Lesson 13.1: Sequences and Their Notations

New Concept

A sequence is an ordered list of numbers, essentially a function whose domain is the set of counting numbers. We'll explore two ways to define its terms: explicitly, using the term's position $n$, and recursively, using preceding terms to find the next.

What’s next

Next, you’ll tackle practice cards on writing sequence terms from explicit and recursive formulas, and then master factorial notation with interactive examples.

Property

A sequence is a function whose domain is the set of positive integers. A finite sequence is a sequence whose domain consists of only the first $n$ positive integers. The numbers in a sequence are called terms. The variable $a$ with a number subscript is used to represent the terms in a sequence and to indicate the position of the term in the sequence.

We call $a_1$ the first term of the sequence, $a_2$ the second term of the sequence, $a_3$ the third term of the sequence, and so on. The term $a_n$ is called the $n$th term of the sequence, or the general term of the sequence. An explicit formula defines the $n$th term of a sequence using the position of the term. A sequence that continues indefinitely is an infinite sequence.

To write the first $n$ terms of a sequence given an explicit formula:

1. Substitute $n=1$ into the formula to find the first term, $a_1$.
2. Substitute $n=2$ to find the second term, $a_2$.
3. Continue this process for all $n$ terms.

Examples

• For the sequence $a_n = 4n - 5$, the first three terms are: $a_1 = 4(1) - 5 = -1$, $a_2 = 4(2) - 5 = 3$, and $a_3 = 4(3) - 5 = 7$. The sequence is $\{-1, 3, 7, \ldots\}$.
• For the sequence $a_n = 10 - 2n$, the first three terms are: $a_1 = 10 - 2(1) = 8$, $a_2 = 10 - 2(2) = 6$, and $a_3 = 10 - 2(3) = 4$. The sequence is $\{8, 6, 4, \ldots\}$.
• For the sequence $a_n = n^2 + 1$, the first three terms are: $a_1 = 1^2 + 1 = 2$, $a_2 = 2^2 + 1 = 5$, and $a_3 = 3^2 + 1 = 10$. The sequence is $\{2, 5, 10, \ldots\}$.

Property

An alternating sequence is one where the terms alternate in sign (positive, negative, positive, etc.). This is often achieved by including a factor of $(-1)$ raised to a power involving $n$, such as $(-1)^n$ or $(-1)^{n+1}$.

To write the terms of an alternating sequence:

1. Substitute each value of $n$ into the formula, starting with $n=1$ for the first term.
2. The factor $(-1)^k$ determines the sign. If the exponent $k$ is odd, the result is negative. If $k$ is even, the result is positive.
3. Continue for all required terms.

Examples

• For the sequence $a_n = (-1)^n(2n)$, the first three terms are: $a_1 = (-1)^1(2(1)) = -2$, $a_2 = (-1)^2(2(2)) = 4$, and $a_3 = (-1)^3(2(3)) = -6$. The sequence is $\{-2, 4, -6, \ldots\}$.
• For the sequence $a_n = \frac{(-1)^{n+1}}{n}$, the first three terms are: $a_1 = \frac{(-1)^2}{1} = 1$, $a_2 = \frac{(-1)^3}{2} = -\frac{1}{2}$, and $a_3 = \frac{(-1)^4}{3} = \frac{1}{3}$. The sequence is $\{1, -\frac{1}{2}, \frac{1}{3}, \ldots\}$.
• For the sequence $a_n = \frac{n}{(-2)^n}$, the first three terms are: $a_1 = \frac{1}{(-2)^1} = -\frac{1}{2}$, $a_2 = \frac{2}{(-2)^2} = \frac{2}{4} = \frac{1}{2}$, and $a_3 = \frac{3}{(-2)^3} = -\frac{3}{8}$. The sequence is $\{-\frac{1}{2}, \frac{1}{2}, -\frac{3}{8}, \ldots\}$.

Property

A piecewise explicit formula uses different rules for different terms, depending on whether $n$ meets a certain condition. Each condition defines a subsection of the sequence with its own unique formula.

To write the terms of a piecewise sequence:

1. For each term, identify which condition the value of $n$ satisfies.
2. Use the formula corresponding to that condition to calculate the term's value.
3. Continue this process for each term you need to find.

Examples

• For $a_n = \begin{cases} n^2 & \text{if } n \text{ is odd} \\ 2n & \text{if } n \text{ is even} \end{cases}$, the first four terms are: $a_1=1^2=1$, $a_2=2(2)=4$, $a_3=3^2=9$, $a_4=2(4)=8$. The sequence is $\{1, 4, 9, 8, \ldots\}$.
• For $a_n = \begin{cases} 10 & \text{if } n \text{ is a multiple of } 3 \\ n & \text{if not} \end{cases}$, the first five terms are: $a_1=1$, $a_2=2$, $a_3=10$, $a_4=4$, $a_5=5$. The sequence is $\{1, 2, 10, 4, 5, \ldots\}$.
• For $a_n = \begin{cases} n+5 & \text{if } n \le 3 \\ n-5 & \text{if } n > 3 \end{cases}$, the first five terms are: $a_1=1+5=6$, $a_2=2+5=7$, $a_3=3+5=8$, $a_4=4-5=-1$, $a_5=5-5=0$. The sequence is $\{6, 7, 8, -1, 0, \ldots\}$.

Property

Given the first few terms of a sequence, you can find an explicit formula by looking for a pattern. This involves working in reverse to determine the rule that generates the sequence.

To find an explicit formula:

1. Look for a pattern among the terms (e.g., arithmetic, geometric, powers).
2. If terms are fractions, analyze patterns in the numerators and denominators separately.
3. If signs alternate, include a factor like $(-1)^n$ or $(-1)^{n+1}$.
4. Write a formula for $a_n$ in terms of $n$ and test it for the first few terms.

Examples

• For the sequence $\{3, 5, 7, 9, \ldots\}$, each term is one more than twice its position. The explicit formula is $a_n = 2n + 1$.
• For the sequence $\{-\frac{1}{2}, \frac{1}{4}, -\frac{1}{8}, \frac{1}{16}, \ldots\}$, the signs alternate and the denominators are powers of 2. The explicit formula is $a_n = \frac{(-1)^n}{2^n}$.
• For the sequence $\{\frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \ldots\}$, the numerator is $n+1$ and the denominator is $n+2$. The explicit formula is $a_n = \frac{n+1}{n+2}$.

Property

A recursive formula is a formula that defines each term of a sequence using preceding term(s). Recursive formulas must always state the initial term, or terms, of the sequence.

To write the terms of a sequence given a recursive formula:

1. Identify the given initial term, $a_1$.
2. To find the second term, $a_2$, substitute the value of the initial term ($a_1$) into the formula for $a_{n-1}$.
3. To find the third term, $a_3$, substitute the value of the second term ($a_2$) into the formula. Repeat for all subsequent terms.

Examples

• Given $a_1 = 5$ and $a_n = a_{n-1} + 4$ for $n \ge 2$, the first four terms are: $a_1=5$, $a_2=5+4=9$, $a_3=9+4=13$, $a_4=13+4=17$. The sequence is $\{5, 9, 13, 17, \ldots\}$.
• Given $a_1 = 2$ and $a_n = 3a_{n-1}$ for $n \ge 2$, the first four terms are: $a_1=2$, $a_2=3(2)=6$, $a_3=3(6)=18$, $a_4=3(18)=54$. The sequence is $\{2, 6, 18, 54, \ldots\}$.
• Given $a_1 = 100$ and $a_n = a_{n-1} - 7$ for $n \ge 2$, the first four terms are: $a_1=100$, $a_2=100-7=93$, $a_3=93-7=86$, $a_4=86-7=79$. The sequence is $\{100, 93, 86, 79, \ldots\}$.

Property

$n$ factorial, denoted $n!$, is the product of the positive integers from 1 to $n$. It is defined for a positive integer $n$ as:

The special case $0!$ is defined as $0! = 1$. The factorial of any whole number $n$ can also be expressed recursively as $n! = n(n-1)!$.

Examples

• To calculate $6!$, you multiply $6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 720$.
• For the sequence $a_n = \frac{n!}{2}$, the first three terms are: $a_1 = \frac{1!}{2} = \frac{1}{2}$, $a_2 = \frac{2!}{2} = \frac{2}{2} = 1$, and $a_3 = \frac{3!}{2} = \frac{6}{2} = 3$. The sequence is $\{\frac{1}{2}, 1, 3, \ldots\}$.
• For the sequence $a_n = \frac{10}{(n+1)!}$, the first two terms are: $a_1 = \frac{10}{(1+1)!} = \frac{10}{2!} = \frac{10}{2} = 5$, and $a_2 = \frac{10}{(2+1)!} = \frac{10}{3!} = \frac{10}{6} = \frac{5}{3}$. The sequence is $\{5, \frac{5}{3}, \ldots\}$.

Explanation

Factorial, written with an exclamation mark ($n!$), means to multiply a number by every whole number smaller than it, all the way down to 1. It is a shorthand for a long multiplication problem that appears in many formulas.