13.7 确定有限几何序列总和

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• 选择节查找精度几何序列的和

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  查找精度几何序列的和

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  Each person has two parents, and each of those parents has two parents, and so on. How many ancestors does a person have if we go back 10 generations? 
  ::每个人有两个父母,每个父母都有两个父母,等等。 如果我们再过10代,一个人有多少祖先?

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  This is a geometric . We cover how to find the sum of a geometric sequence in this section. 
  ::这是几何。我们研究如何在这个部分找到几何序列的总和。

  

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  Finite Sums of Geometric Sequences
  ::几几何序列的有限总和

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  As with arithmetic series,  there is a specific rule that can be used to find the sum of a geometric sequence algebraically. Let's look at a finite geometric sequence and derive this rule.
  ::与计算序列一样,有一个具体的规则可以用来找到几何序列的代数总和。让我们看看一个有限的几何序列并得出这一规则。

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  For a geometric sequence, we are given the formula  $\begin{align*}a_n=a_1r^{n-1}\end{align*}$ . The sum of the 1st $\begin{align*}n\end{align*}$ terms of a geometric sequence is:

  $$\begin{align*}S_n&=a_1+a_1r+a_1r^2+a_1r^3+ \ldots +a_1r^{n-2}+a_1r^{n-1}\\ \end{align*}$$
  ::对于几何序列,我们被给定公式 an=a1rn-1. 几何序列第一条 n 条件的总和是: Sn=a1+a1r+a1r2+a1r2+a1r3+...+a1r_2+a1rn_1

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  Now, factor out $\begin{align*}a_1\end{align*}$ to get $\begin{align*}a_1(1+r^2+r^3+ \ldots +r^{n-2}+r^{n-1})\end{align*}$ . For what remains in parentheses, if we multiply this by $\begin{align*}(1-r)\end{align*}$ as shown below we can simplify the sum:
  ::现在, 乘以 a1 以获得 a1 (1+r2+r3+...+rn- 2+rn- 1) 。 对于括号中保留的内容, 如果我们将括号乘以 1 乘以 (1-r) 如下所示, 我们可以简化总和 :

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  $$\begin{align*}&(1-r)(1+r+r^2+r^3+ \ldots +r^{n-2}+r^{n-1}) \\ &= (1+r+r^2+r^3+ \ldots +r^{n-2}+r^{n-1}-r-r^2-r^3-r^4- \ldots -r^{n-1}-r^n) \\ &= 1+(r-r)+\left(r^2-r^2\right)+\left(r^3-r^3\right)+ \ldots\ldots +\left(r^{n-2}-r^{n-2}\right)+\left(r^{n-1}-r^{n-1}\right) -r^n \\ &= (1-r)^n\end{align*}$$
  :1-r)(1+r+r+r2+r3+...)+(1+r+r+r2+r_2+r3+...)=(1+r+r+r+r+r2+r2+r3+r3+r3+r3+r3+...)+(1+r+r+r+r+r+r+r2+r2+r2+r3+r1+...)+(r2-r+r2+r2+r_1)-rn=1+(r2-r_r__rn)+(r2+r+(r2-r2+r2+r2+r3+r3+...+...+...+(r-r__2+_2+_2+...)+(rn+(r-r-r_1)-rn=1+(r_rn)+(r_1-r+1+_r+_____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

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  By multiplying the sum by $\begin{align*}1-r,\end{align*}$ we were able to cancel out all the middle terms. However, we cannot just multiply by a factor of $\begin{align*}1-r\end{align*}$ , but we can multiply by 1 or   $\begin{align*}\frac{1-r}{1-r}\end{align*}$ . 
  ::通过将总和乘以1-r,我们得以取消所有中间条件。然而,我们不能仅仅乘以1-r,但我们可以乘以1或1-r1-r。

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  So  the sum of a finite geometric sequence is
  ::所以一个有限的几几何序列的总和是

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  $$\begin{align*}S_n&=a_1+a_1r+a_1r^2+a_1r^3+ \ldots +a_1r^{n-2}+a_1r^{n-1} \\ &=a_1(1+r^2+r^3+ \ldots +r^{n-2}+r^{n-1}) \frac{1-r}{1-r}\\ &=\frac{a_1(1-r^n)}{1-r}\end{align*}$$
  ::Sn=a1+a1+a1r+a1r2+a1r3+...+a1rn-2+a1rn_1=a1(1+r2+r3+...+r2+r3+...+rn2+rn_1)1_r1-r=a1(1-rn)1-r=a1-rr

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     Finite Sum of a Geometric Sequence
  ::几何序列精度总和

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  The sum, $\begin{align*}S_n\end{align*}$ , of the 1st n terms of a geometric sequence is 
  ::几何序列第一 n 条件的和Sn是

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  $$\begin{align*}S_n=\frac{a_1(1-r^n)}{1-r},\end{align*}$$ where  $\begin{align*}n\end{align*}$  is the number of terms,  $\begin{align*}a_1\end{align*}$  is the 1st term , and  $\begin{align*}r\end{align*}$  is the common ratio, and $\begin{align*}r \ne 1\end{align*}$ .
  ::Sn=a1(1-rn)1-r,其中n为术语数,a1为第一学期,r为共同比率,r为第1学期。

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  Example 1
  ::例1

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  Find $\begin{align*}\sum\limits_{n=1}^{10} \frac{1}{32}(\text{-}2)^{n-1}\end{align*}$ .
  ::查找 10n=1132(-2-2)-n-1。

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  Solution: Using the formula, $\begin{align*}a_1=\frac{1}{32}\end{align*}$ , $\begin{align*}r=\text{-}2\end{align*}$ , and $\begin{align*}n=10\end{align*}$ .
  ::解决办法:使用公式,a1=132,r=2,n=10。

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  $$\begin{align*}S_{10}=\frac{\frac{1}{32}(1-(\text{-}2)^{10})}{1-(\text{-}2)} = \frac{\frac{1}{32}(1-1024)}{3} = \text{-}\frac{341}{32}\end{align*}$$
  ::S10=132(1-2)-(-2)10-1-2-2=132(1)-10243=-34132

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  Example 2
  ::例2

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  Evaluate $\begin{align*}\sum\limits_{n=3}^8 2(\text{-}3)^{n-1}\end{align*}$ .
  ::评价8n=32(-3-)n-1。

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  Solution:  Since we are asked to find the sum of the $\begin{align*}3^{rd}\end{align*}$ through $\begin{align*}8^{th}\end{align*}$ terms, we will consider $\begin{align*}a_3\end{align*}$ as the 1st term. The 3rd term is $\begin{align*}a_3=2(\text{-}3)^2=2(9)=18\end{align*}$ . Since we are starting with term three, we will be summing 6 terms, $\begin{align*}a_3+a_4+a_5+a_6+a_7+a_8\end{align*}$ , in total. We can use the rule for the sum of a geometric sequence now, with $\begin{align*}a_1=18\end{align*}$ , $\begin{align*}r=\text{-}3,\end{align*}$ and $\begin{align*}n=6\end{align*}$ to find the sum:
  ::解答:既然我们被要求找到第三至第八学期的总和, 我们将考虑将a3作为第一学期。 第三个学期为a3=2( 3)2=2( 9)=18。 由于我们从三学期开始, 我们总共将总结6学期, a3+a4+a5+a5+a6+a7+a8。 我们现在可以使用规则来计算几何序列的总和, 以a1=18, r=3和n=6来查找总和 :

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  $$\begin{align*}\sum_{n=3}^8 2(\text{-}3)^{n-1}=\frac{18(1-(\text{-}3)^6)}{1-(\text{-}3)}=\text{-}3,\!276\end{align*}$$
  ::8=32(-3)-(三)-(一)-18(一)-((三)-6)-1-(三)=3,276

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  Example 3
  ::例3

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  Each person has two parents, and each of those parents has two parents, and so on. How many ancestors does a person have if we go back 10 generations?
  ::每个人有两个父母,每个父母都有两个父母,等等。 如果我们再过10代,一个人有多少祖先?

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  Solution: We start with 2 parents , so that is  $\begin{align*}a_1\end{align*}$ . The common ratio is also 2, and we are considering 10 generations—that is, $\begin{align*}n=10\end{align*}$ .
  ::解决办法:我们从两个父母开始,所以这就是a1. 共同比率也是2,我们考虑的是10代人,即n=10。

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  $$\begin{align*}S_{10}=\frac{2(1-2^{10})}{1-2}=\frac{\text{-}2,\!046}{\text{-}1}=2,\!046\end{align*}$$
  ::S10=2(1-1-210)1-2=2,046-1=2,046

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  Each person has 2,046 ancestors going back 10 generations.   
  ::每个人有2,046个祖先可追溯到10代。

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  by Mathispower4u introduces geometric series and demonstrates how to use the formula. 
  ::由 Mathispower4u 介绍几何序列并演示如何使用公式。

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  Example 4
  ::例4

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  Find the 1st term and the $\begin{align*}n^{th}\end{align*}$ term rule for a geometric sequence in which the sum of the 1st 5 terms is 242, and the common ratio is 3.
  ::查找一个几何序列的第一个学期和 n 学期规则,其中第一个学期和第一个学期的总和是 242, 共同比率是 3。

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  Solution:   S ubstitute  what we know into the formula for the sum, and solve for the 1st term:
  ::解决方案: 将我们所知道的东西 替换到总和的公式中, 并解决第一个学期:

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  $$\begin{align*}242 &= \frac{a_1(1-3^5)}{1-3} \\ 242 &= \frac{a_1(\text{-}242)}{\text{-}2} \\ 242 &= 121a_1 \\ a_1 &= 2\end{align*}$$
  ::242=a1(1-35)1-3242=a1(-242)-2242=121a1a1=2

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  The 1st term is 2 and $\begin{align*}a_n=2(3)^{n-1}\end{align*}$ .
  ::第一个学期是2年,一个=2(3)n-1。

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  Example 5
  ::例5

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  If the sum of the 1st seven terms in a geometric sequence is $\begin{align*}\frac{215}{8}\end{align*}$ and $\begin{align*}r=\text{-}\frac{1}{2}\end{align*}$ , find the 1st term and the $\begin{align*}n^{th}\end{align*}$ term rule.
  ::如果几何序列中第一七个条件的总和为2158和r=12,则找到第一个条件和nth条件规则。

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  Solution: We can substitute what we know into the formula for the sum of a geometric sequence and solve for $\begin{align*}a_1\end{align*}$ .
  ::解答:我们可以用我们所知道的, 来替代公式中的几何序列和a1的解答。

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  $$\begin{align*}\frac{215}{8} &= \frac{a_1 \left(1-\left(\text{-}\frac{1}{2}\right)^7\right)}{1- \left(\text{-}\frac{1}{2}\right)} \\ \frac{215}{8} &= a_1 \left(\frac{43}{64}\right) \\ a_1 &= \left(\frac{64}{43}\right)\left(\frac{215}{8}\right)=40\end{align*}$$
  ::2158=a1(1-(-12)7)1-(-12)2158=a1(4364)a1=(6443)(2158)=40

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  The $\begin{align*}n^{th}\end{align*}$ term rule is $\begin{align*}a_n=40 \left(\text{-}\frac{1}{2}\right)^{n-1}\end{align*}$
  ::nth 术语规则为 an=40(-12)n- 1

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  Example 6
  ::例6

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  Charlie deposits $1,000 on the 1st of each year into his investment account. The account grows at a rate of 8% per year. How much money is in the account on the 1st day of the $\begin{align*}11^{th}\end{align*}$ year?
  ::Charlie每年1月1日将1 000美元存入投资账户。该账户每年增长8%。第11年第1天的账户里有多少钱?

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  Solution: First, consider what is happening here on the 1st day of each year. On the 1st day of the 1st year, $1,000 is deposited. On the 1st day of the 2nd year, $1,000 is deposited and the previously deposited $1,000 earns 8% interest or grows by a factor of 1.08. On the 1st day of the 3rd year, another $1,000 is deposited, the previous year's deposit earns 8% interest, and the original deposit earns 8% interest for two years (we multiply by $\begin{align*}1.08^2\end{align*}$ ).
  ::解决办法:首先,考虑每年第1天这里发生的情况。第一年第1天,存入1 000美元。第二年第1天,存入1 000美元,以前存入的1 000美元赚取8%的利息或增加1.08倍。第三年第1天,再存入1 000美元,上一年的存款赚取8%的利息,最初存款在两年里赚取8%的利息(我们乘以1.082)。

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  $$\begin{align*}&\text{Sum Year} \ 1: 1,\!000\\ &\text{Sum Year} \ 2:1,\!000 + 1,\!000(1.08)\\ &\text{Sum Year} \ 3: 1,\!000 + 1,\!000(1.08) + 1,\!000(1.08)^2\\ &\text{Sum Year} \ 4: 1,\!000 + 1,\!000(1.08) + 1,\!000(1.08)^2 + 1,\!000(1.08)^3\\ & \qquad \qquad \qquad \qquad \qquad \vdots\\ &\text{Sum Year} \ 11: 1,\!000 + 1,\!000(1.08) + 1,\!000(1.08)^2 \\ &+ 1,\!000(1.08)^3 + \ldots + 1,\!000(1.08)^9 + 1,\!000(1.08)^{10}\end{align*}$$
  ::总年数 1: 1 000-08+1 000(1 000)+1 000(1.08)+1 000(1 000)+1 000(1.082)+1 000(1.082)+1 000(1.008)+1 000(1.083)+1 000(1 000)+1 000(1 000)+1 000(1 000)+1 000(1.083+1 000)+1 000(1.083+...+1 000)+1 000(1.089+1 000)10

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  There are 11 terms in this series because on the 1st day of the $\begin{align*}11^{th}\end{align*}$ year, we make our final deposit and the original deposit earns interest for 10 years.
  ::这个系列有11个条件 因为11年的第1天 我们做最后的存款 原始的存款能赚10年的利息

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  This sequence is geometric. The 1st term is 1,000, the common ratio is 1.08, and $\begin{align*}n=11\end{align*}$ . Now we can calculate the sum using the formula, and determine the value of the investment account at the start of the $\begin{align*}11^{th}\end{align*}$ year.
  ::这是几何序列。 第一个学期是 1 000, 共同比率是 1. 08 和 n= 11。 现在我们可以用公式来计算总和, 并确定第 11 年开始时投资账户的价值 。

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  $$\begin{align*}S_{11}=\frac{1,\!000 \left(1-1.08^{11}\right)}{1-1.08}=16,\!645.48746 \approx \$16,\!645.49\end{align*}$$
  ::S11=1,000(1-0811-1)1-1.08=16 645.48746 16 645.49美元

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  Summary
  ::摘要

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  • The sum of a finite number of terms of a geometric sequence is  $\begin{align*}S_n=\frac{a_1(1-r^n)}{1-r},\end{align*}$   where $\begin{align*}n \end{align*}$  is the number of terms, $\begin{align*}a_1\end{align*}$  is the 1st term, and $\begin{align*}r\end{align*}$  is the common ratio.
    ::几何序列的限定术语数总和是 Sn=a1(1-rn)1-r, 其中n为术语数,a1为第一个术语,r为共同比率。

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  Review
  ::回顾

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  Use the formula for the sum of a geometric sequence to find the sum of the 1st five terms.
  ::使用几何序列总和的公式来找到第一五个条件的总和。

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  1.  $\begin{align*}a_n=36 \left(\frac{2}{3}\right)^{n-1}\end{align*}$
  ::1. an=36(23)n-1

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  2.  $\begin{align*}a_n=9(\text{-}2)^{n-1}\end{align*}$
  ::2. an=9(-2-2)-1

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  3.  $\begin{align*}a_n=5(\text{-}1)^{n-1}\end{align*}$
  ::3. an=5(-1-1-1)

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  4.  $\begin{align*}a_n=\frac{8}{25} \left(\frac{5}{2}\right)^{n-1}\end{align*}$
  ::4. a=825(52)n-1

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  5.  $\begin{align*}a_n=\frac{2}{3} \left(\text{-}\frac{3}{4}\right)^{n-1}\end{align*}$
  ::5. an=23(-34-34n-1)

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  Evaluate the following series:
  ::评估以下系列:

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  6.  $\begin{align*}\sum\limits_{n=1}^4(\text{-}1) \left(\frac{1}{2}\right)^{n-1}\end{align*}$
  ::6. 4n=1(-1)(12-1)

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  7.  $\begin{align*}\sum\limits_{n=2}^8(128) \left(\frac{1}{4}\right)^{n-1}\end{align*}$
  ::7. 8=2(128)(14-1)

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  8.  $\begin{align*}\sum\limits_{n=2}^7 \frac{125}{64}\left(\frac{4}{5}\right)^{n-1}\end{align*}$
  ::8. 7n=212564(45n-1)

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  9.  $\begin{align*}\sum\limits_{n=5}^{11} \frac{1}{32}(\text{-}2)^{n-1}\end{align*}$
  ::9. 11n=5132(-1-2)-1

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  Given the sum and the common ratio, find the $\begin{align*}n^{th}\end{align*}$ term rule for each sequence .
  ::根据总和和共同比率,为每个序列找到nth术语规则。

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  10.  $\begin{align*}\sum\limits_{n=1}^6 a_n=\text{-}63\end{align*}$ and $\begin{align*}r=\text{-}2\end{align*}$
  ::10. 6n=1an=-63和r=2

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  11.  $\begin{align*}\sum\limits_{n=1}^4 a_n=671\end{align*}$ and $\begin{align*}r=\frac{5}{6}\end{align*}$
  ::11. 4n=1an=671和r=56

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  12.  $\begin{align*}\sum\limits_{n=1}^5 a_n=122\end{align*}$ and $\begin{align*}r=\text{-}3\end{align*}$
  ::12. 5n=1an=122和r=3

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  13.  $\begin{align*}\sum\limits_{n=2}^7 a_n=\text{-}\frac{63}{2}\end{align*}$ and $\begin{align*}r=\text{-}\frac{1}{2}\end{align*}$
  ::13. 7n=2an=632和r=12

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  Explore More
  ::探索更多

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  1. Sapna's grandparents deposit $1,200 into a college savings account on her $\begin{align*}5^{th}\end{align*}$ birthday. They continue to make this birthday deposit each year until making the final deposit on her $\begin{align*}18^{th}\end{align*}$ birthday. If the account earns 5% interest annually, how much is there after the final deposit?
  ::1. 萨普拉的祖父母在她五岁生日时将1 200美元存入大学储蓄账户,他们每年继续将这一生日存款存入大学储蓄账户,直到她十八岁生日最后存款。如果该账户每年赚取5%的利息,在最后存款之后有多少?

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  2. Jeremy wants to have saved $10,000 in five years. If he makes annual deposits on the 1st of each year, and the account earns 4.5% interest annually, how much should he deposit each year to have $10,000 in the account after the final deposit on the 1st of the $\begin{align*}6^{th}\end{align*}$ year? Round your answer to the nearest $100.
  ::2. Jeremy想在五年内节省10 000美元,如果他每年在第一年存款,账户每年赚取4.5%的利息,那么他每年应在第六年第一期最后存款之后存入10 000美元多少?

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  3. You are saving. On the 1st of each month, you deposit $100  into your savings account. The account grows at a rate of 0.5% per month. How much money is in your account on the 1st day of the $\begin{align*}9^{th}\end{align*}$ month?
  ::3. 储蓄:每个月一日,将100美元存入储蓄账户,每月增长0.5个百分点,9个月第1日的存款有多少?

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  4. The population of a town was 2,400 in 2010. From 2010 to 2015, the population increased 3% per year. What was the population of the town in 2015? 
  ::4. 2010年,一个城镇的人口为2,400人,2010年至2015年,人口每年增长3%,2015年该城镇的人口是多少?

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  5. In 1995, the total box office revenue in U.S. movie theaters was about $5.02 billion. From 1995 to 2008, box office revenue increased by about 5.9% per year. What was the total box office revenue between 1995 and 2008? Round your answer to the nearest billion dollars.
  ::5. 1995年,美国电影院的票面办公室总收入约为50.2亿美元,1995年至2008年,票面办公室收入每年增长约5.9%,1995年至2008年的票面办公室收入总额是多少?

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  6. A fisherman harvested 350 kilograms of fish on Monday. From Monday through Friday, the amount of fish he harvested each day increased by 10%. What was the total amount of fish he harvested during these five days?
  ::6. 星期一,一名渔民捕捞350公斤鱼,从星期一到星期五,每天捕捞的鱼量增加了10%,这五天捕捞的鱼总量是多少?

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  7. A new website got 4,000 page views on the 1st day. During the next four days, the page views increased by 30% each day. What was the total amount of page views in the 1st five days?
  ::7. 一个新的网站在第一天的网页浏览量为4,000页,在接下来的四天中,网页浏览量每天增加30%,头五天的页面浏览量是多少?

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  8.  Sam deposits $50 on the 1st of each month into an account that earns 0.5% interest each month. To the nearest dollar, how much is in the account after Sam makes his last deposit on the 1st day of the 5th year (the $\begin{align*}49^{th}\end{align*}$ month)?
  ::8. Sam每月1月1日将50美元存入每月赚取0.5 % 利息的帐户,至最近的美元,在Sam在5年(第49个月)第1天最后存款之后,帐户内有多少钱?

  播放段落

  Answers for Review and Explore More Problems
  ::回顾和探讨更多问题的答复

  播放段落

  Please see the Appendix.
  ::请参看附录。