求等差数列

(Video) Arithmetic Progressions

by Eddie Woo

Here's some videos by Eddie Woo where he explains arithmetic sequences and series.

Recall that a set is a collection of numbers, a sequence is a set with order, and a series is the sum of a sequence.

In this video he goes over: $$\\\begin{align} &\circ\text{ Arithmetic Sequences}\\ &\circ\text{ Sum of an Arithmetic Sequence}\\ &\circ\text{ General Features}\\ &\circ\text{ Odd Number of Terms}\\ &\circ\text{ Proving it's Arithmetic}\\ \end{align}$$

Arithmetic Sequence

This type of sequence contains terms that have a common difference. The difference between the later and previous term is called the common difference.

An example would be the sequence of even numbers. $$2,4,6,\cdot\cdot\cdot$$ The common difference between each term is $2$. To get from term $1$ to term $2$ you just add two. The odd numbers are also an arithmetic sequence. They have the same difference, just a different starting point.

If we wanted to write the sequence of even numbers as a general algebraic expression we would look at how the term number differs from the terms. $$\begin{array}{cccc} 2,&4,&6,&\cdot\cdot\cdot\\ \uparrow&\uparrow&\uparrow\\ a_1&a_2&a_3 \end{array}$$ We can see that the term is double the term number. So our general algebraic expression would be: $$a_n=2n$$

What we find is that every arithmetic sequence can be written as some constant times $n$ plus another constant. In other words, they'll always have this form: $$a_n=kn+c$$ Here, $c$ tells you where you begin and $k$ tells you how far you progress when you move from term to term.

For example, the sequence of odd numbers would look like: $$a_n=2n-1$$

It would be really easy for us to define $T_n$ for the multiples of three now. We know they progress by three each time. $$a_n=3n$$

Sum of an Arithmetic Sequence

Suppose we wanted to find the sum of the sequence of odd numbers under $100$. $$1+3+5+7+\cdot\cdot\cdot+93+95+97+99$$ How would we do this? We could brute force it and just add every term but I personally don't want to do that. Instead, because of the orderliness of the series, there's a pattern that can be exploited.

Because the arithmetic series is defined by this common difference, you can take term $1$ and term $50$ (because $a_n=2n-1$ and if we plug in $50$ it gives us $99$) and add them together to make $100$. $${\color{#8925AE}{1}}+3+5+7+\cdot\cdot\cdot+93+95+97+{\color{#8925AE}{99}}\\ 1+99=100$$ Now if we go up one term from the first, we add two. And if we go down one term from the last term, we subtract two. Therefore, these two can be added together to create another $100$. $$1+{\color{#8925AE}{3}}+5+7+\cdot\cdot\cdot+93+95+{\color{#8925AE}{97}}+99\\ 3+97=100$$ You can keep on doing this! $$1+3+{\color{#8925AE}{5}}+7+\cdot\cdot\cdot+93+{\color{#8925AE}{95}}+97+99\\ 1+3+5+{\color{#8925AE}{7}}+\cdot\cdot\cdot+{\color{#8925AE}{93}}+95+97+99$$ But now the question becomes, how many times can you do this? How many of these $100$s are you going to get?

To get $100$ you need to combine two of the terms. You need a pair. So how many pairs can you make with $50$? $25$ pairs! In our sum we have the sum of these first $50$ terms equal to the $25$ pairs that add up to $100$. $$S_{50}=25\cdot100=2500$$

(Video) Arithmetic Progressions

by Eddie Woo

General Features of an Arithmetic Sequence/Series

Can we generalize this? Suppose we didn't have actual numbers. Instead, our first term was $a_1$ and the common difference between every number is $d$. So we could write this sequence as: $$\begin{array}{cccc} a_1,&a_1+d,&a_1+2d,&\cdot\cdot\cdot\\ \uparrow&\uparrow&\uparrow\\ a_1&a_2&a_3 \end{array}$$ So we can define $a_n$ as: $$a_n=a_1+(n-1)d$$ Now we need to find the sum of the whole thing, $S_n$. In the previous example, we combined together the first and the last term. Then we multiplied this by the amount of terms we had, $\frac{n}2$. $$S_n=\frac{n}2(a_1+a_n)$$ Here, $a_n$ stands for the last term. These were the numbers that, combined together, gave us the sum that kept repeating. In the previous example, that sum was $100$.

We know what the last term is because we defined what $a_n$ is. So we can simplify this. $$\begin{align} S_n&=\frac{n}2(a_1+a_n)\\ &=\frac{n}2(a_1+\color{#8925AE}{a_1+(n-1)d})\\ S_n&=\frac{n}2[2a_1+(n-1)d] \end{align}$$

The first equation, $S_n=\frac{n}2(a_1+a_n)$, is useful for when you know the first and the last term. The final equation, $S_n=\frac{n}2[2a_1+(n-1)d]$, is useful if you know the difference between each of the terms.

(Video) Arithmetic Progressions

by Eddie Woo

Odd Number of Terms

In the examples above, we had sequences/series with an even number of terms, and generalized the summation formula to be $$S_n=\frac{n}2[2a_1+(n-1)d]$$ Does this work when you have an odd number of terms?

Actually, yes! If we look at the same series from before, the odd numbers under $100$, but changed it to $98$, we would get an odd number of of terms. $$\begin{array}{cccccc} 1,&3,&5,&\cdot\cdot\cdot,&95,&97\\ \uparrow&\uparrow&\uparrow&&\uparrow&\uparrow\\ a_1&a_2&a_3&&a_{48}&a_{49} \end{array}$$

This seems like it wouldn't work. If we plug $n=49$ into our equation we get $\frac{49}2$ out front. We've introduced a fraction into this. How could the sum of whole numbers give us a fraction? But actually, let's look closer at the equation. $$S_n=\frac{49}{2}[{\color{#8925AE}{2a_1}}+(n-1)d]$$ Our $a_1$ term is being multiplied by $2$. Therefore, no matter the $a$, this will come out to be an even number. $$S_n=\frac{49}{2}[2a_1+{\color{#8925AE}{(n-1)d}}]$$ Since we have an odd number of terms, an odd number minus one gives us an even number. No matter the value of $d$, this part will stay even. Then, an even plus and even gives us an even.

Since we have an even number being divided by $2$, it doesn't cause us any fractions. Therefore, an odd number of terms still works for this equation.

Proving a Series/Sequence is Arithmetic

Often you'll be given a sequence or series and are asked to prove that it is arithmetic. To do this we want to show that every term minus the previous term is equal to some constant. $$a_n-a_{n-1}=d$$

If you were given the sequence $$1,3,5,7,\cdot\cdot\cdot,97$$ Then you would need to show a few $$a_2-a_1=2\\ a_3-a_2=2\\ a_4-a_3=2$$

If you were given the general algebraic expression for the $n^{\text{th}}$ term $$a_n=2n-1$$ Then you find the difference between this and the previous term. $$\begin{align} a_n-a_{n-1}&=(2n-1)-(2(n-1)-1)\\ &=(2n-1)-(2n-2-1)\\ &=(2n-1)-(2n-3)\\ &=2n-1-2n+3\\ &=2 \end{align}$$

(Video) A Formula for the $n^{\text{th}}$ Term

by PatrickJMT

This video by the PatrickJMT explains how to find the nth term of an arithmetic sequences.

$\large{\text{Summary}}$

An arithmetic sequence is one that increases by the same amount each term.

The formula is: $$\boxed{a_n=a_1+(n-1)d}$$ Where $a_1=\text{first term}$ and $d=\text{common difference}$.

How we Got This Formula

Let's use: $$3,8,13,18,23,28,\cdots$$ This sequence is arithmetic because each term differs by $5$ from the adjacent terms. So $d=5$.

Let's write our terms down. $$a_1=1$$ For $a_2$ we could write $8$ but let's incorporate our $d$. $$a_2=3+5$$ For $a_3$ we add $5$ a second time. $$a_3=3+2\cdot5$$ To get the $4$th term, we will have to add three $5$s. $$a_4=3+3\cdot5$$ This continues. $$a_5=3+4\cdot5\\ a_6=3+5\cdot5\\ a_7=3+6\cdot5$$

We can see we start with $3$, which is our $a_1$, and then add on $5$, which is our $d$, multiplied by one less than the position we're in. If we were to write a formula for the $n$th term it would be: $$a_n=a_1+(n-1)\cdot d$$

Example 1

Suppose we have: $$3,8,13,18,23,28,33,\cdots$$ Find $a_{10}$ and $a_{202}$.

Let's break out our handy new formula. The first term is, $a_1=3$ and the common difference is, $d=5$. Therefore, our formula is: $$\begin{align} a_n&=a_1+(n-1)d\\ a_n&=3+(n-1)5\\ a_n&=3+5n-5\\ a_n&=5n-2 \end{align}$$

Now that we have the formula, we can just plug in $n=10$ and $n=202$. $$\begin{align} a_{10}&=5(10)-2\\ &=50-2\\ &=48 \end{align}$$ $$\begin{align} a_{202}&=5(202)-2\\ &=1010-2\\ &=1008 \end{align}$$

Example 2

Suppose you have a sequence where $a_8=25$ and $a_{14}=43$. Find $a_1$ and $d$.

First, we want to write each of these given terms using the general formula. $$\begin{align} a_n&=a_1+(n-1)d\\ a_8&=a_1+(8-1)d\\ 25&=a_1+7d \end{align}$$ $$\begin{align} a_n&=a_1+(n-1)d\\ a_{14}&=a_1+(14-1)d\\ 43&=a_1+13d \end{align}$$

Next, take either formula and solve for $a_1$. $$\begin{align} 25&=a_1+7d\\ 25-7d&=a_1 \end{align}$$

Plug this into the other formula and solve for $d$. $$\begin{align} 43&=a_1+13d\\ 43&=(25-7d)+13d\\ 43&=25+6d\\ 18&=6d\\ 3&=d \end{align}$$

Now that we know $d$, we can plug this back into either formula. $$\begin{align} 25-7d&=a_1\\ 25-7(3)&=a_1\\ 25-21&=a_1\\ 4&=a_1 \end{align}$$

There are two main types of sequences we care about, because they have very noticeable patterns.

But back up for a second, what IS a sequence? It's a collection of numbers written in some order, where the order is important for identifying a pattern, if any. Here's an example:

$$\{0, 0, 0, 0, 0, 0\}$$

Well, that was boring. A sequence can have lots of different numbers. Here's another one:

$$\{0, 1, 4, 9, 16, 25\}$$

This sequence is the collection of the first six perfect squares. While this a commonly seen one, we really care about two more fundamental patterns in sequences.

An arithmetic sequence is one for which you can keep adding (or subtracting) the same number to get from one element of the sequence to the next. For instance,

$$\{2, 4, 6, 8, 10\}$$

is an arithmetic sequence. This is because $2 + 2 = 4,$ $4 + 2 = 6,$ and so on. We add $2$ to one term of the sequence to get the next term.

Yep, this certainly is an arithmetic sequence, meaning we can add a number to get from one term of the sequence to the next.

Actually, the sequence $\{1, 2, 4, 8, 16, 32\}$ is very special even though it's not an arithmetic sequence. We'll get back to that in a bit.

A geometric sequence is a sequence for which you can keep multiplying the same number to each term in the sequence to get the next one. Our sequence $\{1, 2, 4, 8, 16, 32\}$ is special because we can multiply each number by $2$ to get the next number. In particular,

$$1 \times 2 = 2$$ $$2 \times 2 = 4$$ $$4 \times 2 = 8$$

and so on.

Maximum Depth Exceeded: Fun with Sequences and Python, Part 2

To read part 1 of this story, go here.

Our Python code lets us generate a sequence based on a recursive formula, which is shown below:

What happens if we plug in $a(n)$ for the first $20$ values of $n$ (starting with $n = 0$)? Here's what we get:

This is an example of an arithmetic sequence, which is a sequence where the consecutive terms differ by the same amount. Since our recursive formula has a $-1,$ that difference is $-1.$

To read part 3 of this story, go here.