13.5 再基因系列

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
• 选择节亚理学系列

  折叠 展开

  亚理学系列

  播放段落

  Introduction
  ::导言

  播放段落

  Imagine you visit the Grand Canyon and drop a rock off the edge of a cliff. The distance the rock falls is 16 feet the 1st second, 48 feet the next second, 80 feet the 3rd second, and so on. The distance the rock falls can be modeled with an arithmetic series.  While it is possible to add arithmetic series one term at a time, it is not feasible or efficient when there are a large number of terms. 
  ::想象一下您访问大峡谷,然后从悬崖边缘扔下一块岩石。 岩石坠落的距离是第1秒16英尺, 第2秒48英尺, 第3秒80英尺, 等等。 岩石坠落的距离可以用一个算术序列来模拟。 虽然可以一次添加一个术语的算术序列, 但当有大量术语时, 它既不可行也不有效 。

  

  播放段落

  How can we use arithmetic series to devise a clever way to add up all the whole numbers between 1 and 100?
  ::我们怎样才能用算术序列 来设计一个聪明的方法 来将1到100之间的所有数字加起来?

  播放段落

  Arithmetic Series
  ::亚理学系列

  播放段落

  The key to adding up a finite arithmetic series is to pair up the 1st term with the last term, the 2nd term with the 2nd-to-last term, and so on. The sum of each pair will be equal. Consider a arithmetic  series:
  ::将有限算术序列相加的关键是将第一个学期与最后一个学期相配,第二个学期与第二个学期与最后一个学期相配,等等。每对的和将相等。考虑一个算术序列:

  播放段落

  $$\begin{array}{r}\underset{i=1}{\overset{n}{\sum }}{a}_i=a_1+a_2+a_3+\cdots a_n.\end{array}$$

  When we pair the 1st and the last terms and note that  $\begin{array}{r}{a}_n=a_1+(n-1)k\end{array}$ , the sum is
  ::“i= 1nai=a1+a2+a3+a3+a3+_aan.当我们对齐第一和最后一个条件并注意 a=a1+(n-1)k时,总和是

  播放段落

  $$\begin{array}{r}{a}_1+a_n=a_1+a_1+(n-1)k=2a_1+(n-1)k.\end{array}$$

  
  ::a1+an=a1+a1+(n-1)k=2a1+(n-1)k。

  播放段落

  When we pair up the 2nd and the 2nd-to-last terms, we obtain the same sum:
  ::当我们配对第二和最后第二至最后条件时, 我们得到同样的金额:

  播放段落

  $$\begin{array}{r}{a}_2+a_{n-1}=(a_1+k)+(a_1+(n-2)k)=2a_1+(n-1)k.\end{array}$$

  
  ::a2+an-1=(a1+k)+(a1+(n-2k)=2a1+(n-1)k。

  播放段落

  The next logical question to ask is: How many pairs are there? If there are  $\begin{array}{r}n\end{array}$ terms total, then there are exactly  $\begin{array}{r}\frac{n}{2}\end{array}$ pairs. If  $\begin{array}{r}n\end{array}$ happens to be even, then every term will have a partner, and  $\begin{array}{r}\frac{n}{2}\end{array}$ will be a whole number. If  $\begin{array}{r}n\end{array}$ happens to be odd, then every term but the middle one will have a partner, and  $\begin{array}{r}\frac{n}{2}\end{array}$ will include a  $\begin{array}{r}\frac{1}{2}\end{array}$ pair that represents the middle term with no partner.
  ::要问的下一个逻辑问题是:有多少对?如果有 n 条件,那么有确切的 n2 对。如果有 n 条件,那么每个条件都会有一个伴侣,n2 将是一个完整的数字。如果有 很奇怪,那么每个条件但中间条件都会有一个伴侣,n2 将包括一个12 对,代表中期条件,没有伴侣。

  播放段落

  G eneral Formula for Arithmetic Series
  ::亚性系列一般公式

  $$\begin{array}{r}\underset{i=1}{\overset{n}{\sum }}{a}_i=\frac{n}{2}\left(2a_1+(n-1)k\right),\end{array}$$

  where  $\begin{array}{r}k\end{array}$ is the common difference for the terms in the series.

  播放段落

  The following video  introduces arithmetic series:
  ::以下录像介绍算术系列:

   

  播放段落

  Play, Learn, and Explore with Arithmetic Series: 
  ::与自学系列一起玩、学习和探索:

  播放段落

  Examples
  ::实例

  播放段落

  Example 1
  ::例1

  播放段落

  Add up the numbers between 1 and 10 (inclusive) in two ways.
  ::以两种方式将1至10(包括)之间的数字相加。

  播放段落

  Solution:  
  ::解决方案 :

  播放段落

  One way to add up lists of numbers is to pair them up for easier mental arithmetic.
  ::将数字列表加在一起的方法之一是将它们配对,以便更容易地进行心理算术。

  $$\begin{array}{rl}1+2+3+4+5+6+7+8+9+10& =3+7+11+15+19\\ & =10+26+19\\ & =36+19\\ & =55\end{array}$$

  播放段落

  Another way is to note that $\begin{array}{r}1+10=2+9=3+8=4+7=5+6=11\end{array}$ . There are 5 pairs of 11, which total 55.
  ::另一种方式是注意到 1+10=2+9=3+8=4+7=5+6=11。 共有5对11,共55对。

  播放段落

  Example 2
  ::例2

  播放段落

  Evaluate the following sum:
  ::评估以下金额:

  播放段落

  $$\begin{array}{r}\underset{k=0}{\overset{5}{\sum }}5k-2.\end{array}$$

  
  ::k=055k-2。

  播放段落

  Solution:
  ::解决方案 :

  播放段落

  The 1st term is -2, the last term is 23, and there are 6 terms making 3 pairs. A common mistake is to forget to count the 0 index .
  ::第一个学期是 - 2,最后一个学期是 23, 还有六个学期是三对。 一个常见的错误是忘记计算 0 指数 。

  播放段落

  $$\begin{array}{r}\underset{k=0}{\overset{5}{\sum }}5k-2=\frac{6}{2}\cdot (-2+23)=3\cdot 21=63\end{array}$$

  
  ::k=055k-2=62(-2+23)=321=63

  播放段落

  Example 3
  ::例3

  播放段落

  Does the technique for finding the sum of an arithmetic series work for finding the sum of a geometric series?
  ::查找算术序列总和的技术是否用于查找几何序列总和?

  $$\begin{array}{r}\frac{1}{8}+\frac{1}{2}+2+8+32\end{array}$$

  播放段落

  Solution:
  ::解决方案 :

  播放段落

  The real sum is  $\begin{array}{r}\frac{341}{8}.\end{array}$
  ::实际金额是3418美元。

  播放段落

  When you try to use the technique used for , you get  $\begin{array}{r}3\left(\frac{1}{8}+32\right)=\frac{771}{8}.\end{array}$
  ::尝试使用 3 (18+32)=7718 的技术时,可获得 3 (18+32)=7718。

  播放段落

  It is important to know that geometric series have their own method for summing. The method learned in this concept works only for arithmetic series.
  ::重要的是,要知道几何序列有其自身的总结方法。 这一概念中学习的方法只适用于算术序列。

  播放段落

  Example 4
  ::例4

  播放段落

  Recall the problem from the Introduction: H ow can we use arithmetic series to devise a clever way to add up all the whole numbers between 1 and 100?
  ::回顾导言中的问题:我们如何利用算术序列来设计一个聪明的方法来将1至100之间的全部数字相加?

  播放段落

  Solution:
  ::解决方案 :

  播放段落

  Carl Friedrich Gauss was a mathematician who lived hundreds of years ago. According to an anecdote about him, as a young boy in school h e misbehaved  and his teacher asked him to add up all the numbers between 1 and 100. Within a few seconds, Gauss stated 5,050.
  ::卡尔·弗里德里希·高斯(Carl Friedrich Gaus)是一位数以百计的数学家,几百年前就曾住过他。 根据关于他的一个故事,当他上学时,他行为不端,老师要求他把数字加起来,在1到100之间。 高斯在几秒钟内说5 050人。

  播放段落

  You should notice that  $\begin{array}{r}1+100=2+99=\cdots =101,\end{array}$ and that there are exactly 50 pairs that sum to be 101. $\begin{array}{r}50\cdot 101=5,050\end{array}$ .
  ::您应该注意到 1+100=2+99101, 并且有50对, 总共是101 50101=5 050。

  播放段落

  Example 5
  ::例5

  播放段落

  Sum the 1st 15 terms of the following arithmetic :
  ::下列算术的第15项和第15项条件相加:

  $$\begin{array}{r}-1,\frac{2}{3},\frac{7}{3},4,\frac{17}{3}\dots \end{array}$$

  播放段落

  Solution:
  ::解决方案 :

  播放段落

  The initial term is -1, and the common difference is $\begin{array}{r}\frac{5}{3}\end{array}$ .
  ::初始任期为-1,共同差为53。

  播放段落

  $$\begin{array}{rl}\underset{i=1}{\overset{n}{\sum }}{a}_i& =\frac{n}{2}(2a_1+(n-1)k)\\ & =\frac{15}{2}\left(2(-1)+(15-1)\frac{5}{3}\right)\\ & =\frac{15}{2}\left(-2+14\cdot \frac{5}{3}\right)\\ & =160\end{array}$$

  
  ::i=1nai=n2( 2a1+(n-1))=152(2(-1)+(15-1)53=152(-2+1453)=160

  播放段落

  Example 6
  ::例6

  播放段落

  Sum the 1st 100 terms of the following arithmetic sequence:
  ::下列算术序列的第100项条件合计:

  $$\begin{array}{r}-7,-4,-1,2,5,8,\dots \end{array}$$

  播放段落

  Solution:
  ::解决方案 :

  播放段落

  The initial term is -7, and the common difference is 3.
  ::初始任期为7年,共同差为3年。

  播放段落

  $$\begin{array}{rl}\underset{i=1}{\overset{n}{\sum }}{a}_i& =\frac{n}{2}(2a_1+(n-1)k)\\ & =\frac{100}{2}(2(-7)+(100-1)3)\\ & =14150\end{array}$$

  
  ::*i=1nai=n2( 2a1+(n-1)k)=1002(2(-7)+(100-1)3=14150

  播放段落

  Example 7
  ::例7

  播放段落

  Evaluate the following sum:
  ::评估以下金额:

  播放段落

  $$\begin{array}{r}\underset{i=0}{\overset{500}{\sum }}2i-312.\end{array}$$

  
  ::i=050002i-312。

  播放段落

  Solution:
  ::解决方案 :

  播放段落

  The initial term is -312, and the common difference is 2.
  ::最初的学期是312个,共同差数是2个。

  播放段落

  $$\begin{array}{rl}\underset{i=0}{\overset{500}{\sum }}2i-312& =\frac{501}{2}(2(-312)+(501-1)2)\\ & =94188\end{array}$$

  
  ::i=050002i-312=5012(2-312)+(501-1)2=94188

  播放段落

  Summary
  ::摘要

  播放段落
  • An arithmetic series is a sum of numbers whose consecutive terms form an arithmetic sequence.
    ::算术序列是数字的总和,其连续术语构成算术序列。
  播放段落
  • The general form for an arithmetic series can be represented by  $\begin{array}{r}\underset{i=1}{\overset{n}{\sum }}{a}_i=\frac{n}{2}(2a_1+(n-1)k),\end{array}$ where k is the common difference for the terms in the series.
    ::算术序列的一般格式可以用“i=1nai=n2(2a1+(n-1)k)表示,K是该序列术语的常见差异。

  播放段落

  Review
  ::回顾

  播放段落

  1. Sum the 1st 24 terms of the sequence $\begin{array}{r}1,5,9,13,\dots \end{array}$
  ::1. 1,5,9,13,...

  播放段落

  2. Sum the 1st 102 terms of the sequence $\begin{array}{r}7,9,11,13,\dots \end{array}$
  ::2. 序号7,9,11,13,...

  播放段落

  3. Sum the 1st 85 terms of the sequence $\begin{array}{r}-3,-1,1,3,\dots \end{array}$
  ::3. 序列顺序的第85个条件 - 3, - 1,1,1,3,...

  播放段落

  4. Sum the 1st 97 terms of the sequence $\begin{array}{r}\frac{1}{3},\frac{2}{3},1,\frac{4}{3},\dots \end{array}$
  ::4. 顺序13,23,1,43...

  播放段落

  5. Sum the 1st 56 terms of the sequence  $\begin{array}{r}-\frac{2}{3},\frac{1}{3},\frac{4}{3},\dots \end{array}$
  ::5. 顺序的第56个条件 - 23,13,43,...

  播放段落

  6. Sum the 1st 91 terms of the sequence $\begin{array}{r}-8,-4,0,4,\dots \end{array}$
  ::6. 序号8 - 4,0,4,...

  播放段落

  Evaluate the following sums:
  ::评估以下金额:

  播放段落

  7. $\begin{array}{r}\underset{i=0}{\overset{300}{\sum }}3i+18\end{array}$
  ::7. i=03003i+18

  播放段落

  8. $\begin{array}{r}\underset{i=0}{\overset{215}{\sum }}5i+1\end{array}$
  ::8. i=02155i+1

  播放段落

  9. $\begin{array}{r}\underset{i=0}{\overset{100}{\sum }}i-15\end{array}$
  ::9. i=0100i-15

  播放段落

  10. $\begin{array}{r}\underset{i=0}{\overset{85}{\sum }}-13i+1\end{array}$
  ::10. i=0.85-13i+1

  播放段落

  11. $\begin{array}{r}\underset{i=0}{\overset{212}{\sum }}-2i+6\end{array}$
  ::11. i=0212-2i+6

  播放段落

  12. $\begin{array}{r}\underset{i=0}{\overset{54}{\sum }}6i-9\end{array}$
  ::12. i=0546i-9

  播放段落

  13. $\begin{array}{r}\underset{i=0}{\overset{167}{\sum }}-5i+3\end{array}$
  ::13. i=0167-5i+3

  播放段落

  14. $\begin{array}{r}\underset{i=0}{\overset{341}{\sum }}6i+102\end{array}$
  ::14. i=03416i+102

  播放段落

  15. $\begin{array}{r}\underset{i=0}{\overset{452}{\sum }}-7i-\frac{5}{2}\end{array}$
  ::15. i=0452-7i-52

  播放段落

  Review (Answers )
  ::回顾(答复)

  播放段落

  Please see the Appendix.
  ::请参看附录。