Course: 8.10 几何序列和指数函数 | The Way To Learn

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几何序列和指数函数
Geometric Sequences and Exponential Functions
::几何序列和指数函数

Consider the following question:
::审议下列问题:

Which would you prefer, being given one million dollars, or one penny the first day, double that penny the next day, and then double the previous day's pennies and so on for a month?
::你会更喜欢哪一种, 得到一百万美元, 或者第一天一分钱, 第二天翻倍的一分钱, 然后在前一天翻倍的零钱,等等一个月?

At first glance it’s easy to say "Give me the million!" But why don’t we do a few calculations to see how the other choice stacks up?
::但为何我们不做一些计算, 看看其他选择是如何堆叠起来的呢?

You start with a penny the first day and keep doubling each day. Doubling means that we keep multiplying by 2 each day for one month (30 days).
::从第一天开始,你只用一分钱,每天翻一番。翻倍意味着我们一个月(30天)每天以2倍的乘以2倍的乘以。

On the first day, you get 1 penny, or $\begin{aligned} 2^{0} \end{aligned}$ pennies.
::第一天,你得到一便士, 或20便士。

On the second day, you get 2 pennies, or $\begin{aligned} 2^{1} \end{aligned}$ pennies.
::第二天,你得到两个便士, 或21便士。

On the third day, you get 4 pennies, or $\begin{aligned} 2^{2} \end{aligned}$ pennies. Do you see the pattern yet?
::第三天,你得到4个便士,或22个便士。你看到模式了吗?

On the fourth day, you get 8 pennies, or $\begin{aligned} 2^{3} \end{aligned}$ pennies. Each day, the exponent is one less than the number of that day.
::第四天,你得到8个便士或23个便士, 每天比那天少一个。

So on the thirtieth day, you get $\begin{aligned} 2^{29} \end{aligned}$ pennies, which is 536,870,912 pennies, or $5,368,709.12. That’s a lot more than a million dollars, even just counting the amount you get on that one day!
::因此,在第30天,你得到了229个便士,也就是536,870,912个便士,或者5,368,709.12美元。这个数字超过100万美元,甚至只是计算一天你得到的金额而已!

This problem is an example of a geometric sequence . In this section, we’ll find out what a geometric sequence is and how to solve problems involving geometric sequences.
::这个问题是几何序列的一个例子。在本节中,我们会发现几何序列是什么,以及如何解决几何序列的问题。

Identify a Geometric Sequence
::识别几何序列

A geometric sequence is a sequence of numbers in which each number in the sequence is found by multiplying the previous number by a fixed amount called the common ratio. In other words, the ratio between any term and the term before it is always the same. In the previous example the common ratio was 2, as the number of pennies doubled each day.
::几何序列是一个数字序列,通过将先前数字乘以一个固定数额,即共同比率来查找序列中的每个数字。换句话说,任何术语和之前的术语之间的比率总是相同的。在前一个例子中,共同比率是2,因为每天的便士数翻了一番。

The common ratio, $\begin{aligned} r \end{aligned}$ , in any geometric sequence can be found by dividing any term by the preceding term.
::在任何几何序列中,通过将任何术语除以前一术语即可找到共同比率r。

Here are some examples of geometric sequences and their common ratios.
::以下是几何序列及其共同比率的一些例子。

$\begin{aligned} 4, 16, 64, 256, \dots r = 4 (divide 16 by 4 to get 4) \\ 15, 30, 60, 120, \dots r = 2 (divide 30 by 15 to get 2) \\ 11, \frac{11}{2}, \frac{11}{4}, \frac{11}{8}, \frac{11}{16}, \dots r = \frac{1}{2} (divide \frac{11}{2} by 11 to get \frac{1}{2}) \\ 25, - 5, 1, - \frac{1}{5}, \frac{1}{25}, \dots r = - \frac{1}{5} (divide 1 by -5 to get - \frac{1}{5}) \end{aligned}$

::4,16,64,256,...r=4 (16乘4,得到4)15,30,60,120,r=2 (30乘15,得到2)11,112,114,118,1116...r=12(112乘11,得到12)25,-5,1,-15,125,...r=15(1乘15)

If we know the common ratio $\begin{aligned} r \end{aligned}$ , we can find the next term in the sequence just by multiplying the last term by $\begin{aligned} r \end{aligned}$ . Also, if there are any terms missing in the sequence, we can find them by multiplying the term before each missing term by the common ratio.
::如果我们知道共同比率 r, 我们就可以在顺序中找到下一个术语, 只需将最后一个术语乘以r。另外,如果在顺序中缺少任何术语, 我们可以通过将每个缺失术语之前的术语乘以共同比率来找到这些术语。

Filling in Missing Terms
::填补未填补的任期

Fill in the missing terms in each geometric sequence.
::填充每个几何序列中缺失的术语。

a) 1, ___, 25. 125, ___
:a) 1,___,25,125,___

First we can find the common ratio by dividing 125 by 25 to obtain $\begin{aligned} r = 5 \end{aligned}$ .
::首先,通过125除以25再除以25来获得r=5,我们就能找到共同比率。

To find the first missing term, we multiply 1 by the common ratio: $\begin{aligned} 1 \cdot 5 = 5 \end{aligned}$
::为了找到第一个缺失的学期,我们乘以1乘以共同比率:1:5=5

To find the second missing term, we multiply 125 by the common ratio: $\begin{aligned} 125 \cdot 5 = 625 \end{aligned}$
::为了找到第二个缺失的学期,我们乘以125乘以共同比率:1255=625

Sequence (a) becomes: 1, 5, 25, 125, 625,...
::序列(a)变成 1,5,25,125,625...

b) 20, ___, 5, ____. 1.25
:b) 20, ____, 5, ____. 1.25

We need to find the common ratio first, but how do we do that when we have no terms next to each other that we can divide?
::我们需要首先找到共同比率,但是,当我们彼此之间没有可以分割的条件时,我们如何做到呢?

Well, we know that to get from 20 to 5 in the sequence we must multiply 20 by the common ratio twice : once to get to the second term in the sequence, and again to get to five. So we can say $\begin{aligned} 20 \cdot r \cdot r = 5 \end{aligned}$ , or $\begin{aligned} 20 \cdot r^{2} = 5 \end{aligned}$ .
::我们知道,要按顺序从20乘20乘5,我们就必须将20乘以共同比率的两倍:一次乘以第二学期,另一次乘以五。因此我们可以说20r=5,或20r2=5。

Dividing both sides by 20, we get $\begin{aligned} r^{2} = \frac{5}{20} = \frac{1}{4} \end{aligned}$ , or $\begin{aligned} r = \frac{1}{2} \end{aligned}$ (because $\begin{aligned} \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \end{aligned}$ ).
::双方分20分,我们得到2=520=14,或r=12(因为12=14)。

To get the first missing term, we multiply 20 by $\begin{aligned} \frac{1}{2} \end{aligned}$ and get 10.
::为了获得第一个缺失的学期我们乘以20乘以12 乘以10

To get the second missing term, we multiply 5 by $\begin{aligned} \frac{1}{2} \end{aligned}$ and get 2.5.
::要获得第二个缺失的学期, 我们乘以5乘以12乘以2.5。

Sequence (b) becomes: 20, 10, 5, 2.5, 1.25,...
::序列(b) 变成: 20, 10, 5, 5, 2.5, 1. 25,...

You can see that if we keep multiplying by the common ratio, we can find any term in the sequence that we want—the tenth term, the fiftieth term, the thousandth term.... However, it would be awfully tedious to keep multiplying over and over again in order to find a term that is a long way from the start. What could we do instead of just multiplying repeatedly?
::你可以看到,如果我们继续以共同比率乘以,我们就可以在我们想要的顺序中找到任何术语——第十个术语,第五十个术语,第一千个术语。......然而,为了找到一个从开始起就很长的术语,不断反复地乘以一个数字,将是非常无聊的。我们可以做些什么,而不是反复地重复?

Let’s look at a geometric sequence that starts with the number 7 and has common ratio of 2.
::让我们看看几何序列, 从数字7开始, 共同比率为2。

$\begin{aligned} The 1^{s t} term is: & 7 = 7 \cdot 2^{0} \\ We obtain the 2^{n d} term by multiplying by 2 : & 7 \cdot 2 = 7 \cdot 2^{1} \\ We obtain the 3^{r d} term by multiplying by 2 again: & 7 \cdot 2 \cdot 2 = 7 \cdot 2^{2} \\ We obtain the 4^{t h} term by multiplying by 2 again: & 7 \cdot 2 \cdot 2 \cdot 2 = 7 \cdot 2^{3} \\ We obtain the 5^{t h} term by multiplying by 2 again: & 7 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 7 \cdot 2^{4} \\ The nth term would be: & 7 \cdot 2^{n - 1} \end{aligned}$

::第一个学期是:7=7+20 我们获得第二个学期,乘以2: 72=721 我们获得第三个学期,再乘以2: 72=722 我们获得第四个学期,再乘以2: 722=723 我们获得第五个学期,再乘以2: 72222=724

The nth term is $\begin{aligned} 7 \cdot 2^{n - 1} \end{aligned}$ because the 7 is multiplied by 2 once for the $\begin{aligned} 2^{n d} \end{aligned}$ term, twice for the third term, and so on—for each term, one less time than the term’s place in the sequence. In general, we write a geometric sequence with $\begin{aligned} n \end{aligned}$ terms like this:
::第n个学期为7x2乘以2,第2个学期乘以2,第3个学期乘以2,第3个学期则乘以2,等等——每个学期比该学期在顺序中的位置少一个学期。

$\begin{aligned} a, a r, a r^{2}, a r^{3}, \dots, a r^{n - 1} \end{aligned}$

::a,ar,ar2,ar3,... arn -1,ar2,ar3,...

The formula for finding a specific term in a geometric sequence is:
::在几何序列中找到一个具体术语的公式是:

$\begin{aligned} n^{t h} \end{aligned}$ term in a geometric sequence: $\begin{aligned} a_{n} = a_{1} r^{n - 1} \end{aligned}$
::几何序列中的 nth 术语: an=a1rn- 1

( $\begin{aligned} a_{1} = \end{aligned}$ first term, $\begin{aligned} r = \end{aligned}$ common ratio)
:a1=第一任期,r=共同比率)

Finding a Specific Term in a Sequence
::在一个序列中查找一个特定术语

For each of these geometric sequences, find the eighth term in the sequence.
::对于这些几何序列中的每一序列, 在序列中找到第八个术语。

a) 1, 2, 4,...
:a) 1, 2, 4,...

First we need to find the common ratio: $\begin{aligned} r = \frac{2}{1} = 2 \end{aligned}$ .
::首先,我们需要找到共同比率:r=21=2。

The eighth term is given by the formula $\begin{aligned} a_{8} = a_{1} r^{7} = 1 \cdot 2^{7} = 128 \end{aligned}$
::第八任期由公式A8=a1r7=127=128给出。

In other words, to get the eighth term we start with the first term, which is 1, and then multiply by 2 seven times.
::换句话说,为了获得第八学期,我们从第一个学期开始,第一个学期是1,然后乘以2,7倍。

b) 16, -8, 4, -2, 1,...
:b) 16,8,8,4,2,1,...

The common ratio is $\begin{aligned} r = \frac{- 8}{16} = \frac{- 1}{2} \end{aligned}$
::共同比率为 1816 12

The eighth term in the sequence is $\begin{aligned} a_{8} = a_{1} r^{7} = 16 \cdot {(\frac{- 1}{2})}^{7} = 16 \cdot \frac{(- 1)^{7}}{2^{7}} = 16 \cdot \frac{- 1}{2^{7}} = \frac{- 16}{128} = - \frac{1}{8} \end{aligned}$
::顺序中的第八词是a8=a1r7=16(-12)7=16(-1)727=161271612818。

Let’s take another look at the terms in that second sequence. Notice that they alternate positive, negative, positive, negative all the way down the list. When you see this pattern, you know the common ratio is negative; multiplying by a negative number each time means that the sign of each term is opposite the sign of the previous term.
::让我们再看看第二个序列中的术语。请注意,它们会从列表中一路转换成正、负、正、负。当你看到这个模式时,你知道共同比率是负的;每次乘以负数意味着每个术语的符号与上一个术语的符号相反。

Solve Real-World Problems Involving Geometric Sequences
::解决涉及几何序列的现实世界问题

Let’s solve two application problems involving geometric sequences.
::让我们解决涉及几何序列的两个应用问题。

Real-World Application: Chess
::真实世界应用程序:象棋

A courtier presented the Indian king with a beautiful, hand-made chessboard. The king asked what he would like in return for his gift and the courtier surprised the king by asking for one grain of rice on the first square, two grains on the second, four grains on the third, etc. The king readily agreed and asked for the rice to be brought. (From Meadows et al. 1972, via Porritt 2005) How many grains of rice does the king have to put on the last square?
::国王问他想要什么作为礼物的回报? 国王要求第一广场一粒米,第二广场两粒,第三广场四粒,等等。国王欣然同意并请求将米带来。 (Meadows等人,1972年,通过Porritt,2005年)。

A chessboard is an $\begin{aligned} 8 \times 8 \end{aligned}$ square grid, so it contains a total of 64 squares.
::棋盘是一个8×8平方格, 所以它总共包含64平方。

The courtier asked for one grain of rice on the first square, 2 grains of rice on the second square, 4 grains of rice on the third square and so on. We can write this as a geometric sequence:
::宫廷要求第一广场一粒大米,第二广场两粒大米,第三广场四粒大米等等。

1, 2, 4,...

The numbers double each time, so the common ratio is $\begin{aligned} r = 2 \end{aligned}$ .
::数字每次翻一番,共同比率为r=2。

The problem asks how many grains of rice the king needs to put on the last square, so we need to find the $\begin{aligned} 64^{t h} \end{aligned}$ term in the sequence. Let’s use our formula:
::问题在于国王需要投入多少稻谷才能进入最后一个广场, 所以我们需要找到第64个学期的顺序。让我们使用我们的公式:

$\begin{aligned} a_{n} = a_{1} r^{n - 1} \end{aligned}$ , where $\begin{aligned} a_{n} \end{aligned}$ is the nth term, $\begin{aligned} a_{1} \end{aligned}$ is the first term and $\begin{aligned} r \end{aligned}$ is the common ratio.
::an=a1rn-1, 其中, a1 是第 n 个学期, a1 是第一个学期, r 是共同比率。

$\begin{aligned} a_{64} = 1 \cdot 2^{63} = 9, 223, 372, 036, 854, 775, 808 \end{aligned}$ grains of rice.
::a64=1263=9,223,372,036,854,775,808大米。

The problem we just solved has real applications in business and technology. In technology strategy, the Second Half of the Chessboard is a phrase, coined by a man named Ray Kurzweil, in reference to the point where an exponentially growing factor begins to have a significant economic impact on an organization's overall business strategy.
::我们刚刚解决的问题在商业和技术方面有真正的应用。在技术战略中,切斯板的后半部分是一个短语,由叫雷·库尔兹魏尔的人发明,指的是一个指数增长因素开始对一个组织的总体商业战略产生重大经济影响。

The total number of grains of rice on the first half of the chessboard is $\begin{aligned} 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 \dots + 2, 147, 483, 648, \end{aligned}$ for a total of exactly 4,294,967,295 grains of rice, or about 100,000 kg of rice (the mass of one grain of rice being roughly 25 mg). This total amount is about $\begin{aligned} \frac{1}{1, 000, 000^{t h}} \end{aligned}$ of total rice production in India in the year 2005 and is an amount the king could surely have afforded.
::棋盘前半部分的稻米总产量为1+2+4+8+8+16+32+64+128+256+512+1024...+2,147,483,648,共4,294,967,295大米,即约10万公斤大米(大米质量约为25毫克),相当于2005年印度稻米总产量的11,000万分左右,是国王可以肯定提供的数额。

The total number of grains of rice on the second half of the chessboard is $\begin{aligned} 2^{32} + 2^{33} + 2^{34} \dots + 2^{63} \end{aligned}$ , for a total of 18, 446, 744, 069, 414, 584, 320 grains of rice. This is about 460 billion tons, or 6 times the entire weight of all living matter on Earth. The king didn’t realize what he was agreeing to—perhaps he should have studied algebra!
::棋盘后半部分的稻谷总数是232+233+234...+263, 共18, 446, 744, 069, 414, 584, 320 谷物大米。这大约是460亿吨, 相当于地球上所有生物物质全部重量的6倍。国王没有意识到他同意什么 — — 也许他应该研究代数!

Examples
::实例

A super-ball has a 75% rebound ratio—that is, when it bounces repeatedly, each bounce is 75% as high as the previous bounce.
::超级球的反弹率是75%, 也就是说,当它反复反弹时, 每个反弹率都比前一次反弹高75%。

We can write a geometric sequence that gives the height of each bounce with the common ratio of $\begin{aligned} r = \frac{3}{4} \end{aligned}$ :
::我们可以写出一个几何序列, 以 r=34 的通用比例, 使每个弹跳的高度达到 r=34 :

$\begin{aligned} 20, 20 \cdot \frac{3}{4}, 20 \cdot {(\frac{3}{4})}^{2}, 20 \cdot {(\frac{3}{4})}^{3} \dots \end{aligned}$

When you drop it from a height of 20 feet:
::当你从20英尺高处掉下来的时候

Example 1
::例1

How high does the ball bounce after it strikes the ground for the third time?
::第三次撞击地面后球弹出多高?

The ball starts at a height of 20 feet; after the first bounce it reaches a height of $\begin{aligned} 20 \cdot \frac{3}{4} = 15 f e e t \end{aligned}$ .
::球开始高度为20英尺;第一次弹跳后,球开始高度为20-34=15英尺。

After the second bounce it reaches a height of $\begin{aligned} 20 \cdot {(\frac{3}{4})}^{2} = 11.25 f e e t \end{aligned}$ .
::第二次反弹后,高度达到20(34)(342)=11.25英尺。

After the third bounce it reaches a height of $\begin{aligned} 20 \cdot {(\frac{3}{4})}^{3} = 8.44 f e e t \end{aligned}$ .
::第三次反弹后,高度达到20(343)3=8.44英尺。

Example 2
::例2

How high does the ball bounce after it strikes the ground for the seventeenth time?
::17次撞击地面后球弹出多高?

Notice that the height after the first bounce corresponds to the second term in the sequence, the height after the second bounce corresponds to the third term in the sequence and so on.
::请注意,第一次反弹之后的高度与顺序中的第二个学期对应,第二次反弹之后的高度与顺序中的第三个学期对应,等等。

This means that the height after the seventeenth bounce corresponds to the $\begin{aligned} 18^{t h} \end{aligned}$ term in the sequence. You can find the height by using the formula for the $\begin{aligned} 18^{t h} \end{aligned}$ term:
::这意味着第十七弹跳后的高度与序列中的第十八学期对应。您可以使用第十八学期的公式找到高度 :

$\begin{aligned} a_{18} = 20 \cdot {(\frac{3}{4})}^{17} = 0.15 f e e t . \end{aligned}$

::a18=20(34)17=0.15英尺。

Here is a graph that represents this information. (The heights at points other than the top of each bounce are just approximations.)
::这里的图表代表此信息。 (每次反弹顶部以外的点的高度只是近似值。)

Review
::回顾

Determine the first five terms of each geometric sequence.
::确定每个几何序列的前五个条件。

$\begin{aligned} a_{1} = 2, r = 3 \end{aligned}$
::a1=2,r=3

$\begin{aligned} a_{1} = 90, r = - \frac{1}{3} \end{aligned}$
::a1=90,r=13

$\begin{aligned} a_{1} = 6, r = - 2 \end{aligned}$
::a1=6,r%2

$\begin{aligned} a_{1} = 1, r = 5 \end{aligned}$
::a1=1,r=5

$\begin{aligned} a_{1} = 5, r = 5 \end{aligned}$
::a1=5,r=5

$\begin{aligned} a_{1} = 25, r = 5 \end{aligned}$
::a1=25,r=5

What do you notice about the last three sequences?
::你注意到最后三个序列是什么?

Find the missing terms in each geometric sequence.
::在每个几何序列中找到缺失的术语。

3, __ , 48, 192, __

81, __ , __ , __ , 1

$\begin{aligned} \frac{9}{4}, \underline{}, \underline{}, \frac{2}{3}, \underline{} \end{aligned}$

2, __ , __ , -54, 162

Find the indicated term of each geometric sequence.
::查找每个几何序列的指定术语。

$\begin{aligned} a_{1} = 4, r = 2 \end{aligned}$ ; find $\begin{aligned} a_{6} \end{aligned}$
::a1=4,r=2; 查找 a6

$\begin{aligned} a_{1} = - 7, r = - \frac{3}{4} \end{aligned}$ ; find $\begin{aligned} a_{4} \end{aligned}$
::a17,r34; 找到a4

$\begin{aligned} a_{1} = - 10, r = - 3 \end{aligned}$ ; find $\begin{aligned} a_{10} \end{aligned}$
::a110,r3; 找到 a10

In a geometric sequence, $\begin{aligned} a_{3} = 28 \end{aligned}$ and $\begin{aligned} a_{5} = 112; \end{aligned}$ find $\begin{aligned} r \end{aligned}$ and $\begin{aligned} a_{1} \end{aligned}$ .
::在几何序列中, a3=28 和 a5=112; 查找 r 和 a1 。

In a geometric sequence, $\begin{aligned} a_{2} = 28 \end{aligned}$ and $\begin{aligned} a_{5} = 112; \end{aligned}$ find $\begin{aligned} r \end{aligned}$ and $\begin{aligned} a_{1} \end{aligned}$ .
::在几何序列中, a2=28 和 a5=112; 查找 r 和 a1 。

As you can see from the previous two questions, the same terms can show up in sequences with different ratios.

Write a geometric sequence that has 1 and 9 as two of the terms (not necessarily the first two).
::写一几何序列,以1和9为两个术语(不一定是前两个)。

Write a different geometric sequence that also has 1 and 9 as two of the terms.
::写一个不同的几何序列,其中1和9是两个术语。

Write a geometric sequence that has 6 and 24 as two of the terms.
::写一个几何序列, 以6和24作为两个术语。

Write a different geometric sequence that also has 6 and 24 as two of the terms.
::写一个不同的几何序列,其中6个和24个作为两个术语。

What is the common ratio of the sequence whose first three terms are 2, 6, 18?
::前三个任期为2、6、18的顺序的常见比率是多少?

What is the common ratio of the sequence whose first three terms are 18, 6, 2?
::前三个任期为18、6、2的顺序的共同比率是多少?

What is the relationship between those ratios?
::这些比率之间有什么关系?

::从前两个问题中可以看到,相同的术语可以以不同比率的顺序出现。写一个以1和9为两个术语的几何序列(不一定是前两个)。写一个以1和9为两个术语的不同几何序列;写一个以6和24为两个术语的几何序列;写一个以6和24为两个术语的不同几何序列;写一个以2、6、18为两个术语的不同几何序列。前三个术语为2、6、6、18的顺序的共同比率是什么?前三个术语为18、6、2的顺序的共同比率是什么?这些比率之间的关系是什么?

Anne goes bungee jumping off a bridge above water. On the initial jump, the bungee cord stretches by 120 feet. On the next bounce the stretch is 60% of the original jump and each additional bounce the rope stretches by 60% of the previous stretch.

What will the rope stretch be on the third bounce?
::第三次弹跳时的绳索拉长是什么?

What will be the rope stretch be on the $\begin{aligned} 12^{t h} \end{aligned}$ bounce?
::第12次弹跳时的绳索拉长是多少?

::Anne从水面上的桥上跳了起来。在最初跳跃时,大腿的绳子延展120英尺。在下一个跳跃时,大腿的绳子伸展是最初跳跃的60%,而每额外跳动的绳子伸展是以前跳跃的60%。第三跳跃时,绳子的绳子伸展是多少?第12跳跃时,绳子的绳子伸展是多少?

Review (Answers)
::回顾(答复)

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