned}$ to use the formula to find the sum of the series. We can use the first and last terms and the $\begin{aligned} n^{t h} \end{aligned}$ term to do this.
::在我们的问题中,我们知道第一个和最后一个术语,但有多少术语?我们需要找到n 来使用公式来找到序列的总和。我们可以使用第一个和最后一个术语以及第n个术语来做到这一点。

$\begin{aligned} a_{n} & = a_{1} + d (n - 1) \\ 39 & = 1 + 2 (n - 1) \\ 38 & = 2 (n - 1) \\ 19 & = n - 1 \\ 20 & = n \end{aligned}$
Now the sum is $\begin{aligned} \frac{20 (1 + 39)}{2} = 400 \end{aligned}$ .
::a=a1+d(n-1)39=1+2(n-1)38=2(n-1)19=n-120=n,现为20(1+39)2=400。

Proof of the Arithmetic Sum Formula
::A. 具有自对数和公式的证明

The rule for finding the $\begin{aligned} n^{t h} \end{aligned}$ term of an arithmetic sequence and properties of summations can be used to prove the formula algebraically. First, we will 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 numerous $\begin{aligned} n^{t h} \end{aligned}$ terms ( $\begin{aligned} n \end{aligned}$ of them to be exact) so we will use the index , $\begin{aligned} i \end{aligned}$ , in a summation as shown below:
::查找算术序列 n 和总和属性 n 值的规则可用于证明公式代数。首先,我们将从 n 术语规则 an=a1+(n- 1) d 开始。我们需要找到多个 n 术语的和 , 这样我们就可以使用索引, i , 其总和如下:

$\begin{aligned} \sum_{i = 1}^{n} [a_{1} + (i - 1) d] \end{aligned}$ Keep in mind that $\begin{aligned} a_{1} \end{aligned}$ and $\begin{aligned} d \end{aligned}$ are constants in this expression.
::i=1n[a1+(i-1)d] 记住, a1 和 d 是此表达式中的常数。

We can separate this into two separate summations as shown: $\begin{aligned} \sum_{i = 1}^{n} a_{1} + \sum_{i = 1}^{n} (i - 1) d \end{aligned}$
::我们可以将它分为两个不同的总和,如“i”=1na1=1na1i=1n(i-1)d

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

In the second summation, $\begin{aligned} d \end{aligned}$ can be brought out in front of the summation and the difference inside can be split up as we did with the addition to get: $\begin{aligned} d [\sum_{i = 1}^{n} i - \sum_{i = 1}^{n} 1] \end{aligned}$ . Using rules you have seen before, $\begin{aligned} \sum_{i = 1}^{n} i = \frac{1}{2} n (n + 1) \end{aligned}$ and $\begin{aligned} \sum_{i = 1}^{n} 1 = n \end{aligned}$ . Putting it all together, we can write an expression without any summation symbols and simplify.
::在第二个总和中,d可以在总和之前提出,而内部的差异可以和我们所做的一样分开,加上以下的附加内容:d[i=1nii=1n1]。使用你以前看到的规则,i=1ni=12n(n+1)和i=1n1=n。把它们放在一起,我们可以在没有任何总和符号和简化的情况下写一个表达式。

$\begin{aligned} a_{1} n + d [\frac{1}{2} n (n + 1) - n] \\ = a_{1} n + \frac{1}{2} d n (n + 1) - d n & Distribute d \\ = \frac{1}{2} n [2 a_{1} + d (n + 1) - 2 d] & Factor out \frac{1}{2} n \\ = \frac{1}{2} n [2 a_{1} + d n + d - 2 d] \\ = \frac{1}{2} n [2 a_{1} + d n - d] \\ = \frac{1}{2} n [2 a_{1} + d (n - 1)] & \leftarrow This version of the equation \\ is very useful if you don't know the n^{t h} term . \\ = \frac{1}{2} n [a_{1} + (a_{1} + d (n - 1))] \\ = \frac{1}{2} n (a_{1} + a_{n}) \end{aligned}$

::a1n+d[12n(n+1)-n]=a1n+12dn(n+1)-dn分配 d=12n[2a1+d(n+1)-2d] 乘数除以 12n=12n[2a1+dn+dn+d-d] =12n[2a1+dn-d] =12n[2a1+d+d(n-1)] = 12n[2a1+d+d(n- 1)] = 12n[2a1+d] =12n[2a1+d(n- 1)] =12n[2a1+d] =12n[a1+d(n-an) =该方程式非常有用。=12n[a1+an]

Now, let's find the sum of the first 40 terms in the arithmetic series $\begin{aligned} 35 + 31 + 27 + 23 + \dots \end{aligned}$
::现在,让我们在计算序列35+31+27+23+中找到前40个条件的总和...

For this particular series we know the first term and the common difference, so let’s use the rule that doesn't 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 $\begin{aligned} a_{1} = 35 \end{aligned}$ .
::对于这个特别的系列,我们知道第一个术语和共同的区别,所以让我们使用不要求第一个术语的规则:12n[2a1+d(n-1)],n=40,d4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) 。

$\begin{aligned} a_{40} = 35 + (- 4) (40 - 1) = 35 - 156 = - 121 \end{aligned}$ , so the sum is $\begin{aligned} \frac{1}{2} (40) (35 - 121) = 20 (- 86) = - 1720. \end{aligned}$
::a40=35+(-4(-4(-40-1)=35-156*121),总和为12(40)(35-121)=20(-86)=1720。

Next , given that in an arithmetic series $\begin{aligned} a_{21} = 165 \end{aligned}$ and $\begin{aligned} a_{35} = 277 \end{aligned}$ , let's find the sum of terms 21 to 35.
::接下来,如果在计算序列a21=165和a35=277中,让我们找到术语21至35的总和。

This time we have the “first” and “last” terms of the series, but not the number of terms or the common difference. Since our series 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.
::这一次,我们有了系列中的“第一个”和“最后一个”术语,而不是术语的数量或共同差异。由于我们的系列以21个术语开始,而以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示的金额。

$\begin{aligned} \frac{1}{2} (15) (165 + 277) = 3315 \end{aligned}$

Finally, let's find the sum of the arithmetic series $\begin{aligned} \sum_{i = 1}^{8} (12 - 3 i) \end{aligned}$ .
::最后,让我们来看看算术序列之和 i=18(12-3i)

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 first and last terms as and the use the rule to find the sum.
::从总和符号中,我们知道我们需要加起来8个条件。我们可以用12-3i的表达方式来找到第一个和最后一个条件,以及用规则来找到总和。

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)__________________________________________________________________________________________________________________________________________________________________

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

We could use the calculator in this problem as well: $\begin{aligned} s u m (s e q (12 - 3 x, x, 1, 8)) = - 12 \end{aligned}$
::在这个问题上,我们也可以使用计算器:总(12-3x,x,1,8) 12

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the total number of seats in the theater.
::早些时候,你被要求找到剧院的席位总数。

For this particular series we know the first term and the common difference, so let’s use the rule that doesn't 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 $\begin{aligned} a_{1} = 5. \end{aligned}$
::对于这个特定的系列,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n个术语的规则:12n[2a1+d(n-1)],其中n=30,d=2和a1=5。

$\begin{aligned} \frac{1}{2} (30) [2 (5) + (2) (30 - 1)] = 15 [10 + 58] = 1020 \end{aligned}$

Therefore, there are a total of 1020 seats in the theater.
::因此,剧院共有1020个席位。

Example 2
::例2

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

$\begin{aligned} d = 8 \end{aligned}$ , so
$\begin{aligned} - 105 & = 87 + (- 8) (n - 1) \\ - 192 & = - 8 n + 8 \\ - 200 & = - 8 n \\ n & = 25 \end{aligned}$

::d=8, 所以- 105=87+(-8(n-1)-1)-192_8n+8-8_200_8nn=25

and then use the rule to find the sum is $\begin{aligned} \frac{1}{2} (25) (87 - 105) = - 225 \end{aligned}$
::然后使用规则确定金额为 12(25) (87-105) @%225

Example 3
::例3

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

$\begin{aligned} 10^{t h} \end{aligned}$ 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 sum of the series is $\begin{aligned} \frac{1}{2} (41) (- 60 + 60) = 0 \end{aligned}$ . Note that the calculator is a great option for this problem: $\begin{aligned} s u m (s e q (3 x - 90, x, 10, 50)) = 0 \end{aligned}$
::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(加上1,包括第10个学期)。系列的总和是12(41)(-60+60)=0。请注意,计算器是解决这个问题的一个很好的选择:总和(3x-90,x,10,50)=0。

Example 4
::例4

Find the sum of the first 30 terms in the series $\begin{aligned} 1 + 6 + 11 + 16 + \dots \end{aligned}$
::查找序列 1+6+11+16+中前30个条件的和...

$\begin{aligned} d = 5 \end{aligned}$ , use the sum formula, $\begin{aligned} \frac{1}{2} n (2 a_{1} + d (n - 1)) \end{aligned}$ , to get $\begin{aligned} \frac{1}{2} (30) [2 (1) + 5 (30 - 1)] = 15 [2 + 145] = 2205 \end{aligned}$
::d=5,使用总和公式12n(2a1+d(n-1)),获得12(30)[2(1)+5(30-1)]=15[2+145]=2205]

Review
::回顾

Find the sums of the following arithmetic series.
::查找以下计算序列的数值。

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

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

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

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

Find the sum of the first 25 terms of the series $\begin{aligned} 215 + 200 + 185 + \dots \end{aligned}$
::查找序列 215+200+185+前25个条件的和...

Find the sum of the first 14 terms in the series $\begin{aligned} 3 + 12 + 21 + \dots \end{aligned}$
::查找序列 3+12+21+中前14个条件的和...

Find the sum of the first 32 terms in the series $\begin{aligned} - 70 + - 65 + - 60 + \dots \end{aligned}$
::查找序列中前32个条件的总和- 706560+...

Find the sum of the first 200 terms in $\begin{aligned} - 50 + - 49 + - 48 + \dots \end{aligned}$
::在- 504948+中查找前200个条件的总和...

Evaluate the following summations.
::评估以下总和。

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

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

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

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

$\begin{aligned} a_{7} = 39 \end{aligned}$ and $\begin{aligned} a_{23} = 103 \end{aligned}$
::a7=39和a23=103

$\begin{aligned} a_{8} = 1 \end{aligned}$ and $\begin{aligned} a_{30} = - 43 \end{aligned}$
::a8=1和a30=43

$\begin{aligned} a_{4} = - 15 \end{aligned}$ and $\begin{aligned} a_{17} = 24 \end{aligned}$
::a4+15和a17=24

How many cans are needed to make a triangular arrangement of cans if the bottom row has 35 cans and successive row has one less can than the row below it?
::如果底排有35个罐头,而连续行的罐头比下排少一个罐头,则需要多少罐头才能作出罐头三角安排?

Thomas gets a weekly allowance. The first week it is one dollar, the second week it is two dollars, the third week it is three dollars and so on. If Thomas puts all of his allowance in the bank, how much will he have at the end of one year?
::Thomas每周领取津贴。第一周是一美元,第二周是两美元,第三周是三美元等等。如果Thomas把所有津贴都放入银行,那么他一年结束时会有多少?

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

查找一个精精度亚学系列的和

Sum of a Finite Arithmetic Series ::有限亚理学系列总和

Proof of the Arithmetic Sum Formula ::A. 具有自对数和公式的证明

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)

Sum of a Finite Arithmetic Series
::有限亚理学系列总和

Proof of the Arithmetic Sum Formula
::A. 具有自对数和公式的证明

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)