节: 序列和总和符号符号 | 13.5 系列和总和编号

章节大纲

According to legend, the mathematician Johann Carl Friedrich Gauss (seen below) did not behave in school when he was a child. One day his teacher assigned him a task—to add up all the integer s from 1 to 100. The teacher assumed this would take Gauss awhile, but Gauss came up with the answer in a matter of seconds. Needless to say, the teacher was shocked. ¹
::根据传说,数学家约翰·卡尔·弗里德里希·高斯(下文见此)在小时候在学校没有表现,有一天,他的老师指派他的任务是将1至100的整数加在一起,老师假设这需要高斯一段时间,但高斯在几秒钟内就找到了答案。不用说,这位教师被震惊了。

As we have seen, the 1, 2, 3, ... is an arithmetic sequence. If we add up the terms of a sequence, we have what is called a series . We discuss series and how to denote them in this section.
::正如我们所看到的,1,2,3,......是一个算术序列。如果我们把一个序列的条件加在一起,我们就有所谓的系列。我们讨论系列,并在本节中如何表示它们。

lesson content

Series and Summation Notation
::序列和总和符号符号

A series is the sum of the terms in a sequence. We often express a series with summation notation (also called sigma notation ), which uses the capital Greek letter $\begin{aligned} \sum \end{aligned}$ , sigma. For example, we could write a series that sums the 1st five counting numbers as $\begin{aligned} \sum_{n = 1}^{5} n = 1 + 2 + 3 + 4 + 5 = 15. \end{aligned}$ Here, $\begin{aligned} n \end{aligned}$ is the index of summation . This tells us what terms to add up. The number beneath the sigma is the lower limit, and the number above the sigma is the upper limit. We often use other letters, like $\begin{aligned} i, j \end{aligned}$ , and $\begin{aligned} k \end{aligned}$ , to indicate the index of summation.
::一个序列是术语在顺序中的总和。我们通常用总和符号( 也称为 igma 符号) 来表达一个序列, 它使用希腊大写字母 {, sigma 。例如, 我们可以写一个序列, 将第 5 个计数数乘以 n= 15n= 1+2+3+3+4+5=15。这里, n 是总和指数。这告诉我们要加起来的术语是什么。这告诉我们。 Sigma 下的数字是下限, Sigma 以上的数字是上限。我们经常使用其他字母, 如 i, j 和 k 来表示总和指数。

Example 1
::例1

Write the terms and find the sum: $\begin{aligned} \sum_{n = 1}^{6} 4 n - 1 \end{aligned}$ .
::写下条件并找到金额:n=164n-1。

Solution: Begin by replacing n with the values 1 through 6 to find the terms in the series, and then add them together.
::解决办法:首先用数值n 取代数值n 1至6,以便在序列中找到术语,然后将它们加在一起。

\begin{aligned} \sum_{n = 1}^{6} 4 n - 1 & = (4 (1) - 1) + (4 (2) - 1) + (4 (3) - 1) + (4 (4) - 1) + (4 (5) - 1) + (4 (6) - 1) \\ = 3 + 7 + 11 + 15 + 19 + 23 \\ = 78 \end{aligned}

::n=164n-1=(4(1)-1-1)+(4(2)-1)+(4(2)-1)+(4(3)-1)+(4(4)-1)+(4(4)-1)+(4(5)-1-1)+(4(6)-1)+(4(6)-1)=3+7+11+15+19+23=78

When we add up the 1st n terms of a series, we sometimes denote it as $\begin{aligned} S_{n} . \end{aligned}$ Here, $\begin{aligned} S_{6} = 78. \end{aligned}$
::当我们加起来一系列条件的第1n项时, 我们有时把它称为Sn。在这里,S6=78。

Example 2
::例2

Write the terms and find the sum: $\begin{aligned} \sum_{n = 9}^{11} \frac{n (n - 1)}{2} \end{aligned}$ .
::写下条件并找到总和:n=911n(n-1)2。

Solution: Replace $\begin{aligned} n \end{aligned}$ with the values 9, 10, and 11, and sum the terms in the resulting series.
::解决办法:将n替换为值9、10和11,并将所产生的系列中的术语相加。

\begin{aligned} \sum_{n = 9}^{11} \frac{n (n - 1)}{2} & = \frac{9 (9 - 1)}{2} + \frac{10 (10 - 1)}{2} + \frac{11 (11 - 1)}{2} \\ = 36 + 45 + 55 \\ = 136 \end{aligned}

::n=911n(n-1)2=9(9-1)2+10(10-1)2+11(11-1)2=36+45+55=136

by Diane R Koenig demonstrates how to evaluate expressions with summation notation.
::Diane R Koenig展示了如何用总和符号来评价表达方式。

Example 3
::例3

Write the terms and find the sum: $\begin{aligned} \sum_{n = 3}^{7} 2 (n - 3) \end{aligned}$ .
::写下条件并找到金额:n=372(n-3)。

Solution:

\begin{aligned} \sum_{n = 3}^{7} 2 (n - 3) & = 2 (3 - 3) + 2 (4 - 3) + 2 (5 - 3) + 2 (6 - 3) + 2 (7 - 3) \\ = 2 (0) + 2 (1) + 2 (2) + 2 (3) + 2 (4) \\ = 0 + 2 + 4 + 6 + 8 \\ = 20 \end{aligned}

::溶液:n=372(n-3)=2(3)-3+2(4-3)+2(3-3)+2(5-3)+2(5-3)+2(6-3)+2(3)+2(7-3)=2(0)+(2)(2)+2(2)+2(3)+2(4)=0+2+4+4+6+8=20

Example 4
::例4

Write the terms and find the sum: $\begin{aligned} \sum_{n = 1}^{7} \frac{1}{2} n + 1 \end{aligned}$ .
::写下条件并找到金额:n=1712n+1。

Solution:

\begin{aligned} \sum_{n = 1}^{7} \frac{1}{2} n + 1 & = (\frac{1}{2} (1) + 1) + (\frac{1}{2} (2) + 1) + (\frac{1}{2} (3) + 1) + (\frac{1}{2} (4) + 1) + (\frac{1}{2} (5) + 1) + (\frac{1}{2} (6) + 1) + (\frac{1}{2} (7) + 1) \\ = \frac{1}{2} + 1 + 1 + 1 + \frac{3}{2} + 1 + 2 + 1 + \frac{5}{2} + 1 + 3 + 1 + \frac{7}{2} + 1 \\ = \frac{16}{2} + 13 \\ = 8 + 13 \\ = 21 \end{aligned}

::解析度: _n=1712n+1=( 12(1)+1)+( 12(2)+1)+( 12(2)+1)+( 12(3)+1)+( 12(4)+1)+( 12(5)+1)+( 12(6)+1)+( 12(7)+1)+1+( 12(7)+1)=12(7)+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+72+1=162+13=8+13=8+13=21

Example 5
::例5

Write the terms and find the sum: $\begin{aligned} \sum_{n = 1}^{4} 3 n^{2} - 5 \end{aligned}$ .
::写下条件并找到金额:n=143n2-5。

Solution:

\begin{aligned} \sum_{n = 1}^{4} 3 n^{2} - 5 & = (3 (1)^{2} - 5) + (3 (2)^{2} - 5) + (3 (3)^{2} - 5) + (3 (4)^{2} - 5) \\ = 3 - 5 + 12 - 5 + 27 - 5 + 48 - 5 \\ = 90 - 20 \\ = 70 \end{aligned}

::解决办法:=143n2-5=(3(1)2-5)+(3(2)2-5)+(3(2)-2-5)+(3(3)2-5)+(3(3)2-5)+(3(4)2-5)=3-5+12-5+5+27-5+48-5=5=90-20=70

How To Find the Value of a Series in Desmos

::如何在 Desmos 中查找系列的值

1. To get a sigma, type s-u-m on the keyboard. A sigma should appear with blanks for the limits and the sequence. (You can also find a sigma under the keyboard menu on the bottom left. Click functions and then misc. )
::1. 要在键盘上找到 s-u-m 键盘上的 S-u-m 键盘, 键盘上应该有一个 sigma , 边框和顺序应该有空白。 (您也可以在左下方键盘菜单下找到 sigma 。单击函数, 然后是错误。)

2. Enter the upper and lower limit of the sum and the sequence, and press enter . The sum will appear in the lower right side of the box.
::2. 输入总和和序列的上下限,按下输入。总和将出现在框的右下角。

by Desmos shows how to find the value of a series, and provides an example of a more advanced application.
::Desmos 显示如何找到一系列的值, 并提供了一个更高级应用程序的示例。

How to Find the Value of a Series With a TI-83/84
::如何用TI-83/84 来找到一个系列的值

1. Press 2nd STAT (to get to the List menu) and arrow over to MATH .
::1. 按第二版STAT(获取列表菜单)和向MATH方向箭头。

2. Select option 5: sum(
::2. 选择备选方案5:总(

3. Then return to the List menu, arrow over to OPS and select option 5: seq( . You get sum(seq( on your screen.
::3. 然后返回列表菜单,向上箭头到OPS,并选择选项5:后(.在屏幕上(seq)得到总(seq)。

4. Next, enter in (expression, variable, begin, end), just as we did previously to list the terms in a sequence. By including the sum( command, the calculator will sum the terms in the sequence for us.
::4. 下一步,输入(表达、变量、开始、结束),就象我们先前按照顺序列出术语一样,通过加入总和(命令),计算器将按顺序对术语进行总和。

For example, the series in Example 1 would be
::例如,例例例1中的系列将是:

\begin{aligned} s u m (s e q (4 x - 1, x, 1, 6)) = 78. \end{aligned}

::Sumeq(4x-1,x,1,6)=78。

And the series in Example 2 would be
::例2中的系列

\begin{aligned} s u m (s e q (x (x - 1) / 2, x, 9, 11)) = 136. \end{aligned}

::苏姆(x(x-1)/2,x,9,11)=136。

Special Cases
::特殊情况

There are a few special series that are used in more advanced math classes, such as calculus. In these series, we will use the variable, $\begin{aligned} i \end{aligned}$ (not the complex number), to represent the index and $\begin{aligned} n \end{aligned}$ to represent the upper bound (the total number of terms) for the sum. One special case is
::有少数特殊系列用于较先进的数学类,例如微积分。在这些系列中,我们将使用变量i(不是复杂数字)来代表指数,而 n(代表总和的上限(术语总数))代表总和的上限(术语总数)。

\begin{aligned} \sum_{i = 1}^{n} 1 = n . \end{aligned}

::i=1n1=n。 i=1n1=n。 i=1n1=n。

Let $\begin{aligned} n = 5. \end{aligned}$ Now we have the series $\begin{aligned} \sum_{i = 1}^{5} 1 = 1 + 1 + 1 + 1 + 1 = 5 \end{aligned}$ . Basically, in the series we are adding 1 to itself $\begin{aligned} n \end{aligned}$ times (or calculating $\begin{aligned} n \times 1 \end{aligned}$ ), so the resulting sum will always be $\begin{aligned} n \end{aligned}$ .
::Let n=5. 现在我们有了“ i=151=1+1+1+1+1+1=5” 系列。基本上, 在序列中, 我们给自身添加了 1 的 n 次数( 或计算 nx1) 。因此结果总和总是 n 。

Another special case is
::另一个特殊情况是:

\begin{aligned} \sum_{i = 1}^{n} i = \frac{n (n + 1)}{2} . \end{aligned}

::i=1ni=n(n+1)2。

If we let $\begin{aligned} n = 5 \end{aligned}$ again, we get $\begin{aligned} \sum_{i = 1}^{n} i = 1 + 2 + 3 + 4 + 5 = 15 = \frac{5 (5 + 1)}{2} . \end{aligned}$ This rule is closely related to the rule for the sum of an arithmetic sequence, and will be used to prove the sum formula later in the chapter.
::如果再使用n=5, 我们就会得到 i=1ni=1+2+3+4+5=15=5(5+1)2。这一规则与算术序列总和的规则密切相关, 并将用于证明本章后面部分的总和公式。

We also have a special case for the square of the terms.
::我们也有一个关于条件方方面面的特殊情况。

\begin{aligned} \sum_{i = 1}^{n} i^{2} = \frac{n (n + 1) (2 n + 1)}{6} \end{aligned}

::i=1n2=n(n+1)(2n+1)6

Let $\begin{aligned} n = 5 \end{aligned}$ once more. Using the rule, the sum is $\begin{aligned} \frac{5 (5 + 1) (2 (5) + 1)}{6} = \frac{5 (6) (11)}{6} = 55. \end{aligned}$
::n=5 再说一次。使用此规则,总和为 5(5+1)(2(5)+1)6=5(6)(11)6=55。

If we write the terms in the series and find their sum, we get $\begin{aligned} 1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} = 1 + 4 + 9 + 16 + 25 = 55. \end{aligned}$
::如果我们在系列中写入术语并找到它们的总和, 我们就会得到12+22+32+42+52=1+4+9+16+25=55。

Example 6
::例6

Use one of the rules above to evaluate $\begin{aligned} \sum_{i = 1}^{15} i^{2} \end{aligned}$ .
::使用上述规则之一评估i=1152。

Solution: Using the rule $\begin{aligned} \sum_{i = 1}^{n} i^{2} = \frac{n (n + 1) (2 n + 1)}{6}, \end{aligned}$ we get $\begin{aligned} \frac{15 (15 + 1) (2 (15) + 1)}{6} = \frac{15 (16) (31)}{6} = 1, 240. \end{aligned}$
::解答:使用规则“i=1ni2=n(n+1)(2n+1)6,我们得到15(15+1)(2(2(2(15)+1)6=15(16)(31)6=1,240)。

Example 7
::例7

Gauss was asked by his teacher to determine the sum of the first 100 counting numbers in grade school. Find the sum.
::教师要求Gaus确定小学头100个计数数字的总和。

Solution: Using the formula for the special case $\begin{aligned} \sum_{i = 1}^{n} i = \frac{n (n + 1)}{2}, \end{aligned}$ we have
::解决方案:使用特例“i=1ni=n(n+1)2”的公式。

\begin{aligned} \sum_{i = 1}^{100} i & = \frac{100 (100 + 1)}{2} \\ = \frac{10, 100}{2} \\ = 5, 050 \end{aligned}

::i=1 100i=100(100+1)2=10 1002=5 050

We will see a more detailed discussion of Gauss's approach in the section on Finding the Sum of an Arithmetic Sequence.
::我们将看到在 " 找到一个神学序列的总和 " 一节中更详细地讨论高斯的做法。

Since we are discussing sums, the properties of addition apply with sigma notation as well.
::既然我们讨论的是总和,添加的特性也用污名表示。

Writing Sums With Sigma Notation

::使用 Sigma 符号书写总和

\begin{aligned} \sum_{i = 1}^{n} (a_{i} + b_{i}) & = \sum_{i = 1}^{n} a_{i} + \sum_{i = 1}^{n} b_{i} & commutative and associative properties \\ \sum_{i = 1}^{n} (a_{i} - b_{i}) & = \sum_{i = 1}^{n} a_{i} - \sum_{i = 1}^{n} b_{i} & commutative and associative properties \\ \sum_{i = 1}^{n} c a_{i} & = c \sum_{i = 1}^{n} a_{i} & distributive property \end{aligned}

Summary
::摘要

A series is the sum of the terms in a sequence.
::A系列是术语在顺序中的总和。

We often express a series with summation notation (also called sigma notation), which uses the capital Greek letter sigma, $\begin{aligned} \sum \end{aligned}$ .

::我们经常使用希腊大写字母西格玛(Sigma)表示一系列加号(也称为Sigma),

Review
::回顾

Write out the terms and find the sums.
::写好条件,找到金额

1. $\begin{aligned} \sum_{n = 1}^{5} 2 n \end{aligned}$
::1.n=152n

2. $\begin{aligned} \sum_{n = 5}^{8} n + 3 \end{aligned}$
::2.n=58n+3

3. $\begin{aligned} \sum_{n = 10}^{15} n (n - 3) \end{aligned}$
::3.n=1015n(n-3)

4. $\begin{aligned} \sum_{n = 3}^{7} \frac{n (n - 1)}{2} \end{aligned}$
::4.n=37n(n-1)2

5. $\begin{aligned} \sum_{n = 1}^{6} 2^{n - 1} + 3 \end{aligned}$
::5.=162n-1+3

6. $\begin{aligned} \sum_{n = 10}^{15} \frac{1}{2} n + 3 \end{aligned}$
::6.n=101512n+3

7. $\begin{aligned} \sum_{n = 0}^{50} n - 25 \end{aligned}$
::7.n=0.50n-25

8. $\begin{aligned} \sum_{n = 1}^{5} {(\frac{1}{2})}^{n - 5} \end{aligned}$
::8.n=15(12)n-5

9. $\begin{aligned} \sum_{n = 5}^{12} \frac{n (2 n + 1)}{2} \end{aligned}$
::9.n=512n(2n+1)2

10. $\begin{aligned} \sum_{n = 1}^{100} \frac{1}{2} n \end{aligned}$
::10.n=110012n

11. $\begin{aligned} \sum_{n = 1}^{200} n \end{aligned}$
::11.n=1200n

12.

a. $\begin{aligned} \sum_{n = 1}^{5} 2 n + 3 \end{aligned}$
::a. =152n+3

b. $\begin{aligned} 3 (5) + \sum_{n = 1}^{5} 2 n \end{aligned}$
::b. 3(5)n=152n

13.

a. $\begin{aligned} \sum_{n = 1}^{5} \frac{n (n + 1)}{2} \end{aligned}$
::a. n=15n(n+1)2

b. $\begin{aligned} \frac{1}{2} \sum_{n = 1}^{5} n (n + 1) \end{aligned}$
::b. 12n=15n(n+1)

14.

a. $\begin{aligned} \sum_{n = 1}^{5} 4 x^{3} \end{aligned}$
::a. n=154x3

b. $\begin{aligned} 4 \sum_{n = 1}^{5} x^{3} \end{aligned}$
::b. 4n=15x3

Explore More
::探索更多

1. You are stacking oranges in the shape of a square pyramid with 7 layers. How many oranges are in this pyramid? Assume there is one orange on top.

::1. 你正以7层平方形金字塔的形式堆叠橘子,金字塔内有多少橙子?假设上面有一个橙子。

lesson content

3. Tim earned $105 on the 1st day of work. If he earned 2 times the amount of money earned the day before each day, how much did he earn in the first 7 days?

::3. 蒂姆在第一天工作时挣了105美元,如果比前天挣了2倍于前天挣的钱,头7天挣了多少钱?

4. Use the table below to determine how many tourists visited the museum in a week.
::4. 利用下表确定一周内有多少游客参观博物馆。

Day	Tourists
Monday	240
Tuesday	360
Wednesday	480
. . .	. . .

5. A shoe company makes 750 pairs of shoes in February, 690 pairs of shoes in March, and 630 pairs of shoes in April. Assuming the sequence continues, how many pairs of shoes does the company make from February to June?

::5. 2月份,一家鞋公司制成750双鞋,3月份制成690双鞋,4月份制成630双鞋,假设顺序继续,该公司2月至6月份制成多少双鞋?

6. The number of tagged deer reported to the game commission one Saturday is represented by the sum $\begin{aligned} \sum_{n = 1}^{6} 3 n - 2 \end{aligned}$ . How many tagged deer were reported?
::6. 星期六向游戏委员会报告的贴有标签的鹿数目为n=163n-2。报告了多少个贴有标签的鹿?

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 this interactive that reinforces the concepts explored in this section:
::尝试这一互动,强化本节所探讨的概念:

References
::参考参考资料

1. "Carl Friedrich Gauss," last edited May 10, 2017,
::1. “卡尔·弗里德里希·高斯”,2017年5月10日最后一次编辑,

章节大纲

Series and Summation Notation ::序列和总和符号符号

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

How To Find the Value of a Series in Desmos ::如何在 Desmos 中查找系列的值

How to Find the Value of a Series With a TI-83/84 ::如何用TI-83/84 来找到一个系列的值

Special Cases ::特殊情况

Example 6 ::例6

Example 7 ::例7

Writing Sums With Sigma Notation ::使用 Sigma 符号书写总和

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

References ::参考参考资料

Series and Summation Notation
::序列和总和符号符号

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Example 5
::例5

How To Find the Value of a Series in Desmos

::如何在 Desmos 中查找系列的值

How to Find the Value of a Series With a TI-83/84
::如何用TI-83/84 来找到一个系列的值

Special Cases
::特殊情况

Example 6
::例6

Example 7
::例7

Writing Sums With Sigma Notation

::使用 Sigma 符号书写总和

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

References
::参考参考资料