Introduction to Sequences Lesson

In this lesson, we will introduce the concept of a sequence in math. A sequence is a function that is used to compute an ordered list. As an example, suppose the average family in Fall Springs consumes 2 pounds of meat each day. Let's use a function to represent the pounds of meat consumed by the average family in Fall Springs after n days. $$f(n) = 2n$$ If we let n = 1, 2, 3, 4,...

Days: n	Pounds of Meat Consumed: f(n)
1	2
2	4
3	6
...	...
9	18

Sequence Notation and Definitions

When we work with sequences, we will use a_n instead of the typical function notation f(n). $$f(n) = a_{n}$$

Using Function Notation:

$$f(n) = 5n + 1$$ $$f(1) = 5(1) + 1 = 5 + 1 = 6$$ $$f(2) = 5(2) + 1 = 10 + 1 = 11$$

Using Sequence Notation:

$$a_{n} = 5n + 1$$ $$a_{1} = 5(1) + 1 = 5 + 1 = 6$$ $$a_{2} = 5(2) + 1 = 10 + 1 = 11$$ The "n" is used as a reminder that the domain or set of allowable values for n is the set of Natural Numbers. $$ℕ: \{1, 2, 3, 4,...\}$$ A finite sequence is a function that has a fixed number of elements for the domain. In other words, we would work with the natural numbers but stop at some fixed value. For example, let's say our sequence is an ordered list of the daily high temperatures in Lakeside for the first three days in January.

Day: n	Temperature (in °F): an
1	42
2	45
3	39

An infinite sequence is a function that has the entire set of natural numbers for the domain. For example, let's say our sequence is just the odd natural numbers. To form this sequence, we would start at 1, and then we would keep adding 2 to get each additional term.

n	an
1	1
2	3
3	5
...	...
9	17

Each element in the range of a sequence is known as a term of the sequence. The general term, or what is known as the nth term, is a_n. The elements of both the domain and range of a sequence are ordered. This means the first term of the sequence is found by letting n = 1. Suppose we are given the following sequence: $$a_n = 3n$$ So for the first term of the sequence, we use the notation a₁ and evaluate by plugging in 1 for n. $$a_1 = 3(1) = 3$$ Then, for the second term of the sequence, we use the notation a₂ and evaluate by plugging in 2 for n. $$a_2 = 3(2) = 6$$ We can continue this process to find additional terms. Let's find the third, fourth, and fifth terms. $$a_3 = 3(3) = 9$$ $$a_4 = 3(4) = 12$$ $$a_5 = 3(5) = 15$$ Let's look at some examples.
Example #1: Find the first three terms of the sequence. $$a_n=11 - 20n$$ To find the first three terms, we start with the first natural number, which is 1 (n = 1). $$a_1=11 - 20(1)=-9$$ To find the second term, we plug in the next natural number, which is 2 (n = 2). $$a_2=11 - 20(2)=- 29$$ Lastly, to find the third term, we plug in the next natural number, which is 3 (n = 3). $$a_3=11 - 20(3)=-49$$ Example #2: Find the first five terms of the sequence. $$a_{n} = -\frac{13}{2} - \frac{3}{2}n$$ Just like in the last example, we will start with n = 1, then proceed to n = 2, and so on. $$a_{1} = -\frac{13}{2} - \frac{3}{2}(1) = -\frac{13}{2} - \frac{3}{2} = \frac{-13 - 3}{2} = \frac{-16}{2} = -8$$ $$a_{2} = -\frac{13}{2} - \frac{3}{2}(2) = -\frac{13}{2} - \frac{6}{2} = \frac{-13 - 6}{2} = \frac{-19}{2} = -\frac{19}{2}$$ $$a_{3} = -\frac{13}{2} - \frac{3}{2}(3) = -\frac{13}{2} - \frac{9}{2} = \frac{-13 - 9}{2} = \frac{-22}{2} = -11$$ $$a_{4} = -\frac{13}{2} - \frac{3}{2}(4) = -\frac{13}{2} - \frac{12}{2} = \frac{-13 - 12}{2} = \frac{-25}{2} = -\frac{25}{2}$$ $$a_{5} = -\frac{13}{2} - \frac{3}{2}(5) = -\frac{13}{2} - \frac{15}{2} = \frac{-13 - 15}{2} = \frac{-28}{2} = -14$$

Using a Recursive Definition

Our previous two examples were sequences defined with an explicit formula. This means the formula given does not depend on the previous term. In some cases, our sequence is defined by a recursive definition. This means each term after the first term or the first few terms is defined as some expression that involves the previous term or terms. Let's look at some examples.
Example #3: Find the first three terms of the sequence. $$a_{n + 1}=a_n \cdot 6$$ $$a_1=1$$ Here, we are given the first term: a₁ = 1. For the next term, we need to think about the notation: $$a_{n + 1}$$In this case, our a_{n + 1} would be a₂, which means a_n would be one less or a₁. All this is saying is to plug in the previous term and evaluate. $$a_2=a_1 \cdot 6$$ $$a_2=1 \cdot 6=6$$ To find the third term, now we plug in the second term, which is 6, and apply the same process. $$a_3=a_2 \cdot 6$$ $$a_3=6 \cdot 6=36$$ Example #4: Find the first five terms of the sequence. $$a_{n + 1} = a_{n} + \frac{1}{2}$$ $$a_{1} = -\frac{1}{3}$$ To find a₂, the a_n will be a₁. Again, each time we will use the previous value to plug in for a_n. $$a_{2} = -\frac{1}{3} + \frac{1}{2} = -\frac{2}{6} + \frac{3}{6} = \frac{1}{6}$$ $$a_{3} = \frac{1}{6} + \frac{1}{2} = \frac{1}{6} + \frac{3}{6} = \frac{4}{6} = \frac{2}{3}$$ $$a_{4} = \frac{2}{3} + \frac{1}{2} = \frac{4}{6} + \frac{3}{6} = \frac{7}{6}$$ $$a_{5} = \frac{7}{6} + \frac{1}{2} = \frac{7}{6} + \frac{3}{6} = \frac{10}{6} = \frac{5}{3}$$

Graphing a Sequence

In some cases, we will be asked to graph a sequence. $$a_{n} = 2n$$ To graph our a_n, we will plot points (n, 2n). Keep in mind here that we are just plotting the points. We don't want to connect them, as this graph is a set of discrete points (separate and distinct). When graphing a line, we have arrows that give a visual indication that it continues. Here, we can't really do that since we have these discrete points (separate and distinct). So if you are asked to graph a sequence, just make sure the pattern is clear.

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Skills Check: