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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