7.3 Sigma的标记和属性总和

Section outline

• Select section General

  Collapse Expand

  General

  Collapse all Expand all

  • Select activity 新闻通告

    

    新闻通告 Forum
• Select section Sigma的计算符号和属性

  Collapse Expand

  Sigma的计算符号和属性

  播放段落

  Sayber and Tuscany sell popsicles during the summer for pocket money. One particular weekend, they purchased a package of 30 popsicles from the store.
  ::Sayber和Tuscany在夏天卖冰棒,买零用钱。一个特殊的周末,他们从商店买了30个冰棒。

  播放段落

  Usually they just offer the popsicles for free, 1 per customer, and accept tips. This time, Tuscany wonders if they would make more money by charging $0.50 per popsicle. At the same time, Sayber wonders if he might be able to increase his tips by encouraging customers to "outbid" each other. The two children decide to each take 15 popsicles and see who makes the most.
  ::通常情况下,他们只免费提供冰棒,每个顾客1个,接受小费。 这次,托斯卡尼怀疑他们能否通过每支冰棒收费0.50美元来赚更多的钱。 与此同时,赛伯怀疑他是否能够通过鼓励顾客“出价 ” , 来增加小费。 两个孩子决定每人拿15个冰棒,看看谁赚得最多。

  播放段落

  How could you calculate how much money each of them makes, assuming Sayber gets a $0.10 tip from the first customer, and is able to convince each successive customer to double the previous person's tip?
  ::假设Sayber从第一个客户那里得到0.10美元的小费, 并且能够说服每个连续的客户将前一个人的小费翻一番?

  播放段落

  Sum Notation and Properties of Sigma
  ::Sigma的计算符号和属性

  播放段落

  Consider for example a defined by a _n = 3 n . If we write out the sum of the first 4 terms, we have 3 + 6 + 9 + 12 = 30. But what if we want to write out terms for a larger sum?
  ::例如,考虑一个以 = 3n 定义的定义。如果我们写出前4个条件的总和,我们就有 3 + 6 + 9 + 12 = 30。但如果我们想写一个更大的条件呢?

  播放段落

  Summation notation is a method of writing sums in a succinct form. To write the sum 3 + 6 + 9 + 12 = 30 , we use the Greek letter Sigma, as follows:
  ::计算符号是一种以简洁形式写出金额的方法。要写出3+6+9+9+12=30,我们使用希腊字母Sigma如下:

  			$\begin{align*}\sum_{n=1}^4 3n\end{align*}$

  播放段落

  The expression 3 n is called the summand , the 1 and the 4 are referred to as the limits of the summation, and the n is called the index of the sum. Here we have used a “sigma” to write a sum. We can also read a sigma, and determine the sum. For example, we can read the above sigma notation as “find the sum of the first four terms of the series, where the n ^th term is 3 n .” We always read the limits from the bottom to the top. The bottom number tells you which term to start with, and the top limit tells you which term is the final term to add. We could then write out the sigma above as:
  ::3n 表达式称为 summand, 1 和 4 被称为 和数的限度, n 被称为 和 和 的 索引 。 在这里, 我们用“ igma” 来写一个数额 。 我们也可以读一个 sigma , 并且确定 和 。 例如, 我们可以读上面的 sigma 表示“ 找到该系列前四个条件的总和, 第 n 个术语是 3n ” 。 我们总是从底到顶读这些界限 。 最底的数字表示要从下到顶开始哪个词, 顶数则表示要添加哪个词 。 然后我们可以写上面的 sigma 表示 :

  		$\begin{align*}\sum_{n=1}^4 3n\end{align*}$	= 3(1) + 3(2) + 3(3) + 3(4)
  			= 3 + 6 + 9 + 12 = 30

  播放段落

  In general, we can either rewrite a given series in sigma notation, or we can read sigma notation in order to find the value of the sum.
  ::一般而言,我们可以改写某个序列的污名,或者读出污名,以找到金额的价值。

  播放段落

  Properties of Sigma
  ::Sigma 属性属性

  播放段落

  Notice that we can write the sum 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 19 + 20 as 2(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10). Therefore $\begin{align*}\sum_{n=1}^{10} 2n = 2 \sum_{n=1}^{10} n\end{align*}$ . In general, we can factor a coefficient out of a sum:
  ::请注意,我们可以将2+4+6+8+8+10+12+14+14+16+19+20作为2(1+2+2+3+4+5+5+6+7+8+9+10写入2+4+6+6+6+8+9+10。 因此,n=1102n=2n=110n。

  		$\begin{align*}\sum_{n=1}^{k}ca_n \end{align*}$	$\begin{align*} = c\sum_{n=1}^{k} a_n \end{align*}$

  播放段落

  Examples
  ::实例

  播放段落

  Example 1
  ::例1

  播放段落

  Earlier, you were asked to  express the incomes  of Sayber and Tuscany, who are selling popsicles.
  ::早些时候,你被要求 表达Sayber和托斯卡纳的收入, 他们卖冰棒。

  播放段落

  Tuscany's income can be expressed as: $\begin{align*}\sum_{n=1}^{10} .5 \to .5 \cdot 10 = \$5.00\end{align*}$ .
  ::托斯卡纳的收入可以表示为:"n=110.55.510=5.00美元。

  播放段落

  Sayber's income can be expressed as: $\begin{align*}\sum_{n=1}^{10}.10 \cdot 2^n \to .10 (2^0) + .10 (2^1) + .10 (2^2)... +.10 (2^{10}) \to \$102.30\end{align*}$ .
  ::Sayber的收入可表述为: 'n=110.10_10_2n%10(20)+10(21)+10(22)...+10(210)_______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

  播放段落

  Sayber must be a heck of a salesman to get that last tip!
  ::赛伯一定是个推销员 才能拿到最后那笔小费!

  播放段落

  Example 2
  ::例2

  播放段落

  Write the sum using sigma notation: 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 19 + 20.
  ::使用西格玛符号: 2+ 4+ 6+ 8+ 10+ 12+ 14+ 16+ 19+20来写总和: 2+ 4+ 6+ 8+ 10+ 12+ 14+ 16+ 19+20。

  播放段落

  $\begin{align*}\sum_{n=1}^{10} 2n\end{align*}$
  ::=1102n

  播放段落

  Every term is a multiple of 2. The first term is 2 × 1, the second term is 2 × 2 , and so on. So the summand of the sigma is 2 n . There are 10 terms in the sum. Therefore the limits of the sum are 1 and 10.
  ::每个学期都是2的倍数,第一个学期是2×1,第二个学期是2×2,等等。因此,Sigma的总和是2n。在学期中共有10个学期。因此,学期的限额是1和10。

  播放段落

  Example 3
  ::例3

  播放段落

  Write out the terms of $\begin{align*}\sum_{n=2}^{5}(n+7) \end{align*}$ and evaluate the sum.
  ::写下n=25(n+7)的条件并评估总额。

  播放段落

  Break the sum into two different sums. The sum is 37.
  ::将总和折成两笔不同的金额,数额为37。

  		$\begin{align*}\sum_{n=2}^{5}(n+7) \end{align*}$	=(2 + 7) + (3 + 7) + (4 + 7) + (5 + 7)
  			=2 + 3 + 4 + 5 + 7 + 7 + 7 + 7
  			=2 + 3 + 4 + 5 + 7 × 4 = 14 + 28 = 42

  播放段落

  Notice we could have written $\begin{align*}\sum_{n=2}^{5}(n+7) \end{align*}$ as $\begin{align*}\sum_{n=2}^{5}n+ \sum_{n=2}^{5}7\end{align*}$ . Also, the second sum does not depend on the index of the sum (i.e. it stays at 7 regardless of the index), only that there are 4 terms to add together. Seeing this can make a sum easier to evaluate.
  ::我们本可以将n=25(n+7)写成n=25n@n=257。此外,第二笔总额并不取决于总和的指数(即,不论指数为何,它仍维持在7),只有4个条件可以合并。看到这一点,就可以更容易地评估总和。

  播放段落

  Example 4
  ::例4

  播放段落

  Write out the terms of $\begin{align*}\sum_{n=0}^{4}32\left ( \frac{1}{4} \right )^n\end{align*}$ and evaluate the sum.
  ::写下"n=0432(14)n"的条件并评估金额。

  播放段落

  In general, we can write a sum as a sum of sums: $\begin{align*}\sum_{n=1}^{n}(a_n+b_n) =\sum_{n=1}^{n}(a_n)+\sum_{n=1}^{n}(b_n)\end{align*}$ .
  ::一般而言,我们可以以总和来写一个总和:n=1n(an+bn)\n=1n(an)n=1n(an)n=1n(bn)。

  播放段落

  The sum is $\begin{align*}42\frac{5}{8}\end{align*}$ .
  ::金额是458。

  		$\begin{align*}\sum_{i=0}^{4}32 \left ( \frac{1}{4} \right)^n\end{align*}$	$\begin{align*}= 32 \left ( \frac{1}{4} \right )^0+32 \left ( \frac{1}{4} \right )^1+32 \left ( \frac{1}{4} \right )^2+32 \left ( \frac{1}{4} \right )^3+32 \left ( \frac{1}{4} \right )^4\end{align*}$
  			$\begin{align*}= 32 \cdot 1+32\cdot \frac{1}{4}+32 \cdot \frac{1}{16}+ 32\cdot \frac{1}{64}+32 \cdot \frac{1}{256}\end{align*}$
  			$\begin{align*}= 32+8+2+\frac{1}{2}+\frac{1}{8}=42 \frac{5}{8}\end{align*}$

  播放段落

  Example 5
  ::例5

  播放段落

  Write the sum using sigma notation: 2 + 6 + 18 + 54.
  ::2+6+18+54。

  播放段落

  $\begin{align*}\therefore \sum_{n = 1}^4 2 \left(3^{n - 1} \right)\end{align*}$ or $\begin{align*}\sum_{n = 0}^3 2 \left(3^n \right)\end{align*}$
  ::=142(3n-1) =032(3n)

  播放段落

  Example 6
  ::例6

  播放段落

  Find the series of numbers and the total of those numbers of the arithmetic series represented by the following sigma notation: $\begin{align*}\sum_{n=8}^{14} {3 + \frac{3}{4}(n-1)}\end{align*}$ .
  ::查找以下列西格玛符号表示的算术序列的数字序列和总数:n=8143+34(n-1))。

  播放段落

  To find the series of numbers, we plug in all the numbers between 8 and 14 for :
  ::要找到数字序列, 我们将 8 和 14 之间的所有数字插入到 :

  $\begin{align*} 3 + \frac{3}{4}((8)-1) = \frac{33}{4}\end{align*}$

  $\begin{align*} 3 + \frac{3}{4}((9)-1) = 9\end{align*}$

  播放段落

  We would continue to do this through the number 14. This leaves us with the following series: $\begin{align*} \frac{33}{4} + 9 + \frac{39}{4} + \frac{21}{2} + \frac{45}{4} + 12 + \frac{51}{4}\end{align*}$ .
  ::我们将继续通过14个数字这样做,这使我们只剩下以下系列:334+9+394+212+454+12+514。

  播放段落

  This is fine, if we are just looking for the individual numbers in the sequence, however when asked to evaluate sumations we are being asked to add all the numbers of the series together. It took plenty long enough to find each number, and now we must add them all together. Fortunately there is a formula, not only eliminating our need to find each number, but that lets us also add them all together and arrive at our answer or sum.
  ::这很好,如果我们只是按顺序来寻找单个数字,但是当我们被要求评估总和时,我们被要求将序列中的所有数字加在一起。 需要足够长的时间才能找到每个数字,现在我们必须把它们加在一起。 幸运的是,有一个公式,不仅消除了我们找到每个数字的需要,而且让我们把它们加在一起,并找到答案或总和。

  播放段落

  The formula is $\begin{align*}\frac{k}{2}(a_1 + a_{n})\end{align*}$ works like this:
  ::公式为 k2(a1+an) , 工作方式像这样 :

  播放段落

  We plug in n = 8 to get $\begin{align*}a_8 = \frac{33}{4}\end{align*}$
  ::我们插入 n= 8 以获得 a8= 334

  播放段落

  Then we plug in n = 14, to get $\begin{align*}a_{14} = \frac{51}{4}\end{align*}$ .
  ::然后我们插入n=14,得到a14=514。

  播放段落

  Identify the number of terms $\begin{align*}(k) = 14 - 8 + 1\end{align*}$ so we use $\begin{align*}k=7\end{align*}$ in the formula below.
  ::标明术语(k)=14-8+1, 所以我们在下面的公式中使用 k=7 。

  播放段落

  Now we can use the formula:  $\begin{align*}\frac{k}{2}(a_8 + a_{14})\end{align*}$ and we get:  $\begin{align*} \frac{7}{2}(\frac{33}{4} + \frac{51}{4})=\frac{147}{2}\end{align*}$ .
  ::现在我们可以使用公式 k2(a8+a14) , 我们得到: 72( 334+514)=1472。

  播放段落

  Example 7
  ::例7

  播放段落

  Find the sum of all the numbers in the arithmetic sequence  $\begin{align*}\sum_{n=8}^{28} {9 - 2(n-1)}\end{align*}$ .
  ::查找计算序列 'n=8289-2(n-1)' 中所有数字的总和。

  播放段落

  Use the formula $\begin{align*}\frac{k}{2}(a_1 + a_{n})\end{align*}$ :
  ::使用公式 k2(a1+an) :

  播放段落

  Substitute n = 8 to get $\begin{align*}a_8 = -5\end{align*}$
  ::替代 n = 8 获得 a8 @% 5

  播放段落

  Substitute n = 28 to get $\begin{align*}a_28 = -45\end{align*}$
  ::替代n=28 获得a28=45

  播放段落

  Identify the number of terms as: $\begin{align*}k = 28-8+1\end{align*}$ so we use $\begin{align*}k = 21\end{align*}$ in our formula below:
  ::确定术语数为: k=28-8+1, 所以我们在以下公式中使用 k=21 :

  播放段落

  Now we can use the formula:  $\begin{align*}\frac{k}{2}(a_8 + a_{14})\end{align*}$ and our answer is $\begin{align*}-525\end{align*}$ .
  ::现在我们可以使用公式 k2(a8+a14) , 我们的答案是 -525 。

  播放段落

  Example 8
  ::例8

  播放段落

  Find the series of numbers and the total of those numbers of the geometric series represented by the sigma notation $\begin{align*}\sum_{n=1}^8 7(\frac{-1}{2})^{n-1}\end{align*}$ .
  ::查找以 sigma name = 187(- 12)n-1 表示的几何序列的数号和总数。

  播放段落

  To identify and sum the terms in the geometric series $\begin{align*}\sum_{n=1}^8 7(\frac{-1}{2})^{n-1}\end{align*}$ :
  ::确定和总和几何序列中的术语=187(-12)n-1:

  播放段落

  Find the sequence of numbers the same way as we did in Example 6 , by plugging in the indicated numbers 1-8 for
  ::通过插入的标记编号 1-8, 找到与例6相同的数字序列

  播放段落

  $\begin{align*}\sum_{n=1}^8 7(\frac{-1}{2})^{((1)-1)}\end{align*}$ which gives us: $\begin{align*} 7 \end{align*}$
  ::=187(- 12)((1)-1), 给了我们: 7

  播放段落

  We do this for the remaining numbers and our sequence looks like this: $\begin{align*}7 + \frac{-7}{2} +\frac{7}{4} + \frac{-7}{8} +\frac{7}{16} +\frac{-7}{32} +\frac{7}{64} + \frac{-7}{128}\end{align*}$
  ::我们这样做是为了剩余数字,我们的顺序看起来是: 772+7478+716732+7647128。

  播放段落

  As in prior examples, we are not just being asked to find the numbers, but add them all together. Again, we do not want to have to take the time to find all the numbers and add them all together. Once again we find ourselves lucky, there is a formula!
  ::和前面的例子一样,我们不仅被要求找到数字,而且被要求把它们加在一起。 同样,我们不想花时间来找到所有数字并把它们加在一起。 我们再次发现幸运的是,我们有一个公式!

  播放段落

  The formula $\begin{align*}a_1(\frac{1-r^k}{1-r})\end{align*}$ works like this:
  ::公式a1 (1- rk1- r) 是这样工作的 :

  播放段落

  We plug n=1 into $\begin{align*}a_1\end{align*}$ which gives us 7.
  ::我们将 n=1 插入给7 的 a1 。

  播放段落

  $\begin{align*}k = 8\end{align*}$ , and $\begin{align*}r=\frac{-1}{2}\end{align*}$
  ::k=8, r12

  播放段落

  Substituting our numbers results: $\begin{align*}7(\frac{1-\frac{-1}{2}^8}{1-\frac{-1}{2}})\end{align*}$
  ::替代我们的数字结果:7(1128112)

  播放段落

  Which gives us $\begin{align*}\frac{595}{128}\end{align*}$ .
  ::也就是说我们有595128人

  播放段落

  Example 9
  ::例9

  播放段落

  Find the sum of the terms in: $\begin{align*}\sum_{n=1}^{11} 9(4)^{n-1}\end{align*}$ .
  ::查找条件的总和为:n=1119(4)n-1。

  播放段落

  Let's use the formula for geometric series again:
  ::让我们再次使用几何序列的公式:

  播放段落

  Identify the terms: $\begin{align*}a_1 = 9\end{align*}$ , $\begin{align*} k = 11\end{align*}$ and $\begin{align*} r = 4\end{align*}$
  ::确定术语:a1=9,k=11和r=4

  播放段落

  Substituting into our formula, we have: $\begin{align*}9(\frac{1-4^{11}}{1-4}) = 12,582,909\end{align*}$ .
  ::作为我们公式的替代,我们有:9(1-4111-4)=12,582,909。

  播放段落

  Review
  ::回顾

  播放段落

  Express the sum using sigma notation:
  ::使用污名符号表示的总和:

  1. $\begin{align*}1 + 3 + 5 + 7 + 9\end{align*}$
  2. $\begin{align*}1 + \frac{1} {2} + \frac{1} {3} + \frac{1} {4} + ... + \frac{1} {10}\end{align*}$

  播放段落

  Find the series of numbers indicated by the sigma notation:
  ::查找刻度符号表示的序列数字 :

  播放段落
  3. $\begin{align*}\sum_{n=0}^{14} -2 -\frac{10}{3}(n-1)\end{align*}$
     ::=014-2-103(n-1)
  播放段落
  4. $\begin{align*}\sum_{n=-8}^{14} -7 -3(n-1)\end{align*}$
     ::N814-7-3(n-1)

  播放段落

  Evaluate:
  ::评价:

  播放段落
  5. $\begin{align*}\sum_{n=-10}^{5} 7 -\frac{4}{3}(n-1)\end{align*}$
     ::1057-43(n-1)
  播放段落
  6. $\begin{align*}\sum_{n=-3}^{3} 3 -\frac{1}{3}(n-1)\end{align*}$
     ::333-13(n-1)
  播放段落
  7. $\begin{align*}\sum_{n=-5}^{1} -6 +\frac{4}{3}(n-1)\end{align*}$
     ::51-6+43(n-1)

  播放段落

  Find the series of numbers indicated by the sigma notation:
  ::查找刻度符号表示的序列数字 :

  播放段落
  8. $\begin{align*}\sum_{n=1}^{2} 3(-\frac{1}{2})^{n-1}\end{align*}$
     ::=123(- 12- 1)
  播放段落
  9. $\begin{align*}\sum_{n=1}^{5} 5(-\frac{4}{3})^{n-1}\end{align*}$
     ::=155(- 43)n-1

  播放段落

  Evaluate:
  ::评价:

  播放段落
  10. $\begin{align*}\sum_{n=1}^{6} 3(\frac{1}{2})^{n-1}\end{align*}$
      ::=163(12)-1
  播放段落
  11. $\begin{align*}\sum_{n=1}^{7} -5(-\frac{1}{2})^{n-1}\end{align*}$
      ::=17-5(- 12- 1)
  播放段落
  12. $\begin{align*}\sum_{n=1}^{11} -7(-\frac{4}{3})^{n-1}\end{align*}$
      ::=111-7(-43)n-1
  播放段落
  13. $\begin{align*}\sum_{n=1}^{6} -7(\frac{1}{4})^{n-1}\end{align*}$
      ::=16 -7(14 -1)
  播放段落
  14. $\begin{align*}\sum_{n=1}^{3} 2(-\frac{3}{2})^{n-1}\end{align*}$
      ::=132(-32-32)n-1

  播放段落

  Review (Answers)
  ::回顾(答复)

  播放段落

  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 。