Sequences

Published: Dec 29th, 2020

The next few lessons will discuss sequences. Your teacher may not cover this subject, but it does become relevant in upper-year math and in some computer science courses.

With that disclaimer out of the way, let's get to it!

A function that takes on natural number input will generate a set of values called a sequence

For example, the function \(f(n) = 3n+5\)\(n \in \mathbb{N}\), generates the set of values \(8, 11, 14, 16, ...\) when \(n\) is set as \(1,2,3,4,...\). Each number in a sequence is called a term

Some sequences are finite. For example, \(1,3,5,7,9,...,15\) is a finite sequence because it has a finite number of terms—eight.

Others are infinite\(1,3,5,7,9,...\) is infinite because it does not have an end term, so it has an infinite number of terms. 

In some cases, a sequence may be defined recursively. In other words, the sequence is defined relative to the term before! Sounds weird, right?

Let me show you an example: \(t_n = t_{n-1} + n\). To get the value \(t_n\), we need to add \(n\) to the value before, \(t_{n-1}\).

Now you might be wondering, "where do we get the very first value from?" Well, that value is called \(t_1\) and is usually already given in the question. 

Alright, with introductions out of the way, let's go through some examples:

Give the first five terms in \(f(n) = n+1\)\(n \in N\)

You might be surprised that we're using \(n\) as a variable rather than \(x\), to which you might have become accustomed. Well, that's a convention used to remind ourselves that our domain should be part of the natural number family, \(\mathbb{N}\)

In this case, for inputs \(1,2,3,4,5\), we'd get the outputs \(2,3,4,5,6\)

Remember that the natural numbers are the counting numbers, and they do not include zero, so we'd start at 1. For more on the different types of numbers, give Lesson #4: Types of Numbers a look.

That was pretty simple, no? Let's move on!

Give the first five terms in \(t_n = 2^{n-1}\)\(n \in N\) 

Whoa, where's the \(f(n)\)? Well, in sequences, we can alternatively use the \(t_n\) notation. This becomes especially useful when we discuss recursive sequences. I'll get to that shortly.

Let's do a few examples:

\(t_1 = 2^{1-1}\)

\(t_1 = 2^{0}\)

\(t_1 = 1\). So, our first term is 1

\(t_2 = 2^{2-1}\)

\(t_2 = 2^{1}\)

\(t_2 = 2\). The second term is 2. You get the gist.

Therefore, for the inputs \(1,2,3,4,5\), our outputs would be \(1,2,4,8,16\)

Now, recursive sequences. 

For the sequence \(t_1=5\)\(t_n = t_{n-1} + 3n\)\(n \in N\)\(n \gt 1\), find the next 4 terms.

Let's put the recursive definition into words: "the next value of the function is equal to the previous value, plus 3 times the term number."

Alright, now let's try to solve a few:

\(t_2 = t_{2-1} + 3(2)\)

\(t_2 = t_{1} + 6\)

\(t_2 = 5 + 6\)

\(t_2 = 11\). So, the second term is 11. Now, the next term:

\(t_3 = t_{3-1} + 3(3)\)

\(t_3 = t_{2} + 9\)

\(t_3 = 11 + 9\)

\(t_3 = 20\)

Continuing the pattern, we get \(20 + 12 = 32\), and \(32 + 15 = 47\).

In some cases, we may be given the sequence but are asked to find the function from which it came, also called the general term. Let's try a few examples:

For the sequence \(7,14,21,28,35,...\), find the general term

If the pattern here isn't immediately obvious, one way to approach the problem is using first differences, or the differences between consecutive terms.

For example, \(14-7= 7\)\(21-14 = 7\)\(28-21 = 7\), and so on. Note that the first differences are always the same. This means that the function here is linear, and follows the form \(ax+b\)

In this case, we can see that the differences between terms is always 7. We can therefore use this number as \(a\), or the rate of change / slope, giving \(7x + b\).

Let's try without a value for \(b\), or \(b=0\). This gives \(7x\). For inputs \(x=1,2,3,4,5\)\(f(x) = 7x\) gives \(7, 14, 21, 28, 35\). We've got it!

Therefore, our general term is \(t_n = 7n\)

For the sequence \(\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5},...\), find the general term

This pattern is a bit more complicated; let's express it in tabular format:

\(n\) \(f(n)\)
\(1\) \(\frac{1}{2}\)
\(2\) \(\frac{2}{3}\)
\(3\) \(\frac{3}{4}\)
\(4\) \(\frac{4}{5}\)

Notice that the numerator of each fraction is equal to \(n\).

What about the denominator? It's one more than \(n\), or \(n+1\).

Therefore, we can express the general term as \(f(n) = \frac{n}{n+1}\).

With that, we've covered the basics of sequences. Be sure to try some practice problems. 

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