A sequence can often be defined using an equation or rule, where the term number is the input and the output is the term or term value. Typically, the term number for the first term is 1, but it does not have to be. We often use subscript notation:

\displaystyle a_n

\bm{a_n}

The term value

\bm{n}

The term number

The subscripts tell us how terms relate to one another. For example, for the sequence:

\displaystyle a_{n-1}, \, a_n , \, a_{n+1}

\bm{a_n}

Current term

\bm{a_{n-1}}

Previous term (term before a_n)

\bm{a_{n+1}}

Next term (term after a_n)

The recursive formula for a sequence tell us how to go from one term to the next. Along with the formula, we must also be given the first term, which can be term 1 or term 0 depending on the context.

The explicit formula for a sequence tells us how to calculate the term value given the term number.

To write a formula for the sequence: -1, 4, 9, 14, \ldots we can either write in about how we get from one term to the next (recursive) or how we get the term value from the term number (explicit).

Recursive formula: a_n=a_{n-1}+5 and a_1=-1
- In words: To get the term value, add 5 to the previous term. The first term is -1.
Explicit formula: a_n=-6+5n
- In words: To get the term value, multiply the term number by 5 and add -6.

We can convert between the different formulas directly by noticing some key features of each or by using one formula to write out the sequence and then using that to write the other formula.

Finite sequence

A sequence that has a countable or limited number of terms

Example:

2, 4, 8, 16, 32 which is a_n=2^n with term numbers: \\n\in \left\{1, 2, 3, 4, 5\right\}

Infinite sequence

A sequence that has an infinite number of terms

Example:

2, 4, 8, 16, 32... which is a_n=2^n with term numbers: n\in \Z^{+}

Worked examples

Example 1

Consider the sequence:

1, 4,9, 16, 25\ldots

Identify a_1 and a_4.

Approach

The subscript tells us the term number, so this is asking for the first and fourth terms.

Solution

The first term, a_1 of the sequence is 1.

The fourth term, a_4 of the sequence is 16.

Reflection

We should assume that the first term has a term number of 1 unless we are explicitly told otherwise.

Find the next two terms in the sequence.

Approach

We want to identify the pattern and then use it to continue the sequence.

Solution

It can be observed that the terms in the sequence are perfect squares. The first term is 1^2=1, the 2nd term is 2^2=4 and the 5th term is 5^2=25. The next two terms, a_{6} and a_{7} should be 6^2=36 and 7^2=49 respectively.

The next two terms of the sequence are 36 and 49.

Write an explicit rule for the nth term of the sequence.

Solution

Because the terms are perfect squares, the nth term of the sequence can be found by raising each term to the 2nd power. The explicit formula for the sequence is

a_n= n^2

Reflection

We can check that our rule is correct by substituting in the first few values and making sure it matches the sequence.

n	a_n
1	a_1=1^2=1
2	a_1=2^2=4
3	a_3=3^2=9

Example 2

Find the first four terms of the sequence described by:

a_n = \dfrac{3 n - 1}{n^{2} + 4}

Solution

For a sequence, the first term a_1 is when n=1. The second term a_2 is when n=2, and so on.

The value of each term is determined by substituting n into the formula.

\displaystyle a_1	\displaystyle =	\displaystyle \dfrac{3 (1) - 1}{(1)^{2} + 4}	a_1=\dfrac{2}{5}
\displaystyle a_2	\displaystyle =	\displaystyle \dfrac{3 (2) - 1}{(2)^{2} + 4}	a_2=\dfrac{5}{8}
\displaystyle a_3	\displaystyle =	\displaystyle \dfrac{3 (3) - 1}{(3)^{2} + 4}	a_3=\dfrac{8}{13}
\displaystyle a_4	\displaystyle =	\displaystyle \dfrac{3 (4) - 1}{(4)^{2} + 4}	a_4=\dfrac{11}{20}

The first four terms of the sequence are \dfrac{2}{5}, \dfrac{5}{8}, \dfrac{8}{13}, and \dfrac{11}{20}.

Example 3

Consider each of the following formulas for sequences.

Convert the explicit formula t_n=3(2)^n to a recursive formula.

Approach

We can either use the explicit formula to generate the sequence and use that to write the recursive formula, or we can use prior knowledge about exponential models to determine the recursive formula.

Solution

We will generate the sequence and then use our prior knowledge to check.

Letting the first term number be n=1 we get the sequence: 6, 12, 24, 48, 96, \ldots

This sequence starts with term 6 and we multiply by 2 to get the next term. As a recursive formula we get: t_n=2t_{n-1}, \, t_1=6

An exponential model of the form t(n)=a(b)^n has a constant factor of b, so for t_n=3(2)^n we multiply by 2 between terms which is reflected in our recursive equation. We can just substitute n=1 to get the first term.

Reflection

There are many correct recursive formulas using different notation, including:

t_n=2t_{n-1}, \, t_0=3
t_{n+1}=2t_{n}, \, t_0=3
t_{n+1}=2t_{n}, \, t_1=6

Convert the recursive formula a_n=a_{n-1}+7, with a_1=-3 to an explicit formula.

Approach

We can see from the recursive formula that in words the pattern is "add 7 to go from one term to the next." This means it will be related to linear growth where 7 is the rate of change. We can use this or write out the sequence to convert.

Solution

Using that we have a constant rate of change of 7, we can start with an explicit formula of the form a_n=7n+b.

\displaystyle a_n	\displaystyle =	\displaystyle 7n+b	Using rate of change
\displaystyle a_1	\displaystyle =	\displaystyle 7(1)+b	Substitute n=1 since we are given a_1
\displaystyle -3	\displaystyle =	\displaystyle 7+b	Substitute a_1=-3 and simplify
\displaystyle -10	\displaystyle =	\displaystyle b	Subtraction property of equality

This means that our explicit formula will be: a_n=7n-10

Reflection

We can check this by generating the sequence using both formulas and ensuring they are the same.

Recursive: Start with -3 and add 7 each time gives: -3, 4, 11, 18, \ldots

Explicit: Substitute in n= 1, 2, 3, 4 gives:

a_1=7(1)-10=-3
a_2=7(2)-10=4
a_3=7(3)-10=11
a_4=7(4)-10=18

These match, so we have converted correctly.

Outcomes

M1.F.BF.A.2

Define sequences as functions, including recursive definitions, whose domain is a subset of the integers. Write explicit and recursive formulas for arithmetic and geometric sequences in context and connect them to linear and exponential functions.*

M1.MP1

Make sense of problems and persevere in solving them.

M1.MP2

Reason abstractly and quantitatively.

M1.MP3

Construct viable arguments and critique the reasoning of others.

M1.MP6

Attend to precision.

M1.MP7

Look for and make use of structure.

M1.MP8

Look for and express regularity in repeated reasoning.

7.01 Introduction to sequence notation

Concept summary

Worked examples

Example 1

Approach

Solution

Reflection

Approach

Solution

Solution

Reflection

Example 2

Solution

Example 3

Approach

Solution

Reflection

Approach

Solution

Reflection

Outcomes

M1.F.BF.A.2

M1.MP1

M1.MP2

M1.MP3

M1.MP6

M1.MP7

M1.MP8

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