课程： 11.4 系列和总和编号

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选择节序列和总和符号符号

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

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

A series is the sum of the terms in a . A series is often expresses in summation notation (also called sigma notation) which uses the capital Greek letter $\begin{align*}\sum\end{align*}$ , sigma. Example: $\begin{align*}\sum\limits_{n=1}^5 n=1+2+3+4+5=15\end{align*}$ . Beneath the sigma is the index (in this case $\begin{align*}n\end{align*}$ ) which tells us what value to plug in first. Above the sigma is the upper limit which tells us the upper limit to plug into the rule.
::序列是 a 中术语的总和。一个序列通常以总和符号( 也称为 Sigma 符号) 表示, 该符号使用希腊大写字母 {, sigma 。例如 : {n= 15n= 1+2+3+4+5=15。低于 sigma 是指数( 在本案中 n) , 它告诉我们先插入什么值。高于 sigma 是上限, 它告诉我们插入规则的上限。

Let's write the terms and find the sum of the following series.
::让我们写下条款找到以下系列的总和

$\begin{align*}\sum\limits_{n=1}^6 4n+1\end{align*}$
::=164n+1

Begin by replacing n with the values 1 through 6 to find the terms in the series and then add them together.
::开始以数值 n 取代值 1 至 6, 以查找序列中的术语, 然后将它们加在一起。

$\begin{aligned} (4 (1) - 1) + (4 (2) - 1) + (4 (3) - 1) + (4 (4) - 1) + (4 (5) - 1) + (4 (6) - 1) \\ 3 + 7 + 11 + 15 + 19 + 23 \\ = 78 \end{aligned}$
$\begin{align*}& \left(4(1)-1\right)+\left(4(2)-1\right)+\left(4(3)-1\right)+\left(4(4)-1\right)+\left(4(5)-1\right)+\left(4(6)-1\right) \\ & 3+7+11+15+19+23 \\ & =78\end{align*}$

Calculator: The graphing calculator can also be used to evaluate this sum. We will use a compound function in which we will sum a sequence. Go to $\begin{align*}2^{nd}\end{align*}$ STAT (to get to the List menu) and arrow over to MATH . Select option 5: sum( then return to the List menu, arrow over to OPS and select option 5: seq( to get sum(seq( on your screen. Next, enter in (expression, variable, begin, end) to list the terms in a sequence. By including the sum( command, the calculator will sum the terms in the sequence for us. For this particular problem the expression and result on the calculator are:
::计算器 : 图形计算器也可以用来评估此总和。我们将使用一个复合函数来计算一个序列。转到第二STAT( 转到列表菜单) 和箭头到 MATH 。选择选项 5 : sum( 然后返回列表菜单, 箭头转到 OPS 和选择选项 5 : 后( 获得 suq( 在屏幕上 ) 。下一步, 输入( 表达式、变量、开始、结束) 来按序列列出术语。通过包含总和( 命令) , 计算器将计算我们序列中的术语。对于这个特定的问题, 计算器的表达式和结果是 :

$\begin{aligned} s u m (s e q (4 x - 1, x, 1, 6)) = 78 \end{aligned}$
$\begin{align*}sum(seq(4x-1,x,1,6))=78\end{align*}$
::suph( 4x-1, x, 1, 6) =78

To obtain a list of the terms, just use $\begin{align*}seq\left(4x-1,x,1,6\right)=\{3 \ \ 7 \ \ 11 \ \ 15 \ \ 19 \ \ 23\}\end{align*}$
::为了获得术语清单,请使用以下术语(4x-1,x,1,6)+3 7 11 15 19 23}

$\begin{align*}\sum\limits_{n=9}^{11} \frac{n(n+1)}{2}\end{align*}$
::n=911n(n+1)2

Replace $\begin{align*}n\end{align*}$ with the values 9, 10 and 11 and sum the resulting series.
::n 改为数值9、10和11,并加之所产生的序列。

$\begin{aligned} \frac{9 (9 - 1)}{2} + \frac{10 (10 - 1)}{2} + \frac{11 (11 - 1)}{2} \\ 36 + 45 + 55 \\ 136 \end{aligned}$
$\begin{align*}& \frac{9(9-1)}{2}+\frac{10(10-1)}{2}+\frac{11(11-1)}{2} \\ & \qquad \qquad \qquad 36+45+55 \\ & \qquad \qquad \qquad \qquad 136\end{align*}$

Using the calculator: $\begin{align*}sum(seq(x(x-1)/2,x,9,11))=136\end{align*}$ .
::使用计算器:Sumeq(x(x-1)/2,x,9,11)=136。

There are a few special series which are used in more advanced math classes, such as calculus. In these series, we will use the variable, $\begin{align*}i\end{align*}$ , to represent the index and $\begin{align*}n\end{align*}$ to represent the upper bound (the total number of terms) for the sum.
::有一些特殊系列用于较先进的数学课程,例如微积分。在这些系列中,我们将使用变量(i)来表示指数, n 来表示总和的上限(术语总数)。

$\begin{align*}\sum\limits_{i=1}^n 1=n\end{align*}$
::*i=1n1=n

Let $\begin{align*}n = 5\end{align*}$ , now we have the series $\begin{align*}\sum\limits_{i=1}^5 1=1+1+1+1+1=5\end{align*}$ . Basically, in the series we are adding 1 to itself $\begin{align*}n\end{align*}$ times (or calculating $\begin{align*}n\times1\end{align*}$ ) so the resulting sum will always be $\begin{align*}n\end{align*}$ .
::Let n=5, 现在我们有序列“ i=151=1+1+1+1+1+1=5 ” 。基本上, 在序列中, 我们给自己增加一个 n 次( 或计算 nx1) 数, 所以结果总和总是 n 。

$\begin{align*}\sum\limits_{i=1}^n i=\frac{n(n+1)}{2}\end{align*}$
::i=1ni=n(n+1)2

If we let $\begin{align*}n=5\end{align*}$ again we get $\begin{align*}\sum\limits_{i=1}^n i=1+2+3+4+5=15=\frac{5(5+1)}{2}\end{align*}$ . This one is a little harder to derive but can be illustrated using different values of $\begin{align*}n\end{align*}$ . This rule is closely related to the rule for the sum of an arithmetic series and will be used to prove the sum formula later on.
::如果我们再次让 n= 5, 我们就会得到“ i= 1ni= 1+2+3+4+5=15=5( 5+1) 2 。这个比较难取出, 但可以用 n 的不同值来说明。此规则与计算序列总和的规则密切相关, 以后将用来证明总和公式。

$\begin{align*}\sum\limits_{i=1}^n i=\frac{n(n+1)(2n+1)}{6}\end{align*}$
::i=1ni=n(n+1)(2n+1)6

Let $\begin{align*}n=5\end{align*}$ once more. Using the rule, the sum is $\begin{align*}\frac{5(5+1)(2(5)+1)}{6}=\frac{5(6)(11)}{6}=55\end{align*}$
::n=5 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 $\begin{align*}1^2+2^2+3^2+4^2+5^2=1+4+9+16+25=55\end{align*}$ .
::如果我们在系列中写下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。

The derivation of this rule is beyond the scope of this course.
::这一规则的推理超出了本程序的范围。

Now, let's use one of the rules above to evaluate $\begin{align*}\sum\limits_{i=1}^{15} i^2\end{align*}$ .
::现在,让我们用上面的规则之一来评估"i=1152"。

Using the rule $\begin{align*}\sum\limits_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6}\end{align*}$ , we get $\begin{align*}\frac{15(15+1)(2(15)+1)}{6}=\frac{15(16)(31)}{6}=1240\end{align*}$
::使用细则“i=1n2=n(n+1)(2n+1)6,我们得到15(15+1)(2(2(2(15)+1)6=15(16)(31)6=1240)

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the number of tagged deer that were reported.
::早些时候,有人要求你找到报告的贴有标签的鹿数目。

Begin by replacing n with the values 1 through 6 to find the terms in the series and then add them together.
::开始以数值 n 取代值 1 至 6, 以查找序列中的术语, 然后将它们加在一起。

$\begin{aligned} (3 (1) - 2) + (3 (2) - 2) + (3 (3) - 2) + (3 (4) - 2) + (3 (5) - 2) + (3 (6) - 2) \\ 1 + 4 + 7 + 10 + 13 + 16 \\ = 51 \end{aligned}$
$\begin{align*}& \left(3(1)-2\right)+\left(3(2)-2\right)+\left(3(3)-2\right)+\left(3(4)-2\right)+\left(3(5)-2\right)+\left(3(6)-2\right) \\ & 1+4+7+10+13+16 \\ & =51\end{align*}$

Therefore, 51 deer were reported.
::因此,报告有51头鹿。

Evaluate the following. First without a calculator, then use the calculator to check your result.
::评估以下内容。首先没有计算器, 然后使用计算器检查您的计算结果。

Example 2
::例2

Find the sum: $\begin{align*}\sum\limits_{n=3}^7 2(n-3)\end{align*}$ .
::查找总和:n=372(n-3)。

$\begin{aligned} \sum_{n = 3}^{7} 2 (n - 3) & = 2 (3 - 3) + 2 (4 - 3) + 2 (5 - 3) + 2 (6 - 3) + 2 (7 - 3) \\ = 2 (0) + 2 (1) + 2 (2) + 2 (3) + 2 (4) \\ = 0 + 2 + 4 + 6 + 8 \\ = 20 \end{aligned}$
$\begin{align*}\sum\limits_{n=3}^7 2(n-3) &=2(3-3)+2(4-3)+2(5-3)+2(6-3)+2(7-3) \\ &=2(0)+2(1)+2(2)+2(3)+2(4) \\ &=0+2+4+6+8 \\ &=20\end{align*}$
::=372=372-3=2(3-3)+3(3-3)+2(4-3)+2(5-3)+2(3)+2(5-3)+2(6-3)+2(3)+2(3)+2(2)(2)(7-3)=2(0)+2(2(2)+2(2)+2(3)+2(4)=0+2+4+6+8=20

$\begin{align*}sum(seq(2(x-3),x,3,7)=20\end{align*}$
::苏姆(2x-3,x,3,7)=20

Example 3
::例3

Find the sum: $\begin{align*}\sum\limits_{n=1}^7 \frac{1}{2}n+1\end{align*}$ .
::查找总和: 'n=1712n+1。

$\begin{aligned} \sum_{n = 1}^{7} \frac{1}{2} n + 1 & = \frac{1}{2} (1) + 1 + \frac{1}{2} (2) + 1 + \frac{1}{2} (3) + 1 + \frac{1}{2} (4) + 1 + \frac{1}{2} (5) + 1 + \frac{1}{2} (6) + 1 + \frac{1}{2} (7) + 1 \\ = \frac{1}{2} + 1 + 1 + 1 + \frac{3}{2} + 1 + 2 + 1 + \frac{5}{2} + 1 + 3 + 1 + \frac{7}{2} + 1 \\ = \frac{16}{2} + 13 \\ = 8 + 13 \\ = 21 \end{aligned}$
$\begin{align*}\sum\limits_{n=1}^7 \frac{1}{2}n+1 &=\frac{1}{2}(1)+1+\frac{1}{2}(2)+1+\frac{1}{2}(3)+1+\frac{1}{2}(4)+1+\frac{1}{2}(5)+1+\frac{1}{2}(6)+1+\frac{1}{2}(7)+1 \\ &=\frac {1}{2}+1+1+1+\frac{3}{2}+1+2+1+\frac{5}{2}+1+3+1+\frac{7}{2}+1 \\ &=\frac{16}{2}+13 \\ &=8+13 \\ &=21\end{align*}$
::nn=1712n+1+1=12(1)+1+12(2)+1+12(3)+1+12(4)+1+12(4)+1+12(4)+12(5)+1+12(6)+1+12(7)+1=12+1+1+1+1+1+1+1+1+1+32+1+1+1+1+1+1+52+1+3+1+1+1+1+1+1+1+1+721=162+13=8+13=21

$\begin{align*}sum(seq(1/2x+1,x,1,7)=21\end{align*}$
::等值(1/2x+1,x,1,7)=21

Example 4
::例4

Find the sum: $\begin{align*}\sum\limits_{n=1}^4 3n^2-5\end{align*}$ .

$\begin{aligned} \sum_{n = 1}^{4} 3 n^{2} - 5 & = 3 (1)^{2} - 5 + 3 (2)^{2} - 5 + 3 (3)^{2} - 5 + 3 (4)^{2} - 5 \\ = 3 - 5 + 12 - 5 + 27 - 5 + 48 - 5 \\ = 90 - 20 \\ = 70 \end{aligned}$
$\begin{align*}\sum\limits_{n=1}^4 3n^2-5 &=3(1)^2-5+3(2)^2-5+3(3)^2-5+3(4)^2-5 \\ &=3-5+12-5+27-5+48-5 \\ &=90-20 \\ &=70 \end{align*}$
::查找总和:%n=143n2-5。%n=143n2-5=3(1)2-5+3(2)2-5+3(3)-2-5+3(3)-2-5+3(4)-2-5=3-5+12-5+27-5+48-5=90-20=70

$\begin{align*}sum(seq(3x^2-5,x,1,4)=70\end{align*}$
::和( 3x2 - 5, x, 1, 4) =70

Review
::回顾

Write out the terms and find the sum of the following series.
::写出术语并找到以下系列的总和。

$\begin{align*}\sum\limits_{n=1}^5 2n\end{align*}$
::=152n

$\begin{align*}\sum\limits_{n=5}^8 n+3\end{align*}$
::=58n+3=58n+3

$\begin{align*}\sum\limits_{n=10}^{15} n(n-3)\end{align*}$
::n=1015n(n-3)

$\begin{align*}\sum\limits_{n=3}^7 \frac{n(n-1)}{2}\end{align*}$
::n=37n(n-1)2

$\begin{align*}\sum\limits_{n=1}^6 2^{n-1}+3\end{align*}$
::=162n-1+3 = 162n-1+3

Use your calculator to find the following sums.
::使用您的计算器查找以下金额。

$\begin{align*}\sum\limits_{n=10}^{15} \frac{1}{2}n+3\end{align*}$
::=101512n+3

$\begin{align*}\sum\limits_{n=0}^{50} n-25\end{align*}$
::=050n-25 = 050n-25

$\begin{align*}\sum\limits_{n=1}^5 \left(\frac{1}{2}\right)^{n-5}\end{align*}$
::Nn=15( 12) - 5

$\begin{align*}\sum\limits_{n=5}^{12} \frac {n(2n+1)}{2}\end{align*}$
::n=512n(2n+1)2

$\begin{align*}\sum\limits_{n-1}^{100} \frac{1}{2}n\end{align*}$
:n-10012n)n-110012n

$\begin{align*}\sum\limits_{n=1}^{200} n\end{align*}$
::=1200n =1200n

In problems 12-14, write out the terms in each of the series and find the sums.
::在问题12-14中,在每个系列中写出术语并找出金额。

.

$\begin{align*}\sum\limits_{n=1}^5 2n+3\end{align*}$
::=152n+3 =152n+3

$\begin{align*}3(5)+\sum\limits_{n=1}^5 2n\end{align*}$
::3(5)n=152n

::=152n+3 3(5)n=152n

.

$\begin{align*}\sum\limits_{n=1}^5 \frac{n(n+1)}{2}\end{align*}$
::n=15n(n+1)2

$\begin{align*}\frac{1}{2}\sum\limits_{n=1}^5 n(n+1)\end{align*}$
::12n=15n(n+1)

::.n=15n(n+1)2 12n=15n(n+1)

.

$\begin{align*}\sum\limits_{n=1}^5 4x^3\end{align*}$
::=154x3 = 154x3

$\begin{align*}4\sum\limits_{n=1}^5 x^3\end{align*}$
::4n=15x3

::.n=154x3 4=15x3

Explain why each pair in questions 12-14 has the same sum.
::解释为什么问题12-14中的每对配对都有相同数额。

What is another way to explain the series in #11?
::解释11号系列的另一种方式是什么?

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

序列和总和符号符号

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

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)

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

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)