节: 系列总和和和高斯公式 | 7.4 总计和高斯公式

章节大纲

EyeScreem Ltd is having a special sale on their ice cream cones. In order to get as many people in the store to try out their new flavors, they have decided to run the following promotion:

The first customer to buy an ice cream will pay $6 for a large cone with chocolate fudge.
::第一个买冰淇淋的顾客将支付6美元一个大锥体巧克力软糖。

The second customer will pay only $5.90.
::第二个客户只付5.90美元。

The third customer will only pay $5.80.
::第三个客户只付5.80美元。

Each successive ice cream lover will be charged $0.10 less than the last, until everyone that comes in gets a free cone!
::每一个连续的冰淇淋情人都会比过去少收0.10美元直到每个来的人都得到免费的甜筒!

How much money will the store bring in during the sale, assuming at least some customers are given free cones?
::商店在出售期间能带多少钱? 假设至少有些顾客可以免费得到甜甜圈?

Series Sums and Gauss's Formula
::系列总和和和高斯公式

Gauss's Formula
::高斯公式

German mathematician C.F. Gauss is often credited with discovering a formula for calculating the sum of a series when he was a young child. The story is likely apocryphal (a legend), but it has been passed down since Gauss lived in the 1700s. According to the story, Gauss’s teacher wanted to occupy students by having them add up large sets of numbers. When Gauss was asked to add up the first 100 integers, he found the sum very quickly, by pairing the numbers:
::德国数学家C.F.Gaus常常在发现计算他年幼时一系列数字之和的公式时被表扬。故事很可能是一个传说,但自从Gaus生活在1700年代以后就被传下来了。根据故事,Gaus的老师想通过让学生们相加大量数字来占据学生。当要求Gaus将最初的100整数加起来时,他很快地发现了这个数字,通过对数字进行配对:

All of the numbers in the sum could be paired to make groups of 101. There are one hundred numbers being added, so there are such fifty pairs. Therefore the sum is 50(101) = 5050.
::所有总和中的数字可以配对成101组, 增加100个数字, 所以有这样的50对, 因此总和是50(101)=5050。

The method Gauss used to solve this problem is the basis for a formula that allows us to add together the first n positive integers:
::解决这个问题所使用的方法,是计算公式的基础,使我们能够将第一个正数整数相加:

%20(n%20%2B%201)%7D%20%7B2%7D"> $\begin{aligned} \sum = \frac{(n) (n + 1)}{2} \end{aligned}$
:n)(n+1)2

Using a Graphing Calculator
::使用图形计算计算器

To generate either a or a series, you can use a graphing calculator. The TI-83/84 series gives you several options. You can, for example, work in sequence mode, which allows you to define a sequence and find terms. If you have an explicit formula for a sequence, you can keep your calculator in function mode. For example, consider the sum $\begin{aligned} \sum_{n = 1}^{9} n^{2} \end{aligned}$ . It would be time consuming to write out the first 9 squares. The calculator is faster. To generate the 9 terms, press [LIST], then select OPS , then option 5, seq (. This takes you back to the main screen. You should see seq (. After this, enter x ^2, x , 1, 9, 1). (The x tells the calculator that x is the input. The 1 and the 9 tell it the limits of the sum. The second 1 tells the calculator to go up in increments of 1.)
::要生成一个或一个序列, 您可以使用图形计算器。 TI- 83/ 84 系列可以给您多个选项。例如, 您可以在序列模式中工作, 允许您定义序列并查找条件。如果您对序列有明确的公式, 您可以将计算器保留在函数模式中。例如, 考虑 $n=19n2 。写出前 9 个方块需要花费时间。计算器更快。要生成9 个条件, 请按 [LIST] , 然后选择 OPS, 然后选择 5 选项 , 后选项( 这可以将您带回主屏幕 ) 。您应该看到续( 在此之后, 输入 x% 2, x, 1, 9 1)。 ( x 告诉计算器 x 是输入的 x。 1 和 9 告诉它总数的限值。第二个 1 告诉计算器以1 递增 ) 。

Press , and you should see the list of squares. Scroll to the right to see all of them. The scrolling will end when you reach 81.
::按下,您应该看到方块列表。滚到右侧看所有方块。滚动将在81岁时结束。

If you want to find the sum of the terms, first store the sequence in a list (see screen below), then [LIST], then select MATH , then option 5, sum(. Then enter the name of the list and press . You should get 285.
::如果您想要找到术语的总和, 请先在列表中存储序列( 见下文屏幕) , 然后 [LIST], 然后选择 MATH, 然后选择 5, sum (. 然后输入列表的名称并按下。您应该得到 285 。

Examples
::实例

Example 1
::例1

Earlier, you were asked to find how much money the store would bring in during its ice cream promotion.
::早些时候,有人要求你找出这家商店在推广冰淇淋期间会带来多少钱

The series sum formula %20(n%20%2B%201)%7D%20%7B2%7D"> $\begin{aligned} \sum = \frac{(n) (n + 1)}{2} \end{aligned}$ is designed for integers, so let's use it to solve for the number of dimes brought in (since that is the unit each term reduces by) and then convert to dollars:
::序列总和公式 +1) 2 是为整数设计的,所以让我们用它来解决引入的硬币数量(因为这是每个术语减少的单位),然后转换成美元:

$\begin{aligned} \sum = \frac{(60) (61)}{2} \end{aligned}$

$\begin{aligned} \sum = \frac{3, 660}{2} \end{aligned}$

$\begin{aligned} \sum = 1830.00 \to $ 183.00 \end{aligned}$

The store will bring in $183.00, which will probably not cover the costs of the day. However, they will certainly get a lot of people through the door to try out the ice cream!
::商店将带来183.00美元,这很可能无法支付日常费用。然而,他们肯定会让很多人通过门来试吃冰淇淋!

Example 2
::例2

Expand the sigma and find the sum: $\begin{aligned} \sum_{n = 1}^{7} (2 n - 3) \end{aligned}$ .
::展开西格玛, 并查找总和: n=17( 2n-3) 。

$\begin{aligned} \sum_{n = 1}^{7} (2 n - 3) \end{aligned}$	$\begin{aligned} = (2 \times 1 - 3) + (2 \times 2 - 3) + (2 \times 3 - 3) + (2 \times 4 - 3) + (2 \times 5 - 3) \\ + (2 \times 6 - 3) + (2 \times 7 - 3) \end{aligned}$
	$\begin{aligned} = (- 1) + (1) + (3) + (5) + (7) + (9) + (11) \end{aligned}$
	$\begin{aligned} = 35 \end{aligned}$

Example 3
::例3

Expand the sigma and find the sum by adding the terms: $\begin{aligned} \sum_{n = 3}^{6} (n^{2} - 5) \end{aligned}$ .
::扩展 sigma , 并添加条件 : n= 36 (n2- 5) 来查找总和。

Why is Gauss's formula not recommended for this question?
::为什么Gaus的公式不推荐给这个问题呢?

$\begin{aligned} \sum_{n = 3}^{6} (n^{2} - 5) \end{aligned}$	$\begin{aligned} = (3^{2} - 5) + (4^{2} - 5) + (5^{2} - 5) + (6^{2} - 5) \end{aligned}$
	$\begin{aligned} = (4) + (11) + (20) + (31) \end{aligned}$
	$\begin{aligned} = 66 \end{aligned}$

There are actually a couple of reasons not to use Gauss's formula here, but the biggest is that the formula assumes you are adding all of the integers from 0 to the last number in the series. In the question, you are asked to only sum from the 3rd to 6th term.
::在此不使用高斯的公式有几个原因, 但最大的原因是公式假设您在将序列中的所有整数从 0 添加到最后一个整数。在这个问题中, 您只需要从第三学期到第六学期的总和。

Example 4
::例4

If the sum of the first n integers is 210, what is n ?
::如果第一个n整数的总和是210, 什么是n?

Using Gauss's formula:
::使用 Gaus 的公式 :

%20(n%20%2B%201)%7D%20%7B2%7D"> $\begin{aligned} \sum = \frac{(n) (n + 1)}{2} \end{aligned}$
:n)(n+1)2

%20(n%20%2B%201)%7D%20%7B2%7D"> $\begin{aligned} 210 = \frac{(n) (n + 1)}{2} \end{aligned}$
::210=(n+1)2

Multiply both sides by 2: (n%20%2B%201)"> $\begin{aligned} 420 = (n) (n + 1) \end{aligned}$
::将两边乘以 2: 420=+1

Distribute: $\begin{aligned} 420 = (n^{2} + n) \end{aligned}$
::分配:420=(n2+n)

Complete the square: $\begin{aligned} 420 = (n^{2} + n + 1 / 4) \end{aligned}$
::完成方形: 420=(n2+n+1/4)

Factor: $\begin{aligned} 420 = (n + 1 / 2)^{2} \end{aligned}$
::系数:420=(n+1/2)2

Square root both sides: $\begin{aligned} 201 / 2 = n + 1 / 2 \end{aligned}$
::双方平方根:201/2=n+1/2

$\begin{aligned} n = 20 \end{aligned}$
::n=20

Example 5
::例5

Express the sum using sigma notation: 1 + 3 + 9 + 27 + ...
::1+3+9+27+.

$\begin{aligned} \sum_{n = 1}^{\infty} 3^{n - 1} \end{aligned}$ or $\begin{aligned} \sum_{n = 0}^{\infty} 3^{n} \end{aligned}$
::=1=3n -1或0=3n

Example 6
::例6

Find the sum of $\begin{aligned} \sum_{n = 0}^{12} \frac{- 11}{3} (n - 1) \end{aligned}$ .
::查找%n=012-113(n-1)之和。

Let's look at both ways of solving this one:
::让我们来看看解决这一难题的两种方法:

We could plug in all the numbers between 0 and 12 to get: $\begin{aligned} \sum_{n = 0}^{12} \frac{- 11}{3} (n - 1) \end{aligned}$ and then add them together to get the sum.
::我们可以在 0 到 12 之间的所有数字中插入 :\n= 012-113 (n-1) , 然后将它们加在一起, 才能得到这个数字。

Using a formula: A slightly modified version of Gauss's formula looks like this: $\begin{aligned} S_{n} = \frac{k}{2} (a_{0} + a_{n}) \end{aligned}$ , where k is the number of terms in the series plus one , and a ₀ and a _n are the first and last terms in the series
::使用公式 : 稍稍修改的 Gaus 公式版本看起来类似 : Sn=k2( a0+an) , k 是序列中的术语数 + 1, a0 和 a 是序列中的第一个也是最后一个条件

\begin{aligned} (\frac{13}{2}) ((\frac{- 11}{3}) (- 1) + (\frac{- 11}{3}) (11)) & = (\frac{13}{2}) (\frac{11}{3} + \frac{- 121}{3}) \\ = (\frac{13}{2}) (\frac{- 110}{3}) \\ = \frac{- 1430}{6} \\ = \frac{- 715}{3} \to - 238 \frac{1}{3} \end{aligned}

Example 7
::例7

Use Gauss's formula to find the sum of the first 200 positive integers.
::使用 Gaus 的公式查找第一个200正整数的总和。

Gauss's formula:
::高斯的公式 :

%20(n%20%2B%201)%7D%20%7B2%7D"> $\begin{aligned} = \frac{(n) (n + 1)}{2} \end{aligned}$
::=(n+1)2

n = 200
::n=200

$\begin{aligned} = \frac{(200) (200 + 1)}{2} \end{aligned}$

Sum = 20,100
::总计=20 100

Review
::回顾

Calculate the sums of the given series, you may use addition of individual terms, or a series sum formula. You may use a graphing tool for any 3 of them. Try to use each method at least once.
::计算给定序列的计算总和, 您可以使用添加单个术语或序列总和公式。您可以为其中任何 3 个使用图形工具。尝试使用每种方法至少一次。

$\begin{aligned} \sum_{n = 0}^{16} - 10 + 3 (n - 1) \end{aligned}$
::=016-10+3(n-1)
$\begin{aligned} 64 + 72 + 80 + . . . + 200 \end{aligned}$

$\begin{aligned} \sum_{n = 0}^{6} 6 - \frac{1}{2} (n - 1) \end{aligned}$
::n=066-12(n-1)
$\begin{aligned} 2 + 4 + 6 + . . . 26 \end{aligned}$
$\begin{aligned} - 2 - 1 + 0 + . . . + 12 \end{aligned}$

$\begin{aligned} \sum_{n = 7}^{20} - 1 + (n - 1) \end{aligned}$
::n=720-1+(n-1)

$\begin{aligned} \sum_{n = - 6}^{22} 5 + 3 (n - 1) \end{aligned}$
::6225+3(n-1)

$\begin{aligned} \sum_{n = - 2}^{13} - 5 - \frac{3}{2} (n - 1) \end{aligned}$
:n)213-5-32(n-1)

Consider the sums $\begin{aligned} \sum_{n = 1}^{5} (n + 1) \end{aligned}$ and $\begin{aligned} \sum_{n = 1}^{5} (n - 4) \end{aligned}$ .
::考虑 n=15(n+1)和n=15(n-4)。

What is the product of $\begin{aligned} (\sum_{n = 1}^{5} (n + 1)) (\sum_{n = 1}^{5} (n - 4)) \end{aligned}$
:=15(n+1))(=15(n-4))的产物是什么?

What is the sum of $\begin{aligned} \sum_{n = 1}^{5} (n + 1) (n - 4) \end{aligned}$ ?
::+n=15(n+1)(n-4)的和是多少?

Look closer at the last two problems, what does this tell you about rules for working with sums?
::仔细看看最后两个问题这能告诉你什么与金额合作的规则?

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。

章节大纲

Series Sums and Gauss's Formula ::系列总和和和高斯公式

Gauss's Formula ::高斯公式

Using a Graphing Calculator ::使用图形计算计算器

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Example 6 ::例6

Example 7 ::例7

Review ::回顾

Review (Answers) ::回顾(答复)