Course: 7.10 无限几何序列总和

Section outline

Select section General

Collapse Expand
General

Collapse all Expand all
- Select activity 新闻通告
  
  新闻通告 Forum

无限几何系列总和

Sayber's mom told him to clean his room on Saturday morning.
::赛伯的妈妈让他周六早上去打扫房间

"But, MOM! It's gonna take forever!" said Sayber.
::"但是,妈妈,要花一辈子的时间!" 赛伯说:

"Oh, don't be overly dramatic," said mom.
::妈妈说"哦,别太戏剧化了"

"I am NOT being dramatic!" Sayber said.
::"我不是戏剧性的!" 赛伯说

"If I start right now, it is going to take me at LEAST an hour to clean this half alone, then it will take another half hour to clean half of the remainder, and 15 mins to clean half of THAT remainder... since I will always have half left, I will never be done!"
::"如果我现在就开始,我需要一个小时的时间来清理这一半, 然后再花半小时的时间来清理剩下的一半, 还有15分钟的时间来清理剩下的一半... 因为我永远只剩下一半了, 我永远不会完成!"

Do you agree with Sayber? Will Sayber be stuck with a vacuum in his hand forever? Tune in next week...
::你同意赛伯的看法吗?赛伯会永远被吸尘器困在手上吗?

Sums of Infinite Geometric Series
::无限几何系列总和

Let’s return to the situation in the introduction: Poor Sayber is stuck cleaning his room. He cleans half of the room in 60 mins. Then he cleans half of what is left, 30 more minutes, half again for 15 more. If he keeps cleaning half of the remaining area, how will he ever finish the room?
::让我们回到引言中的情况:可怜的Sayber在清理自己的房间。他在60分钟内清理了一半的房间。然后他清理了剩下的一半,再清理了30分钟,再清理了15分钟。如果他继续清理剩下的一半地区,他将如何完成这个房间?

We know that the pieces have to add up to some finite time period (no matter what it feels like, Sayber CAN get the room clean), but how is it possible for the sum of an infinite number of terms to be a finite number?
::我们知道,这几个字必须加在一起,加到一些有限的时间段(不管它是什么感觉,Sayber可以把房间清理干净),但是,一个无限数目的术语的总和怎么可能是一个有限的数字?

To find the sum of an infinite number of terms, we should consider some . Three partial sums, relatively early in the series, could be: $\begin{align*}S_2 = 90, S_3 = 105,\end{align*}$ and $\begin{align*}S_6 = 118.125\end{align*}$ or $\begin{align*}118 \frac{1}{8}\end{align*}$
::为了找到无限条件的总和,我们应考虑一些条件。在本系列中相对较早的三部分可以是:S2=90、S3=105和S6=118.125或11818。

Now let’s look at larger values of $\begin{align*}n:\end{align*}$
::现在让我们来看看更大的 n 值 :

$\begin{align}S_7\end{align}$	$\begin{align}= \frac{60 \left(1 - (\frac{1} {2})^7 \right)} {1 - \frac{1} {2}} \approx 119.06\end{align}$ minutes
$\begin{align}S_8\end{align}$	$\begin{align}= \frac{60 \left(1 - (\frac{1} {2})^8 \right)} {1 - \frac{1} {2}} \approx 119.5\end{align}$ minutes
$\begin{align}S_{10}\end{align}$	$\begin{align}= \frac{60 \left(1 - (\frac{1} {2})^{10} \right)} {1 - \frac{1} {2}} \approx 119.9\end{align}$ minutes

As n approaches infinity, the value of S _n seems to approach 120 minutes. In terms of the actual sums, what is happening is this: as n increases, the n ^th term gets smaller and smaller, and so the n ^th term contributes less and less to the value of S _n . We say that the series converges , and we can write this with a limit:
::当 n 接近无穷时, Sn 的值似乎接近120分钟。从实际总和来看,正在发生的是: 随着n 的增加, n 值越来越小, 所以 n 值对 Sn 值的贡献越来越小。我们说, 序列会趋同, 我们可以用一个限度来写:

$\begin{align}\lim_{n \rightarrow \infty} S_n\end{align}$	$\begin{align}= \lim_{n \rightarrow \infty} \left(\frac{60 \left(1 - (\frac{1} {2})^{n} \right)} {1 - \frac{1} {2}} \right )\end{align}$
	$\begin{align}= \lim_{n \rightarrow \infty} \left(\frac{60 \left(1 - (\frac{1} {2})^{n} \right)} {\frac{1} {2}} \right )\end{align}$
	$\begin{align}= \lim_{n \rightarrow \infty} \left(120 \left(1 - \left(\frac{1} {2} \right)^{n} \right ) \right) \end{align}$

As n approaches infinity, the value of $\begin{align*}\left(\frac{1} {2} \right)^{n}\end{align*}$ gets smaller and smaller. That is, the value of this expression approaches 0. Therefore the value of $\begin{align*}1 - \left(\frac{1} {2} \right)^{n}\end{align*}$ approaches 1, and $\begin{align*}120 \left(1 - \left(\frac{1} {2} \right)^n \right)\end{align*}$ approaches 120(1) = 120.
::由于 n 方法的宽度, (12) 值越来越小。也就是说, 这个表达式的值是 0 。因此, 1 - (12) 方法 1 和 120 (1 - (12) 方法 120 (1) = 120。

Therefore, no matter how long the process continues, Sayber will not spend more than 2hrs cleaning the room. Of course, it may SEEM like a lot more!
::因此,无论这一过程持续多久,Sayber不会花费超过2小时的时间来打扫房间。当然,它可能比这要长得多!

We can do the same analysis for the general case of a geometric series, as long as the terms are getting smaller and smaller. This means that the common ratio must be a number between -1 and 1: |r| < 1.
::只要条件越来越小,我们可以对几何序列的一般情况进行同样的分析。这意味着共同比率必须是-1:1和1: @r < 1之间的数字。

$\begin{align}\lim_{n \rightarrow \infty} S_n\end{align}$	$\begin{align}= \lim_{n \rightarrow \infty} \left (\frac{a_1(1 - r^n)} {1 - r} \right)\end{align}$
	$\begin{align}=\frac{a_1} {1 - r},\end{align}$ as $\begin{align}(1 - r^n) \rightarrow 1\end{align}$

Therefore, we can find the sum of an infinite geometric series using the formula $\begin{align*}S = \frac{a_1} {1 - r}\end{align*}$ .
::因此,我们可以用S=a11-r公式找到无限几何序列的总和。

When an infinite sum has a finite value, we say the sum converges . Otherwise, the sum diverges . A sum converges only when the terms get closer to 0 after each step, but that alone is not a sufficient criterion for convergence. For example, the sum $\begin{align*}\sum_{n = 1}^\infty \frac{1} {n} = 1 + \frac{1} {2} + \frac{1} {3} + \frac{1} {4} + ....\end{align*}$ does not converge .
::当一个无限总和有一定的值时,我们说总和会趋同。否则,总和会不同。只有当条件每步后接近0时,总和才会趋同,但单是这一点不足以作为趋同的标准。例如,总和 n=11n=1+12+13+14+.没有趋同。

Examples
::实例

Example 1
::例1

Find the sum of the convergent series: $\begin{align*}40 + -20 + 10 + -5 + ...\end{align*}$
::查找集合序列的总和: 40+20+10+5+...

The common ratio is $\begin{align*}\frac{-1} {2}\end{align*}$ . Therefore the sum converges to:
::共同比率是-12。因此,总和一致到:

	$\begin{align}\frac{40} {1 - \left (\frac{-1} {2} \right)} = \frac{40} {\frac{3} {2}} = 40 \left(\frac{2} {3} \right) = \frac{80} {3}\end{align}$

Example 2
::例2

Determine if the series converges. If it converges, find the sum.
::确定序列是否相交。如果相交,请找到总和。

a. $\begin{align}1 + \frac{1} {3} + \frac{1} {9} + \frac{1} {27}+ ... \end{align}$	b. $\begin{align}3 + -6 +12 + -24 + ... \end{align}$

$\begin{align*}1 + \frac{1} {3} + \frac{1} {9} + \frac{1} {27} + ... \end{align*}$ converges. The common ratio is (1/3) . Therefore the sum converges to:
::1+13+19+127+... convents. 通用比率为(1/3)。因此,总和会和到 :

	$\begin{align}\frac{1} {1 - \frac{1} {3}} = \frac{1} {\frac{2} {3}} = \frac{3} {2}\end{align}$

The series 3 + -6 + 12 + -24 + ... does not converge, as the common ratio is -2.
::3+-6+12+ -24+...没有趋同,因为共同比率是-2。

Remember that the idea of an infinite sum was introduced in the context of a realistic situation, albeit a paradoxical one. We can in fact use infinite geometric series to model other realistic situations. Here we will look at another example: the total vertical distance traveled by a bouncing ball.
::记住无限总和的理念是在现实形势的背景下引入的,尽管这是一个自相矛盾的局面。事实上,我们可以使用无限几何序列来模拟其他现实形势。在这里,我们将看到另一个例子:弹球所穿行的总垂直距离。

Example 3
::例3

A ball is dropped from a height of 20 feet. Each time it bounces, it reaches 50% of its previous height. What is the total vertical distance the ball travels?
::A球从20英尺高处掉下来,每次弹跳,都达到前一个高度的50%。球的总垂直距离是多少?

We can think of the total distance as the distance the ball travels down + the distance the ball travels back up. The downward bounces form a geometric series:
::我们可以想出球向下行进的距离 + 球向上行进的距离。向下弹跳形成几何序列 :

20 + 10 + 5 +...

The upward bounces form the same series, except the first term is 10.
::上调形成相同的序列,但第一个学期为10个学期除外。

So the total distance is: $\begin{align*}\sum_{n=1}^\infty 20 \left ( \frac{1}{2} \right )^{n-1}+\sum_{n=1}^\infty 10 \left ( \frac{1}{2} \right )^{n-1}\end{align*}$ .
::总距离是: @n=120( 12)n- 1n=110( 12)n- 1。

Each sum converges, as the common ratio is (1/2). Therefore the total distance is:
::每个总和相趋同,因为共同比率(1/2)。因此,总距离是:

			$\begin{align}\frac {20}{1-\frac {1}{2}}+\frac {10}{1-\frac {1}{2}}=\frac {20}{\frac {1}{2}}+\frac {10}{\frac {1}{2}}=40+20=60\end{align}$

So the ball travels a total vertical distance of 60 feet.
::所以球的垂直距离是60英尺

Example 4
::例4

Determine if the following series converges or diverges. If it converges, find the sum.
::确定以下序列相交或相差。如果相交,请查找总和。

240 + 60 + 15 + ...

The sum converges. S = 320.
::总和汇合。 S=320。

Example 5
::例5

In this lesson, we proved the formula for the sum of a geometric series, $\begin{align*}S_n=\frac {a_1(1-r^n)}{1-r}\end{align*}$ using induction.
::在这个教训中,我们证明了一个几何序列之和的公式,Sn=a1(1-rn)1-r使用感应。

Prove this formula without induction:
::证明这个配方没有上岗:

Step 1: Let S _n = a ₁ + a ₁ r + a ₁ r ² + ... + a ₁ r ^n-1
::第1步:让 Sn = a1 + a1 + a1r + a1r2+...+ a1rn-1

^{$\begin{align*} S_n = a_1 + a_1{r} + a_1{r^2} + ... + a_1{r^{n-1}}\end{align*}$}
::Sn=a1+a1r+a1r2+...+a1rn-1

Step 2: Multiply S _n by r to obtain a second equation
::步骤2:将Sn乘以r以获得第二个方程

$\begin{align*}r{S_n} = a_1{r} + a_1{r^2} + a_1{r^3} + ... + a_1{r^n}\end{align*}$
::rSn=a1r+a1r2+a1r3+...+a1nn

Step 3: Subtract the equations and solve for S _n .
::第3步: 减号方程和SN的解答。

$\begin{align*} S_n - rS_n = a_1 - a_1{r^n}\end{align*}$
::Sn-rSn=a1 - a1-a1nn

$\begin{align*}\Rightarrow S_n(1 - r) = a(1 - r^n)\end{align*}$
::Sn(1-r)=a(1-rn)

$\begin{align*}\Rightarrow S_n = \frac {a(1 - r^n)}{(1 - r)}\end{align*}$
::Sn=a(1 - rn(1 - r)

Example 6
::例6

A ball is dropped from a height of 40 feet, and each time it bounces, it reaches 25% of its previous height.
::一个球从40英尺高处掉下来, 每次弹跳, 都达到前一个高度的25%。

Find the total vertical distance the ball travels, using the method used in the lesson.
::使用课中使用的方法,查找球的垂直距离。

$\begin{align*}\sum_{n=1}^\infty 40 \left ( \frac{1}{4} \right )^{n-1}+\sum_{n=1}^\infty 20 \left ( \frac{1}{4} \right )^{n-1}=66\frac {2}{3}\end{align*}$
::n=140(14)n-1n=120(14)n-1=6623

Find the total vertical distance the ball travels using a single series.
::使用单个序列查找球的垂直总距离。

$\begin{align*}\sum_{n=1}^\infty 50 \left ( \frac{1}{4} \right )^{n-1}=66\frac {2}{3}\end{align*}$
::n=150(14)n-1=6623

(Hint: write out several terms for each bounce. For example, the first bounce is: 40 feet down + 10 feet up = 50 feet traveled.)
:提示: 为每次弹出写几个条件。例如, 第一弹出是: 向下40英尺 + 向上10英尺 = 50英尺。 )

Example 7
::例7

Below are two infinite series that are not geometric. Use a graphing calculator to examine partial sums. Does either series converge?
::下面是两个不几何的无限序列。使用图形计算器来检查部分总和。这两种序列是否相容 ?

$\begin{align*}1 + \frac {1}{2} + \frac {1}{3} + \frac {1}{4} + ...\end{align*}$

This series does not converge.
::此序列不趋同。

$\begin{align*}1 + \frac {1}{4} + \frac {1}{9} + \frac {1}{16} + ...\end{align*}$

This series converges around 1.65. (The actual sum is $\begin{align*}\frac {\pi^2}{6}\end{align*}$ )
::这一系列约合1.65左右。 (实际总数是26。 )

Review
::回顾

Find the sum of the first 10 terms of $\begin{align*}\sum_{n = 1}^\infty \left (\frac{1} {5} \right)^n\end{align*}$ using a graphing calculator.
::使用图形计算计算器查找 n=1(15) 前 10 个条件的总和。

Find the sum of the first 20 terms of $\begin{align*}\sum_{n = 1}^\infty \left (\frac{1} {5} \right)^n\end{align*}$ using a graphing calculator.
::使用图形计算器查找 'n=1'1'(15) 前20个条件的总和。

Conjecture on the possible convergence of the series in questions 1 and 2.
::在问题1和2中猜测该系列可能趋同于问题1和2。

Evaluate the infinite sum of each of the following geometric series:
::评估以下每一几何序列的无限总和:

$\begin{align*}-2 + 1 -\frac{1}{2}+...\end{align*}$
$\begin{align*} -6 + \frac{24}{5} -\frac{96}{25}+...\end{align*}$
$\begin{align*} 3 +\frac{3}{2} -\frac{3}{4}+...\end{align*}$
$\begin{align*} -6 + 4 -\frac{8}{3}+...\end{align*}$
$\begin{align*}1 + \frac{1}{2}+\frac{1}{4}+...\end{align*}$

Evaluate the infinite sum of each of the following geometric series:
::评估以下每一几何序列的无限总和:

$\begin{align*}\sum_{n=1}^{\infty} -3(\frac{1}{2})^{(n-1)}\end{align*}$
::n=1-3(12)(n-1)

$\begin{align*}\sum_{n=1}^{\infty} -2(\frac{4}{7})^{(n-1)}\end{align*}$
::n=1-2(47)(n-1)

$\begin{align*}\sum_{n=1}^{\infty} 7(\frac{-4}{5})^{(n-1)}\end{align*}$
::n=17(- 45)(- 1)

$\begin{align*}\sum_{n=1}^{\infty} -9(\frac{-1}{5})^{(n-1)}\end{align*}$
::n=1-9(-15)-(一)

$\begin{align*}\sum_{n=1}^{\infty} 5(\frac{-5}{7})^{(n-1)}\end{align*}$
::Nn=15(-57)(- 1)

$\begin{align*}\sum_{n=1}^{\infty} 6(\frac{1}{5})^{(n-1)}\end{align*}$
::n=16(15) (n-1)

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。

$\begin{align}S_7\end{align}$	$\begin{align}= \frac{60 \left(1 - (\frac{1} {2})^7 \right)} {1 - \frac{1} {2}} \approx 119.06\end{align}$ minutes
$\begin{align}S_8\end{align}$	$\begin{align}= \frac{60 \left(1 - (\frac{1} {2})^8 \right)} {1 - \frac{1} {2}} \approx 119.5\end{align}$ minutes
$\begin{align}S_{10}\end{align}$	$\begin{align}= \frac{60 \left(1 - (\frac{1} {2})^{10} \right)} {1 - \frac{1} {2}} \approx 119.9\end{align}$ minutes

$\begin{align}\lim_{n \rightarrow \infty} S_n\end{align}$	$\begin{align}= \lim_{n \rightarrow \infty} \left(\frac{60 \left(1 - (\frac{1} {2})^{n} \right)} {1 - \frac{1} {2}} \right )\end{align}$
	$\begin{align}= \lim_{n \rightarrow \infty} \left(\frac{60 \left(1 - (\frac{1} {2})^{n} \right)} {\frac{1} {2}} \right )\end{align}$
	$\begin{align}= \lim_{n \rightarrow \infty} \left(120 \left(1 - \left(\frac{1} {2} \right)^{n} \right ) \right) \end{align}$

Section outline

General

无限几何系列总和

Sums of Infinite Geometric Series ::无限几何系列总和

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Example 6 ::例6

Example 7 ::例7

Review ::回顾

Review (Answers) ::回顾(答复)