Lesson 12.1: Sequences

New Concept

A sequence is an ordered list of numbers following a specific rule. We'll learn to write a sequence's terms from its general formula, $a_n$, find the formula from its terms, and use factorial ($n!$) and summation ($Σ$) notations.

What’s next

Now, let's put this into practice. You'll start by writing the first few terms of a sequence with interactive examples and then move on to finding the general formula in a series of practice cards.

Property

A sequence is a function whose domain is the counting numbers. A sequence can also be seen as an ordered list of numbers and each number in the list is a term. A sequence may have an infinite number of terms (infinite sequence) or a finite number of terms (finite sequence). The notation $a_n$ represents the $n$th term of the sequence.

Examples

• Write the first four terms of the sequence with general term $a_n = 3n + 2$. The terms are $a_1 = 3(1)+2=5$, $a_2 = 3(2)+2=8$, $a_3 = 3(3)+2=11$, and $a_4 = 3(4)+2=14$. The sequence is $5, 8, 11, 14, \ldots$.

• Write the first four terms of the sequence with general term $a_n = (-1)^n(n+1)$. The terms are $a_1 = (-1)^1(1+1)=-2$, $a_2 = (-1)^2(2+1)=3$, $a_3 = (-1)^3(3+1)=-4$, and $a_4 = (-1)^4(4+1)=5$. The sequence is $-2, 3, -4, 5, \ldots$.

Property

The general term of the sequence is found from the formula for writing the $n$th term of the sequence. The $n$th term of the sequence, $a_n$, is the term in the $n$th position where $n$ is a value in the domain.

Examples

• Find a general term for the sequence $5, 10, 15, 20, 25, \ldots$. Each term is 5 times its position number, $n$. So, the general term is $a_n = 5n$.

• Find a general term for the sequence $1, -2, 3, -4, 5, \ldots$. The numbers are the position, $n$, but the signs alternate, starting with positive. So, the general term is $a_n = (-1)^{n+1}n$.

Property

If $n$ is a positive integer, then $n!$ is

We define $0!$ as 1, so $0! = 1$.

Examples

• Write the first four terms of the sequence $a_n = \frac{1}{(n-1)!}$. The terms are $a_1 = \frac{1}{0!} = 1$, $a_2 = \frac{1}{1!} = 1$, $a_3 = \frac{1}{2!} = \frac{1}{2}$, and $a_4 = \frac{1}{3!} = \frac{1}{6}$.

• Write the first four terms of the sequence $a_n = \frac{(n+1)!}{n!}$. The terms are $a_1 = \frac{2!}{1!} = 2$, $a_2 = \frac{3!}{2!} = 3$, $a_3 = \frac{4!}{3!} = 4$, and $a_4 = \frac{5!}{4!} = 5$. This simplifies to $a_n = n+1$.

Property

The sum of the first $n$ terms of a sequence whose $n$th term is $a_n$, is written in summation notation as:

The $i$ is the index of summation and the 1 tells us where to start and the $n$ tells us where to end. When we add a finite number of terms, we call the sum a partial sum.

Examples

• Expand and find the value of the partial sum $\sum_{i=1}^{4} (i+2)$. This is $(1+2) + (2+2) + (3+2) + (4+2) = 3 + 4 + 5 + 6 = 18$.

• Write the sum $2+4+6+8+10$ using summation notation. The terms follow the pattern $2i$ where $i$ goes from 1 to 5. So, the notation is $\sum_{i=1}^{5} 2i$.