11.10 查找有限几何系列总和

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  查找精度几何序列的总和

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  You are saving for summer camp. You deposit $100 on the first of each month into your savings account. The account grows at a rate of 0.5% per month. How much money is in your account on the first day on the $\begin{align*}9^{th}\end{align*}$ month?
  ::您正在为夏令营储蓄。 您每月头一个月将100美元存入您的储蓄账户。 该账户以每月0.5%的速率增长。 您在9个月第一天的账户里存了多少钱 ?

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  Sum of Finite Geometric Series
  ::有限几何系列总和

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  We have discussed how to use the calculator to find the sum of any series provided we know the $\begin{align*}n^{th}\end{align*}$ term rule. For a geometric series, however, there is a specific rule that can be used to find the sum algebraically. Let’s look at a finite geometric and derive this rule.
  ::我们已经讨论过如何使用计算器来找到任何序列的总和,只要我们知道 nth 术语规则。 但是,对于几何序列来说,有一条具体规则可以用来找到总代数。 让我们看看一个有限的几何方法并得出这一规则。

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  Given $\begin{align*}a_n=a_1r^{n-1}\end{align*}$ .
  ::给定 an= a1rn- 1 。

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  The sum of the first $\begin{align*}n\end{align*}$ terms of a geometric sequence is: $\begin{align*}S_n=a_1+a_1r+a_1r^2+a_1r^3+ \ldots +a_1r^{n-2}+a_1r^{n-1}\end{align*}$ .
  ::几何序列的第一个 n 条件总和是: Sn=a1+a1r+a1r2+a1r3+...+a1rn-2+a1rn-1。

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  Now, factor out $\begin{align*}a_1\end{align*}$ to get $\begin{align*}a_1(1+r^2+r^3+ \ldots +r^{n-2}+r^{n-1})\end{align*}$ . If we isolate what is in the parenthesis and multiply this sum 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) 。 如果我们将括号中的内容分隔开来, 并将这个总和乘以下文所示的( 1-r) , 就可以简化总和 :

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  $$\begin{align*}&(1-r)S_n=(1-r)(1+r+r^2+r^3+ \ldots +r^{n-2}+r^{n-1}) \\ &= (1+r+r^2+r^3+ \ldots +r^{n-2}+r^{n-1}-r-r^2-r^3-r^4- \ldots -r^{n-1}-r^n) \\ &= (1+r+r^2+r^3+ \ldots +r^{n-2}+r^{n-1}-r-r^2-r^3-r^4- \ldots -r^{n-1}-r^n) \\ &= (1-r)^n\end{align*}$$
  :1-r)Sn=(1+r+r+r2+r3+...+rn2+rn1)=(1+r+r+r2+r3+...+...+r2+r2+r3+r3+r4-rn_r2_r2-r3-r4-...-rn-rn_r3+...+r_r2+r2+r_r2+r_r3_r4_r4__...-rn_1_rn)=(1+r+r+r+r2+r2+r2+r3+r3+r3+..._r_r3_r4__...-rn)=(1-rn)n

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  By multiplying the sum by $\begin{align*}1-r\end{align*}$ we were able to cancel out all of the middle terms. However, we have changed the sum by a factor of $\begin{align*}1-r\end{align*}$ , so what we really need to do is multiply our sum by $\begin{align*}\frac{1-r}{1-r}\end{align*}$ , or 1.
  ::通过将总和乘以1-r,我们得以取消所有中间条件。然而,我们已经将总和换成1-r,所以我们真正需要做的是将总和乘以1-r1或1。

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  $\begin{align*}a_1(1+r^2+r^3+ \ldots +r^{n-2}+r^{n-1}) \frac{1-r}{1-r}=\frac{a_1(1-r^n)}{1-r}\end{align*}$ , which is the sum of a finite geometric series.
  ::a1(1+r2+r3+...+rn-2+rn-1)1-r1-r=a1(1-rn)1-r,是一定几何序列的总和。

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  So, $\begin{align*}S_n=\frac{a_1(1-r^n)}{1-r}\end{align*}$ .
  ::那么,Sn=a1(1-rn)1-r

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  Let's find the sum of the first ten terms of the geometric sequence $\begin{align*}a_n=\frac{1}{32}(-2)^{n-1}\end{align*}$ . This could also be written as, “Let's find $\begin{align*}\sum\limits_{n=1}^{10} \frac{1}{32}(-2)^{n-1}\end{align*}$ .”
  ::让我们找到几何序列 an=132(- 2)n-1的前十个条件的总和。 这也可以写为, “让我们找到n=110132(- 2)n-1 ” 。

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  Using the formula, $\begin{align*}a_1=\frac{1}{32}\end{align*}$ , $\begin{align*}r=-2\end{align*}$ , and $\begin{align*}n=10\end{align*}$ .
  ::使用公式, a1=132, r2, n=10。

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  $$\begin{align*}S_{10}=\frac{\frac{1}{32}(1-(-2)^{10})}{1-(-2)} = \frac{\frac{1}{32}(1-1024)}{3} = -\frac{341}{32}\end{align*}$$
  ::S10=132(1-2)1010-1-2)1-2=132(1-1024)334132

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  We can also use the calculator as shown below.
  ::我们还可以使用下文所示的计算器。

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  $$\begin{align*}sum(seq(1/32(-2)^{x-1}, x, 1, 10))=-\frac{341}{32}\end{align*}$$
  :seq(1/32(-2)x-1,x,1,10)34132

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  Now, let's find the first term and the $\begin{align*}n^{th}\end{align*}$ term rule for a geometric series in which the sum of the first 5 terms is 242 and the common ratio is 3.
  ::现在,让我们来为几何序列找到第一个学期和第n学期规则, 在其中,前五个学期的总和是242, 共同比率是3。

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  Plug in what we know to the formula for the sum and solve for the first term:
  ::插入我们知道的金额公式 并解决第一个任期:

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

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  The first term is 2 and $\begin{align*}a_n=2(3)^{n-1}\end{align*}$ .
  ::第一个任期为2年,一个=2(3)n-1。

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  Finally, let's solve the following problem.
  ::最后,让我们解决以下问题。

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  Charlie deposits $1000 on the first 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 first day on the $\begin{align*}11^{th}\end{align*}$ year.
  ::Charlie每年第一年将1000美元存入投资账户。该账户以每年8%的速度增长。第11年的第一天,账户里有多少钱。

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  First, consider what is happening here on the first day of each year. On the first day of the first year, $1000 is deposited. On the first day of the second year $1000 is deposited and the previously deposited $1000 earns 8% interest or grows by a factor of 1.08 (108%). On the first day of the third year another $1000 is deposited, the previous year’s deposit earns 8% interest and the original deposit earns 8% interest for two years (we multiply by $\begin{align*}1.08^2\end{align*}$ ):
  ::首先,考虑每年第一天这里发生的情况。第一年的第一天,1 000美元被存入。第二年的第一天,1 000美元被存入,以前存入的1 000美元获得8%的利息或增长系数为1.08(108 % ) 。 第三年的第一天,再存入1 000美元,上一年的存款获得8%的利息,原始存款在两年中赚取8%的利息(我们乘以1.082 ):

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  $$\begin{align*}&\text{Sum Year} \ 1: 1000\\ &\text{Sum Year} \ 2:1000 + 1000(1.08)\\ &\text{Sum Year} \ 3: 1000 + 1000(1.08) + 1000(1.08)^2\\ &\text{Sum Year} \ 4: 1000 + 1000(1.08) + 1000(1.08)^2 + 1000(1.08)^3\\ & \qquad \qquad \qquad \qquad \qquad \vdots\\ &\text{Sum Year} \ 11: 1000 + 1000(1.08) + 1000(1.08)^2 + 1000(1.08)^3 + \ldots + 1000(1.08)^9 + 1000(1.08)^{10}\end{align*}$$
  ::11000-000+1000-000(1.08)+1000-000(1.083+...+1000-000(1.08)+1000-20000(1.08)+1000(1.08)+1000(1.08)+1000(1.082+1000)+1000(1.083)+(%0)+Sum 11:1000+1000(1.08)+1000(1.08)+1000(1.082)+1000(1.082)+1000(1.083+...+1000(1.08)+1000(1.089)+1000(1.08)10

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  $\begin{align*}^*\end{align*}$ There are 11 terms in this series because on the first day of the $\begin{align*}11^{th}\end{align*}$ year we make our final deposit and the original deposit earns interest for 10 years.
  ::* 本系列有11个条件,因为在11年的第一天,我们最后存款和原始存款赚取利息达10年之久。

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  This series is geometric. The first term is 1000, 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1000, 共同比率是1.08 和 n=11。 现在我们可以使用公式来计算总和, 并在第十一年开始时确定投资账户的价值 。

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  $$\begin{align*}s_{11}=\frac{1000 \left(1-1.08^{11}\right)}{1-1.08}=16645.48746 \approx \$16,645.49\end{align*}$$
  ::11=1000(1-0811-1)1-1.08=16645.48746 $16 645.49

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked to find how much  money is in your account on the first day of the  $\begin{align*}9^{th}\end{align*}$  month. 
  ::早些时候,你被要求去查一下 9月9日的第一天 你账户里有多少钱

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  There are 9 terms in this series because on the first day of the $\begin{align*}9^{th}\end{align*}$ month you make your final deposit and the original deposit earns interest for 8 months.
  ::这个系列有9个条件 因为9月的第一天 你缴纳了最后的存款 原始的存款在8个月里赚取利息

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  This series is geometric. The first term is 100, the common ratio is 1.005 and $\begin{align*}n=9\end{align*}$ . Now we can calculate the sum using the formula and determine the value of the investment account at the start of the $\begin{align*}9^{th}\end{align*}$ month.
  ::这是几何序列。 第一个学期是100, 共同比率是1.005和n=9。 现在我们可以使用公式计算总和, 并在9月初确定投资账户的价值 。

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  $$\begin{align*}s_{9}=\frac{100 \left(1-1.005^{9}\right)}{1-1.005}= 918.22\end{align*}$$
  ::s9=100(1-1.00591-11.005=918.22)

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  Therefore there is $918.22 in the account at the beginning of the ninth month.
  ::因此,在第九个月初,账户中有918.22美元。

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  Example 2
  ::例2

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  Evaluate $\begin{align*}\sum\limits_{n=3}^8 2(-3)^{n-1}\end{align*}$ .
  ::评价n=382(-3)n-1。

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  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 first term. The third term is $\begin{align*}a_3=2(-3)^2=2(9)=18\end{align*}$ . 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 series now with $\begin{align*}a_1=18\end{align*}$ , $\begin{align*}r=-3\end{align*}$ and $\begin{align*}n=6\end{align*}$ to find the sum:
  ::由于我们被要求找到第三至第八学期的总和,我们将考虑将a3作为第一个学期。第三个学期为a3=2(-3)2=2(9)=18。由于我们从第三学期开始,我们将共学6个学期,a3+a4+a5+a6+a7+a8。我们可以使用这一规则来计算目前为a1=18、r3和n=6的几何序列的总和:

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  $$\begin{align*}\sum_{n=3}^8 2(-3)^{n-1}=\frac{18(1-(-3)^6)}{1-(-3)}=-3276\end{align*}$$
  ::n=382(-3)(-3)n-1=18(1-(-3)6)1-(-3)-(-3)_______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

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  Example 3
  ::例3

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  If the sum of the first seven terms in a geometric series is $\begin{align*}\frac{215}{8}\end{align*}$ and $\begin{align*}r=-\frac{1}{2}\end{align*}$ , find the first term and the $\begin{align*}n^{th}\end{align*}$ term rule.
  ::如果几何序列中前七个术语的总和是2158和r12,则找到第一个术语和nth术语规则。

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  We can substitute what we know into the formula for the sum of a geometric series and solve for $\begin{align*}a_1\end{align*}$ .
  ::我们可以用我们所知道的 来代替公式 和几何序列的总和 然后解决A1

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  $$\begin{align*}\frac{215}{8} &= \frac{a_1 \left(1-\left(-\frac{1}{2}\right)^7\right)}{1- \left(-\frac{1}{2}\right)} \\ \frac{215}{8} &= a_1 \left(\frac{43}{64}\right) \\ a_1 &= \left(\frac{64}{43}\right)\left(\frac{215}{8}\right)=40\end{align*}$$
  ::2158=a1(1-(-)127)1-(-)12-2158=a1(-4364)a1=(6443)(2158)=40

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

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  Example 4
  ::例4

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  Sam deposits $50 on the first of each month into an account which earns 0.5% interest each month. To the nearest dollar, how much is in the account right after Sam makes his last deposit on the first day of the fifth year (the $\begin{align*}49^{th}\end{align*}$ month).
  ::Sam每月头一个月将50美元存入每个月赚取0.5 % 利息的账户。 至最近的美元,在Sam在第五年(第49个月)的第一天(即第49个月)最后一次存款之后,在账户中存了多少钱。

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  The deposits that Sam make and the interest earned on each deposit generate a geometric series,
  ::Sam的存款和每笔存款的利息 产生几何序列

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  $$\begin{align*}& S_{49}=50+50(1.005)^1+50(1.005)^2+50(1.005)^3+ \ldots +50(1.005)^{47}+50(1.005)^{48}, \\ & \qquad \ \ \uparrow \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \uparrow \\ & \quad \text{last deposit} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \text{first deposit}\end{align*}$$
  ::S49=50+50(1.0051)1+50(1.0052)2+50(1.0053+...)+50(1.005)47+50(1.005)48, 最后一次交存第一交存

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  Note that the first deposit earns interest for 48 months and the final deposit does not earn any interest. Now we can find the sum using $\begin{align*}a_1=50\end{align*}$ , $\begin{align*}r=1.005\end{align*}$ and $\begin{align*}n=49\end{align*}$ .
  ::请注意,第一笔存款可赚取48个月的利息,而最后一笔存款不赚取任何利息。 现在我们可以用a1=50、r=1.005和n=49来找到这笔金额。

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  $$\begin{align*}S_{49}=\frac{50(1-(1.005)^{49})}{(1-1.005)} \approx \$2768\end{align*}$$
  ::S49=50(1至(1.005)49(1至1.005)________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

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  Review
  ::回顾

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  Use the formula for the sum of a geometric series to find the sum of the first five terms in each series.
  ::使用几何序列总和的公式,在每个序列中找到前五个术语的总和。

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  1. $\begin{align*}a_n=36 \left(\frac{2}{3}\right)^{n-1}\end{align*}$
     ::an=36( 23n- 1)
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  2. $\begin{align*}a_n=9(-2)^{n-1}\end{align*}$
     ::an=9(--2)n- 1
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  3. $\begin{align*}a_n=5(-1)^{n-1}\end{align*}$
     ::an=5 (- 1) n- 1
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  4. $\begin{align*}a_n=\frac{8}{25} \left(\frac{5}{2}\right)^{n-1}\end{align*}$
     ::an=825(52)n-1
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  5. $\begin{align*}a_n=\frac{2}{3} \left(-\frac{3}{4}\right)^{n-1}\end{align*}$
     ::an=23(- 34n- 1)

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  Find the indicated sums using the formula and then check your answers with the calculator.
  ::使用公式查找指定金额,然后用计算器检查答案。

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  6. $\begin{align*}\sum\limits_{n=1}^4(-1) \left(\frac{1}{2}\right)^{n-1}\end{align*}$
     ::Nn=14(-1)(12)(1)(12)n-1
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  7. $\begin{align*}\sum\limits_{n=2}^8(128) \left(\frac{1}{4}\right)^{n-1}\end{align*}$
     ::=28(128)(14)(14)n-1
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  8. $\begin{align*}\sum\limits_{n=2}^7 \frac{125}{64}\left(\frac{4}{5}\right)^{n-1}\end{align*}$
     ::=2712564(45-1)
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  9. $\begin{align*}\sum\limits_{n=5}^{11} \frac{1}{32}(-2)^{n-1}\end{align*}$
     ::=511132(-2-2)n-1

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  Given the sum and the common ratio, find the $\begin{align*}n^{th}\end{align*}$ term rule for the series.
  ::考虑到总和和共同比率,请找到该系列的第n期规则。

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  10. $\begin{align*}\sum\limits_{n=1}^6 a_n=-63\end{align*}$ and $\begin{align*}r=-2\end{align*}$
      ::=16an\\\\63和r2
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  11. $\begin{align*}\sum\limits_{n=1}^4 a_n=671\end{align*}$ and $\begin{align*}r=\frac{5}{6}\end{align*}$
      ::n=14an=671和r=56
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  12. $\begin{align*}\sum\limits_{n=1}^5 a_n=122\end{align*}$ and $\begin{align*}r=-3\end{align*}$
      ::=1500=122和r3
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  13. $\begin{align*}\sum\limits_{n=2}^7 a_n=-\frac{63}{2}\end{align*}$ and $\begin{align*}r=-\frac{1}{2}\end{align*}$
      ::=27 =632 和 r =12

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  Solve the following word problems using the formula for the sum of a geometric series.
  ::使用几何序列之和的公式解决下列字问题。

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  14. Sapna’s grandparents deposit $1200 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?
      ::萨普拉的祖父母在她五岁生日时将1200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18岁生日最后存款。 如果该账户每年赚取5 % 的利息,那么在最后存款之后会有多少?
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  15. Jeremy wants to have save $10,000 in five years. If he makes annual deposits on the first of each year and the account earns 4.5% interest annually, how much should he deposit each year in order to have $10,000 in the account after the final deposit on the first of the $\begin{align*}6^{th}\end{align*}$ year. Round your answer to the nearest $100.
      ::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,那么他每年应该存入多少,才能在第六年第一年最后存款之后在账户中存10 000美元。请将您对最近100美元的答复四舍五入。

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  Review (Answers)
  ::回顾(答复)

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  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 。