课程： 7.5 以系列总和解决问题

章节大纲

选择节常规

折叠展开
常规

全部折叠展开全部
- 选择活动新闻通告
  
  新闻通告讨论区
选择节以系列总和解决问题

折叠展开
以系列总和解决问题
Katrina just started her new job last week. She is making $60 per day, and saving 10%. Her savings account has been slowly building up at $6 per day, and currently has $73 in it.
::卡特里娜上周刚开始新工作。她每天挣60美元,储蓄10%。她的储蓄账户以每天6美元缓慢增长,目前只有73美元。

How much money will she have after working another 15 days? How much after another 27 days?
::再工作15天她能赚多少钱?再工作27天又能赚多少?

Problem Solving with Series Sums
::以系列总和解决问题

In the previous lesson, we discussed Gauss's formula for finding the total of the numbers in a series. In this lesson, we continue to practice that, and also to make use of the modified version of Gauss's formula that is commonly seen in other texts:
::在以往的教训中,我们讨论了高斯在一系列数字中寻找总数的方法。在这个教训中,我们继续实践这个方法,并使用其他案文中常见的经修改的高斯公式:

The sum of the first n terms in an arithmetic series is $\begin{aligned} S_{n} = \frac{n (a_{1} + a_{n})}{2} \end{aligned}$ .
::计算序列中第一个 n 术语的总和是 Sn=n(a1+an) 2。

Examples
::实例

Example 1
::例1

Find the sum of the first 50 terms of an arithmetic series if the first term is 5 and the common difference is 3.
::如果第一个学期是5,而共同差数是3,则查找算术序列头50个学期的总和。

$\begin{aligned} S_{50} = \frac{50 (a_{1} + a_{n})}{2} \end{aligned}$ ..... Substitute the applicable number of terms in the
::S50=50(a1+an)2...。替代S50=50(a1+an)2...。

$\begin{aligned} \frac{50 (5 + (5 + 49 \times 3))}{2} \end{aligned}$ ..... Substitute the values for the first and 50th terms
::50( 5+( 5+49x3)) 2...。取代第 1和第 50 个条件的值

$\begin{aligned} \frac{50 (157)}{2} = (25) (127) \end{aligned}$ ..... Simplify
::50(157)2=( 25) (127).... 简化

$\begin{aligned} 3, 175 \end{aligned}$

This method is clearly much easier than writing out and adding 50 numbers!
::这个方法显然比写出和增加50个数字容易得多!

Example 2
::例2

Find the sum of the first 40 terms of an arithmetic series in which the first term is 8 and the common difference is 5.
::查找算术序列头40个条件的总和,其中第一个任期为8个,共同差数为5个。

Use the formula $\begin{aligned} S_{n} = \frac{n (a_{1} + a_{n})}{2} \end{aligned}$
::使用公式 Sn=n(a1+an)2

$\begin{aligned} S_{40} = \frac{40 (8 + 39 \cdot 5)}{2} \end{aligned}$ ..... Substitute in the number of terms, and the value of the first and last terms
::S40=40(8+395)2...。代替条件数和第一个和最后一个条件值

$\begin{aligned} S_{40} = \frac{40 (8 + 195)}{2} \end{aligned}$ ..... Simplify
::S40=40( 8+1952) 2..... 简化

$\begin{aligned} S_{40} = \frac{8120}{2} \end{aligned}$ ..... Simplify
::S40=81202.简化

$\begin{aligned} S_{40} = 4060 \end{aligned}$
::S40=4060

Example 3
::例3

Consider the series: -13 + -3 + 7 + 17 + 27 + ...
::考虑此序列: -13 + -3 + 7 + 17 + 27 +...

What is the 25 ^th term?
::第25个学期是什么?

This is an arithmetic series where each successive term is 10 more than the last.
::这是一个算术序列,每个连续任期比上一个任期多10个。

$\begin{aligned} - 13 + 24 (10) = 227 \end{aligned}$

What is the sum of the first 25 terms?
::前25个学期的总和是多少?

Use the formula $\begin{aligned} S_{n} = \frac{n (a_{1} + a_{n})}{2} \end{aligned}$ .
::使用公式 Sn=n(a1+an) 2。

This is the sum of the first 25 terms of an arithmetic series with a common difference of 10
::这是算术序列中共同差数为10的最初25个条件的总和。

$\begin{aligned} S_{25} = \frac{25 (- 13 + 227)}{2} \end{aligned}$ ..... Substitute in the number of terms, and the value of the first and last terms
::S25=25(- 13+227) 2...。替代条件的数量和第一个和最后一个条件的价值

$\begin{aligned} S_{25} = \frac{25 \cdot 214}{2} \end{aligned}$ ..... Simplify
::S25=252142.... 简化

$\begin{aligned} S_{25} = \frac{5350}{2} \end{aligned}$ ..... Simplify
::S25 = 53502.... 简化

$\begin{aligned} S_{25} = 2675 \end{aligned}$
::S25=2675

Example 4
::例4

A student needed to find the sum of the first 10 terms of the series 4 + 12 + 36 + ... and so he wrote the following:
::一名学生需要找到系列4+12+36+.的前10个学期的总和,因此他写道:

$\begin{aligned} S_{10} = \frac{10 (4 + 4 \times 3^{9})}{2} = \frac{10 (78, 736)}{2} = 393, 680 \end{aligned}$
::S10=10(4+4x39)2=10(78,736)2=393,680

Do you agree with the student’s work? Explain.
::您是否同意学生的工作?请解释。

Because the series is geometric, this formula is not appropriate. The work here does not represent the sum of the first 10 terms. Using a graphing calculator, you can find that the sum is 118,096.
::由于序列是几何, 此公式不合适。这里的工程不代表前 10 个术语的总和。使用图形计算器, 您可以发现总计是 118 096 。

Example 5
::例5

Given an arithmetic sequence $\begin{aligned} (a_{n}) \end{aligned}$ determined by $\begin{aligned} a_{1} = 143 \end{aligned}$ and $\begin{aligned} d = - 3 \end{aligned}$ :
::a1=143和d3确定的算术序列(an):

What is the 220th number of the sequence?
::序列的第220号是多少?

We find the solution by using the following formula: $\begin{aligned} a_{n} = a + (n - 1) d \end{aligned}$ .
::我们通过使用以下公式来找到解决办法:a=a+(n-1)d。

$\begin{aligned} a_{220} = a_{1} + 219 d \end{aligned}$
::a220=a1+219d

$\begin{aligned} = 143 + 219 (- 3) \end{aligned}$

$\begin{aligned} = - 514 \end{aligned}$

Example 6
::例6

How do we find the sum of the first 220 numbers of the sequence, given the same information as in Example 5 above?
::根据与上文例5相同的信息,我们如何找到序列头220个数字的总和?

We use the following formula: "> $\begin{aligned} s_{n} = \frac{a_{1} + a_{n}}{2} (n) \end{aligned}$ . We set n = 220.
::我们使用以下公式:sn=a1+an2。我们设定n=220。

To find the sum of the first 220 numbers:
::要找到前220个数字的总和:

$\begin{aligned} s_{220} = \frac{a_{1} + a_{220}}{2} (220) \end{aligned}$
::s220=a1+a2202(220)

$\begin{aligned} s_{220} = \frac{143 + (- 514)}{2} (220) \end{aligned}$
::220=143+(-514)(220)

$\begin{aligned} s_{2} 20 = \frac{- 371}{2} \cdot 220 \end{aligned}$
::2203712220

The sum of the first 220 numbers of the sequence is $\begin{aligned} - 40, 810 \end{aligned}$ .
::序列的前220个数字总和是-40,810。

Review
::回顾

Use the arithmetic series formula to solve the following problems:
::使用计算序列公式解决下列问题:

Given the arithmetic series of numbers: 1, 4, 7, 10, 13... a)Find the 200th number in the sequence b)find the sum of the first 200 numbers.
::根据数字的算术序列:1、4、7、10、13... (a) 查找序列中的第200个数字 b) 查找前200个数字的总和。

The sum of the first five numbers of an arithmetic sequence is -45. What is the value of the third number? ( hint: find $\begin{aligned} a_{3} \end{aligned}$ if $\begin{aligned} s_{5} = - 45 \end{aligned}$ )
::算术序列的前五个数字总和是 -45。第三个数字的值是多少? (提示: 如果 s545, 则查找 a3)

You have an arithmetic series of numbers defined by: $\begin{aligned} a_{1} = 45 \end{aligned}$ and $\begin{aligned} d = - 5 \end{aligned}$ a)Determine $\begin{aligned} a_{1} 50 \end{aligned}$ b)Identify the sum of: $\begin{aligned} a_{1} + a_{2} + . . . + a_{150} \end{aligned}$
::您的计算序列由以下数字组成: a1=45 和 d=5 a) 确定 a150 b) 确定总和: a1+a2+...+a150

The sum of the first three numbers in an arithmetic sequence is 219. The sum of the first nine numbers in the same sequence is 603. What is the 143rd number of the sequence.
::算术序列中前三个数字的总和是 219。同一序列中前九个数字的总和是 603。序列中的第143个数字是多少。

The first eight numbers of an arithmetic sequence add up to 604. The next eight numbers added up equal 156. Find the first number and the common difference in the sequence.
::算术序列的前八个数字加起来等于604。接下来的八个数字加起来等于156。在序列中查找第一个数字和常见的差别。

The first number in an arithmetic sequence is 80. Find the common difference if we also know that $\begin{aligned} s_{9} \end{aligned}$ is eighteen times $\begin{aligned} a_{11} \end{aligned}$
::算术序列中的第一个数字是80 如果我们知道S9是每1118倍的话找出共同的差别

If $\begin{aligned} a_{n} \end{aligned}$ is an arithmetic sequence with $\begin{aligned} a_{1} = 1 \end{aligned}$ . Find the second number if we know that the sum of the first five numbers is one-fourth of the sum of the next five numbers.
::如果 a 是 a1 =1 的算术序列, 请找到第二个数字, 如果我们知道前五个数字的总和是后五个数字总和的四分之一。

Given $\begin{aligned} (a_{n}) = 78, 75, 72, 69... \end{aligned}$ Find $\begin{aligned} a_{150} \end{aligned}$ and $\begin{aligned} s_{150} \end{aligned}$
:an)=78,75,72,69.Finda150和S150

The following conditions exist within a sequence of numbers: $\begin{aligned} a_{50} = 252 \end{aligned}$ and $\begin{aligned} s_{50} = 2800 \end{aligned}$ . What is the first number of the series, and what is the common difference?
::以下条件存在于数字序列的顺序中:a50=252和s50=2800。序列的第一个数字是多少,共同的区别是什么?

What are the values of $\begin{aligned} a \end{aligned}$ and $\begin{aligned} d \end{aligned}$ , given that $\begin{aligned} (a_{n}) \end{aligned}$ is an arithmetic series of numbers, if we know: $\begin{aligned} a_{15} = 62 \end{aligned}$ and $\begin{aligned} s_{20} = 700 \end{aligned}$ ?
::a和d的值是多少,因为(a)是数字的算术序列,如果我们知道:a15=62和S20=700?

Given the sequence: $\begin{aligned} a_{1} = - 16 \end{aligned}$ and $\begin{aligned} d = \frac{1}{3} \end{aligned}$ Find the values of n, so that $\begin{aligned} s_{n} = 50 \end{aligned}$
::根据序列 : a1 16 和 d=13 查找 n 的值, 所以 sn= 50

If $\begin{aligned} a_{1} = 8 \end{aligned}$ and $\begin{aligned} d = - 3 \end{aligned}$ . What are the values of $\begin{aligned} a_{20} \end{aligned}$ and $\begin{aligned} s_{20} \end{aligned}$ ?
::如果 a1=8 和 d3 。 20 和 s20 的值是多少 ?

If $\begin{aligned} a_{34} = 193 \end{aligned}$ and $\begin{aligned} s_{17} = 306 \end{aligned}$ , find $\begin{aligned} a \end{aligned}$ and $\begin{aligned} d \end{aligned}$ .
::如果a34=193和S17=306,则找到a和d。

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

常规

以系列总和解决问题

Problem Solving with Series Sums ::以系列总和解决问题

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

Review ::回顾

Review (Answers) ::回顾(答复)

Problem Solving with Series Sums
::以系列总和解决问题

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

Review
::回顾

Review (Answers)
::回顾(答复)