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Sequences and Their Notations
Write the terms of a sequence defined by an explicit formula.
Sequences and Their Notations
- Write the first few terms of a sequence (IA 12.1.1)
- Find a formula for the general term (nth term) of a sequence (IA 12.1.2)
A patient takes a 30 mg antibiotic capsule. At the end of that hour, the amount of antibiotic remaining in her body is only 90% of the amount in the beginning of that hour. The 30mg dose is taken at time t = 1 hour. How much of this dose remains at the end of 1 hour? 2hours? 3 hours? 4 hours?
| Time t | Dose remaining after time t |
| 1 | 0.90(30)=27mg |
| 2 | 0.90(27)=24.3mg |
| 3 | 0.90(24.3)=21.87mg |
| 4 | 0.90(21.87)=19.68mg |
This ordered list of numbers 27, 24.3, 21.87, 19.68, … is a sequence. Each number in the list is a term.
A sequence is a function whose domain is the counting numbers. A sequence may have an infinite number of terms or a finite number of terms. Our sequence has three dots (ellipsis) at the end which indicates the list never ends. If the domain is the set of all counting numbers, then the sequence is an infinite sequence.
Often when working with sequences we do not want to write out all the terms. We want a more compact way to show how each term is defined. When we worked with functions, we wrote \(f(x)=2x\) and we said the expression 2x was the rule that defined values in the range.
While a sequence is a function, we do not use the usual function notation. Instead of writing the function as \(f(x)=2x\) , we would write it as \({a}_{n}=2n\) . The \({a}_{n}\) is the nth term of the sequence, the term in the nth position where n is a value in the domain. The formula for writing the nth term of the sequence is called the general term or formula of the sequence.
General sequence terms are denoted as follows:
\(\begin{array}{ll}{a}_{1}-\text{first}\text{term} & \\ {a}_{2}-\text{second}\text{term} & \\ {a}_{3}-\text{third}\text{term} & \\ . & \\ . & \\ . & \\ {a}_{n}-{n}^{th}\text{term} & \\ {a}_{n+1}-(n+1)\text{term} & \\ . & \\ . & \\ . & \end{array}\)
Example
Try it.
Write the first five terms of the sequence whose general term is \({a}_{n}=2n-7\) .
Solution
\(\begin{array}{llllll}n & 1 & 2 & 3 & 4 & 5 \\ {a}_{n} & {a}_{1} & {a}_{2} & {a}_{3} & {a}_{4} & {a}_{5} \\ 2n-7 & 2(1)-7 & 2(2)-7 & 2(3)-7 & 2(4)-7 & 2(5)-7 \\ & -5 & -3 & -1 & 1 & 3\end{array}\)
Write the first few terms of a sequence.
Try it.
Write the first five terms of the sequence whose general term is \({a}_{n}=4n+2\).
| \(n\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) |
| \({a}_{n}\) | \({a}_{1}\) | \({a}_{2}\) | \({a}_{3}\) | \({a}_{4}\) | \({a}_{5}\) |
| \(4n+2\) | |||||
Try it.
Write the first five terms of the sequence whose general term is \({a}_{n}={3}^{n}-1\).
| \(n\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) |
| \({a}_{n}\) | \({a}_{1}\) | \({a}_{2}\) | \({a}_{3}\) | \({a}_{4}\) | \({a}_{5}\) |
| \({3}^{n}-1\) | |||||
Condensed — the full section is in OpenStax College Algebra 2e.
Writing the Terms of a Sequence Defined by an Explicit Formula
One way to describe an ordered list of numbers is as a sequence. A sequence is a function whose domain is a subset of the counting numbers. The sequence established by the number of hits on the website is
\[\{2,4,8,16,32,\ldots \}.\]The ellipsis (…) indicates that the sequence continues indefinitely. Each number in the sequence is called a term. The first five terms of this sequence are 2, 4, 8, 16, and 32.
Listing all of the terms for a sequence can be cumbersome. For example, finding the number of hits on the website at the end of the month would require listing out as many as 31 terms. A more efficient way to determine a specific term is by writing a formula to define the sequence.
One type of formula is an explicit formula, which defines the terms of a sequence using their position in the sequence. Explicit formulas are helpful if we want to find a specific term of a sequence without finding all of the previous terms. We can use the formula to find the nth term of the sequence, where \(n\) is any positive number. In our example, each number in the sequence is double the previous number, so we can use powers of 2 to write a formula for the \(n\text{th}\) term.
The first term of the sequence is \({2}^{1}=2,\) the second term is \({2}^{2}=4,\) the third term is \({2}^{3}=8,\) and so on. The \(n\text{th}\) term of the sequence can be found by raising 2 to the \(n\text{th}\) power. An explicit formula for a sequence is named by a lower case letter \(a,b,c...\) with the subscript \(n.\) The explicit formula for this sequence is
\[{a}_{n}={2}^{n}.\]Now that we have a formula for the \(n\text{th}\) term of the sequence, we can answer the question posed at the beginning of this section. We were asked to find the number of hits at the end of the month, which we will take to be 31 days. To find the number of hits on the last day of the month, we need to find the 31st term of the sequence. We will substitute 31 for \(n\) in the formula.
\[\begin{array}{l}{a}_{31}={2}^{31} \\ \ =\text{2,147,483,648}\end{array}\]If the doubling trend continues, the company will get \(\text{2,147,483,648}\) hits on the last day of the month. That is over 2.1 billion hits! The huge number is probably a little unrealistic because it does not take consumer interest and competition into account. It does, however, give the company a starting point from which to consider business decisions.
| \(n\) | 1 | 2 | 3 | 4 | 5 | \(n\) |
| \(n\text{th}\) term of the sequence, \({a}_{n}\) | 2 | 4 | 8 | 16 | 32 | \({2}^{n}\) |
Condensed — the full section is in OpenStax College Algebra 2e.
Writing the Terms of a Sequence Defined by a Recursive Formula
Sequences occur naturally in the growth patterns of nautilus shells, pinecones, tree branches, and many other natural structures. We may see the sequence in the leaf or branch arrangement, the number of petals of a flower, or the pattern of the chambers in a nautilus shell. Their growth follows the Fibonacci sequence, a famous sequence in which each term can be found by adding the preceding two terms. The numbers in the sequence are 1, 1, 2, 3, 5, 8, 13, 21, 34,…. Other examples from the natural world that exhibit the Fibonacci sequence are the Calla Lily, which has just one petal, the Black-Eyed Susan with 13 petals, and different varieties of daisies that may have 21 or 34 petals.
Each term of the Fibonacci sequence depends on the terms that come before it. The Fibonacci sequence cannot easily be written using an explicit formula. Instead, we describe the sequence using a recursive formula, a formula that defines the terms of a sequence using previous terms.
A recursive formula always has two parts: the value of an initial term (or terms), and an equation defining \({a}_{n}\) in terms of preceding terms. For example, suppose we know the following:
\[\begin{array}{l}{a}_{1}=3 \\ {a}_{n}=2{a}_{n-1}-1\text{for}n\ge 2\end{array}\]We can find the subsequent terms of the sequence using the first term.
\[\begin{array}{l}{a}_{1}=3 \\ {a}_{2}=2{a}_{1}-1=2(3)-1=5 \\ {a}_{3}=2{a}_{2}-1=2(5)-1=9 \\ {a}_{4}=2{a}_{3}-1=2(9)-1=17\end{array}\]So the first four terms of the sequence are \(\{3,\ 5,\ 9,\ 17\}\) .
The recursive formula for the Fibonacci sequence states the first two terms and defines each successive term as the sum of the preceding two terms.
\[\begin{array}{l}{a}_{1}=1 \\ {a}_{2}=1 \\ {a}_{n}={a}_{n-1}+{a}_{n-2}\text{for}n\ge 3\end{array}\]To find the tenth term of the sequence, for example, we would need to add the eighth and ninth terms. We were told previously that the eighth and ninth terms are 21 and 34, so
\[{a}_{10}={a}_{9}+{a}_{8}=34+21=55\]Condensed — the full section is in OpenStax College Algebra 2e.
Using Factorial Notation
The formulas for some sequences include products of consecutive positive integers. \(n\) factorial, written as \(n!,\) is the product of the positive integers from 1 to \(n.\) For example,
\[\begin{array}{l}4!=4⋅3⋅2⋅1=24 \\ 5!=5⋅4⋅3⋅2⋅1=120\end{array}\]An example of formula containing a factorial is \({a}_{n}=(n+1)!.\) The sixth term of the sequence can be found by substituting 6 for \(n.\)
\[{a}_{6}=(6+1)!=7!=7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1=5040\]The factorial of any whole number \(n\) is \(n(n-1)!\) We can therefore also think of \(5!\) as \(5⋅4!\text{.}\)
Example
Try it.
Write the first five terms of the sequence defined by the explicit formula \({a}_{n}=\frac{5n}{(n+2)!}.\)
Solution
Substitute \(n=1,n=2,\) and so on in the formula.
\[\begin{array}{lllll}n=1 & & & & {a}_{1}=\frac{5(1)}{(1+2)!}=\frac{5}{3!}=\frac{5}{3\cdot 2\cdot 1}=\frac{5}{6} \\ n=2 & & & & {a}_{2}=\frac{5(2)}{(2+2)!}=\frac{10}{4!}=\frac{10}{4\cdot 3\cdot 2\cdot 1}=\frac{5}{12} \\ n=3 & & & & {a}_{3}=\frac{5(3)}{(3+2)!}=\frac{15}{5!}=\frac{15}{5\cdot 4\cdot 3\cdot 2\cdot 1}=\frac{1}{8} \\ n=4 & & & & {a}_{4}=\frac{5(4)}{(4+2)!}=\frac{20}{6!}=\frac{20}{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}=\frac{1}{36} \\ n=5 & & & & {a}_{5}=\frac{5(5)}{(5+2)!}=\frac{25}{7!}=\frac{25}{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}=\frac{5}{1\text{,}008}\end{array}\]The first five terms are \(\{\frac{5}{6},\frac{5}{12},\frac{1}{8},\frac{1}{36},\frac{5}{1,008}\}.\)
Key Concepts
- A sequence is a list of numbers, called terms, written in a specific order.
- Explicit formulas define each term of a sequence using the position of the term. See , , and .
- An explicit formula for the \(n\text{th}\) term of a sequence can be written by analyzing the pattern of several terms. See .
- Recursive formulas define each term of a sequence using previous terms.
- Recursive formulas must state the initial term, or terms, of a sequence.
- A set of terms can be written by using a recursive formula. See and .
- A factorial is a mathematical operation that can be defined recursively.
- The factorial of \(n\) is the product of all integers from 1 to \(n\) See .
Sequences and Their Notations
- Write the first few terms of a sequence (IA 12.1.1)
- Find a formula for the general term (nth term) of a sequence (IA 12.1.2)
A patient takes a 30 mg antibiotic capsule. At the end of that hour, the amount of antibiotic remaining in her body is only 90% of the amount in the beginning of that hour. The 30mg dose is taken at time t = 1 hour. How much of this dose remains at the end of 1 hour? 2hours? 3 hours? 4 hours?
| Time t | Dose remaining after time t |
| 1 | 0.90(30)=27mg |
| 2 | 0.90(27)=24.3mg |
| 3 | 0.90(24.3)=21.87mg |
| 4 | 0.90(21.87)=19.68mg |
This ordered list of numbers 27, 24.3, 21.87, 19.68, … is a sequence. Each number in the list is a term.
A sequence is a function whose domain is the counting numbers. A sequence may have an infinite number of terms or a finite number of terms. Our sequence has three dots (ellipsis) at the end which indicates the list never ends. If the domain is the set of all counting numbers, then the sequence is an infinite sequence.
Often when working with sequences we do not want to write out all the terms. We want a more compact way to show how each term is defined. When we worked with functions, we wrote \(f(x)=2x\) and we said the expression 2x was the rule that defined values in the range.
While a sequence is a function, we do not use the usual function notation. Instead of writing the function as \(f(x)=2x\) , we would write it as \({a}_{n}=2n\) . The \({a}_{n}\) is the nth term of the sequence, the term in the nth position where n is a value in the domain. The formula for writing the nth term of the sequence is called the general term or formula of the sequence.
General sequence terms are denoted as follows:
\(\begin{array}{ll}{a}_{1}-\text{first}\text{term} & \\ {a}_{2}-\text{second}\text{term} & \\ {a}_{3}-\text{third}\text{term} & \\ . & \\ . & \\ . & \\ {a}_{n}-{n}^{th}\text{term} & \\ {a}_{n+1}-(n+1)\text{term} & \\ . & \\ . & \\ . & \end{array}\)
Example
Try it.
Write the first five terms of the sequence whose general term is \({a}_{n}=2n-7\) .
Solution
\(\begin{array}{llllll}n & 1 & 2 & 3 & 4 & 5 \\ {a}_{n} & {a}_{1} & {a}_{2} & {a}_{3} & {a}_{4} & {a}_{5} \\ 2n-7 & 2(1)-7 & 2(2)-7 & 2(3)-7 & 2(4)-7 & 2(5)-7 \\ & -5 & -3 & -1 & 1 & 3\end{array}\)
Write the first few terms of a sequence.
Try it.
Write the first five terms of the sequence whose general term is \({a}_{n}=4n+2\).
| \(n\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) |
| \({a}_{n}\) | \({a}_{1}\) | \({a}_{2}\) | \({a}_{3}\) | \({a}_{4}\) | \({a}_{5}\) |
| \(4n+2\) | |||||
Try it.
Write the first five terms of the sequence whose general term is \({a}_{n}={3}^{n}-1\).
| \(n\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) |
| \({a}_{n}\) | \({a}_{1}\) | \({a}_{2}\) | \({a}_{3}\) | \({a}_{4}\) | \({a}_{5}\) |
| \({3}^{n}-1\) | |||||
Condensed — the full section is in OpenStax Precalculus 2e.
Writing the Terms of a Sequence Defined by an Explicit Formula
One way to describe an ordered list of numbers is as a sequence. A sequence is a function whose domain is a subset of the counting numbers. The sequence established by the number of hits on the website is
\[\{2,4,8,16,32,\ldots \}.\]The ellipsis (…) indicates that the sequence continues indefinitely. Each number in the sequence is called a term. The first five terms of this sequence are 2, 4, 8, 16, and 32.
Listing all of the terms for a sequence can be cumbersome. For example, finding the number of hits on the website at the end of the month would require listing out as many as 31 terms. A more efficient way to determine a specific term is by writing a formula to define the sequence.
One type of formula is an explicit formula, which defines the terms of a sequence using their position in the sequence. Explicit formulas are helpful if we want to find a specific term of a sequence without finding all of the previous terms. We can use the formula to find the nth term of the sequence, where \(n\) is any positive number. In our example, each number in the sequence is double the previous number, so we can use powers of 2 to write a formula for the \(n\text{th}\) term.
The first term of the sequence is \({2}^{1}=2,\) the second term is \({2}^{2}=4,\) the third term is \({2}^{3}=8,\) and so on. The \(n\text{th}\) term of the sequence can be found by raising 2 to the \(n\text{th}\) power. An explicit formula for a sequence is named by a lower case letter \(a,b,c...\) with the subscript \(n.\) The explicit formula for this sequence is
\[{a}_{n}={2}^{n}.\]Now that we have a formula for the \(n\text{th}\) term of the sequence, we can answer the question posed at the beginning of this section. We were asked to find the number of hits at the end of the month, which we will take to be 31 days. To find the number of hits on the last day of the month, we need to find the 31st term of the sequence. We will substitute 31 for \(n\) in the formula.
\[\begin{array}{l}{a}_{31}={2}^{31} \\ \ =\text{2,147,483,648}\end{array}\]If the doubling trend continues, the company will get \(\text{2,147,483,648}\) hits on the last day of the month. That is over 2.1 billion hits! The huge number is probably a little unrealistic because it does not take consumer interest and competition into account. It does, however, give the company a starting point from which to consider business decisions.
| \(n\) | 1 | 2 | 3 | 4 | 5 | \(n\) |
| \(n\text{th}\) term of the sequence, \({a}_{n}\) | 2 | 4 | 8 | 16 | 32 | \({2}^{n}\) |
Condensed — the full section is in OpenStax Precalculus 2e.
Writing the Terms of a Sequence Defined by a Recursive Formula
Sequences occur naturally in the growth patterns of nautilus shells, pinecones, tree branches, and many other natural structures. We may see the sequence in the leaf or branch arrangement, the number of petals of a flower, or the pattern of the chambers in a nautilus shell. Their growth follows the Fibonacci sequence, a famous sequence in which each term can be found by adding the preceding two terms. The numbers in the sequence are 1, 1, 2, 3, 5, 8, 13, 21, 34,…. Other examples from the natural world that exhibit the Fibonacci sequence are the Calla Lily, which has just one petal, the Black-Eyed Susan with 13 petals, and different varieties of daisies that may have 21 or 34 petals.
Each term of the Fibonacci sequence depends on the terms that come before it. The Fibonacci sequence cannot easily be written using an explicit formula. Instead, we describe the sequence using a recursive formula, a formula that defines the terms of a sequence using previous terms.
A recursive formula always has two parts: the value of an initial term (or terms), and an equation defining \({a}_{n}\) in terms of preceding terms. For example, suppose we know the following:
\[\begin{array}{l}{a}_{1}=3 \\ {a}_{n}=2{a}_{n-1}-1\text{for}n\ge 2\end{array}\]We can find the subsequent terms of the sequence using the first term.
\[\begin{array}{l}{a}_{1}=3 \\ {a}_{2}=2{a}_{1}-1=2(3)-1=5 \\ {a}_{3}=2{a}_{2}-1=2(5)-1=9 \\ {a}_{4}=2{a}_{3}-1=2(9)-1=17\end{array}\]So the first four terms of the sequence are \(\{3,\ 5,\ 9,\ 17\}\) .
The recursive formula for the Fibonacci sequence states the first two terms and defines each successive term as the sum of the preceding two terms.
\[\begin{array}{l}{a}_{1}=1 \\ {a}_{2}=1 \\ {a}_{n}={a}_{n-1}+{a}_{n-2}\text{for}n\ge 3\end{array}\]To find the tenth term of the sequence, for example, we would need to add the eighth and ninth terms. We were told previously that the eighth and ninth terms are 21 and 34, so
\[{a}_{10}={a}_{9}+{a}_{8}=34+21=55\]Condensed — the full section is in OpenStax Precalculus 2e.
Using Factorial Notation
The formulas for some sequences include products of consecutive positive integers. \(n\) factorial, written as \(n!,\) is the product of the positive integers from 1 to \(n.\) For example,
\[\begin{array}{l}4!=4⋅3⋅2⋅1=24 \\ 5!=5⋅4⋅3⋅2⋅1=120\end{array}\]An example of formula containing a factorial is \({a}_{n}=(n+1)!.\) The sixth term of the sequence can be found by substituting 6 for \(n.\)
\[{a}_{6}=(6+1)!=7!=7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1=5040\]The factorial of any whole number \(n\) is \(n(n-1)!\) We can therefore also think of \(5!\) as \(5⋅4!\text{.}\)
Example
Try it.
Write the first five terms of the sequence defined by the explicit formula \({a}_{n}=\frac{5n}{(n+2)!}.\)
Solution
Substitute \(n=1,n=2,\) and so on in the formula.
\[\begin{array}{lllll}n=1 & & & & {a}_{1}=\frac{5(1)}{(1+2)!}=\frac{5}{3!}=\frac{5}{3\cdot 2\cdot 1}=\frac{5}{6} \\ n=2 & & & & {a}_{2}=\frac{5(2)}{(2+2)!}=\frac{10}{4!}=\frac{10}{4\cdot 3\cdot 2\cdot 1}=\frac{5}{12} \\ n=3 & & & & {a}_{3}=\frac{5(3)}{(3+2)!}=\frac{15}{5!}=\frac{15}{5\cdot 4\cdot 3\cdot 2\cdot 1}=\frac{1}{8} \\ n=4 & & & & {a}_{4}=\frac{5(4)}{(4+2)!}=\frac{20}{6!}=\frac{20}{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}=\frac{1}{36} \\ n=5 & & & & {a}_{5}=\frac{5(5)}{(5+2)!}=\frac{25}{7!}=\frac{25}{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}=\frac{5}{1\text{,}008}\end{array}\]The first five terms are \(\{\frac{5}{6},\frac{5}{12},\frac{1}{8},\frac{1}{36},\frac{5}{1,008}\}.\)
Key Concepts
- A sequence is a list of numbers, called terms, written in a specific order.
- Explicit formulas define each term of a sequence using the position of the term. See , , and .
- An explicit formula for the \(n\text{th}\) term of a sequence can be written by analyzing the pattern of several terms. See .
- Recursive formulas define each term of a sequence using previous terms.
- Recursive formulas must state the initial term, or terms, of a sequence.
- A set of terms can be written by using a recursive formula. See and .
- A factorial is a mathematical operation that can be defined recursively.
- The factorial of \(n\) is the product of all integers from 1 to \(n\) See .
Practice (40)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
-
Write the first five terms of the sequence whose general term is \({a}_{n}=2n-7\) .
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\(\begin{array}{llllll}n & 1 & 2 & 3 & 4 & 5 \\ {a}_{n} & {a}_{1} & {a}_{2} & {a}_{3} & {a}_{4} & {a}_{5} \\ 2n-7 & 2(1)-7 & 2(2)-7 & 2(3)-7 & 2(4)-7 & 2(5)-7 \\ & -5 & -3 & -1 & 1 & 3\end{array}\)
-
Write the first five terms of the sequence whose general term is \({a}_{n}=4n+2\).
\(n\) \(1\) \(2\) \(3\) \(4\) \(5\) \({a}_{n}\) \({a}_{1}\) \({a}_{2}\) \({a}_{3}\) \({a}_{4}\) \({a}_{5}\) \(4n+2\) -
Write the first five terms of the sequence whose general term is \({a}_{n}={3}^{n}-1\).
\(n\) \(1\) \(2\) \(3\) \(4\) \(5\) \({a}_{n}\) \({a}_{1}\) \({a}_{2}\) \({a}_{3}\) \({a}_{4}\) \({a}_{5}\) \({3}^{n}-1\) -
- ⓐ Find a general term for the sequence whose first five terms are shown below:
4, 8, 12, 16, 20... - ⓑ
Find a general term for the sequence whose first five terms are shown below:
\(\frac{1}{3},\frac{1}{9},\frac{1}{27},\frac{1}{81},\frac{1}{243},...\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
ⓐ Look for a pattern in the terms The numbers are all multiples of 4 The general term of the sequence: \({a}_{n}=4n.\) ⓑ Look for a pattern in the terms. The numerators are all 1 and the denominators are powers of 3 The general term of the sequence: \({a}_{n}=\frac{1}{{3}^{n}}\) - ⓐ Find a general term for the sequence whose first five terms are shown below:
-
Find a general term for the sequence whose first five terms are shown:
8, 16, 24, 32, 40, ...Look for a pattern in the terms Terms: ________________ The general term of the sequence: ________________ -
Find a general term for the sequence whose first five terms are shown:
\(\frac{1}{4},\frac{1}{16},\frac{1}{64},\frac{1}{256},\frac{1}{1024},...\)Look for a pattern in the terms Terms: ________________ The general term of the sequence: ________________ -
Write the first five terms of the sequence defined by the explicit formula \({a}_{n}=-3n+8.\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
Substitute \(n=1\) into the formula. Repeat with values 2 through 5 for \(n.\)
\[\begin{array}{llllll}n=1 & & & & & {a}_{1}=-3(1)+8=5 \\ n=2 & & & & & {a}_{2}=-3(2)+8=2 \\ n=3 & & & & & {a}_{3}=-3(3)+8=-1 \\ n=4 & & & & & {a}_{4}=-3(4)+8=-4 \\ n=5 & & & & & {a}_{5}=-3(5)+8=-7\end{array}\]The first five terms are \(\{5,\ 2,\ -1,\ -4,\ -7\}.\)
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Write the first five terms of the sequence defined by the explicit formula \({t}_{n}=5n-4.\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
The first five terms are \(\{1,6,11,16,21\}.\)
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Write the first five terms of the sequence.
\[{a}_{n}=\frac{{(-1)}^{n}{n}^{2}}{n+1}\]ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
Substitute \(n=1,\) \(n=2,\) and so on in the formula.
\[\begin{array}{llll}n=1 & \begin{array}{ll} & \end{array} & {a}_{1}=\frac{{(-1)}^{1}{1}^{2}}{1+1}=-\frac{1}{2} \\ n=2 & \begin{array}{ll} & \end{array} & {a}_{2}=\frac{{(-1)}^{2}{2}^{2}}{2+1}=\frac{4}{3} \\ n=3 & \begin{array}{ll} & \end{array} & {a}_{3}=\frac{{(-1)}^{3}{3}^{2}}{3+1}=-\frac{9}{4} \\ n=4 & \begin{array}{ll} & \end{array} & {a}_{4}=\frac{{(-1)}^{4}{4}^{2}}{4+1}=\frac{16}{5} \\ n=5 & & {a}_{5}=\frac{{(-1)}^{5}{5}^{2}}{5+1}=-\frac{25}{6}\end{array}\]The first five terms are \(\{-\frac{1}{2},\frac{4}{3},-\frac{9}{4},\frac{16}{5},-\frac{25}{6}\}.\)
-
Write the first five terms of the sequence.
\[{a}_{n}=\frac{4n}{{(-2)}^{n}}\]ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
The first five terms are \(\{-2,2,-\frac{3}{2},1,\ -\frac{5}{8}\}.\)
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Write the first six terms of the sequence.
\[{a}_{n}=\{\begin{array}{ll}{n}^{2} & \text{if }n\ \text{is not divisible by 3} \\ \frac{n}{3} & \text{if }n\ \text{is divisible by 3}\end{array}\]ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
Substitute \(n=1,n=2,\) and so on in the appropriate formula. Use \({n}^{2}\) when \(n\) is not a multiple of 3. Use \(\frac{n}{3}\) when \(n\) is a multiple of 3.
\[\begin{array}{lllll}{a}_{1}={1}^{2}=1\begin{array}{llll} & & & \end{array} & \text{1 is not a multiple of 3}\text{. Use }{n}^{2}. \\ {a}_{2}={2}^{2}=4 & \text{2 is not a multiple of 3}\text{. Use }{n}^{2}. \\ {a}_{3}=\frac{3}{3}=1 & \text{3 is a multiple of 3}\text{. Use }\frac{n}{3}. \\ {a}_{4}={4}^{2}=16 & \text{4 is not a multiple of 3}\text{. Use }{n}^{2}. \\ {a}_{5}={5}^{2}=25 & \text{5 is not a multiple of 3}\text{. Use }{n}^{2}. \\ {a}_{6}=\frac{6}{3}=2 & \text{6 is a multiple of 3}\text{. Use }\frac{n}{3}.\end{array}\]The first six terms are \(\{1,\ 4,\ 1,\ 16,\ 25,\ 2\}.\)
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Write the first six terms of the sequence.
\[{a}_{n}=\{\begin{array}{ll}2{n}^{3} & \text{if }n\ \text{is odd} \\ \frac{5n}{2} & \text{if }n\ \text{is even}\end{array}\]ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
The first six terms are \(\{2,\ 5,\ 54,\ 10,\ 250,\ 15\}.\)
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Write an explicit formula for the \(n\text{th}\) term of each sequence.
- ⓐ \(\{-\frac{2}{11},\frac{3}{13},-\frac{4}{15},\frac{5}{17},-\frac{6}{19},\ldots \}\)
- ⓑ \(\{-\frac{2}{25}\text{,}-\frac{2}{125}\text{,}-\frac{2}{625}\text{,}-\frac{2}{3\text{,}125}\text{,}-\frac{2}{15\text{,}625}\text{,}\ldots \}\)
- ⓒ \(\{{e}^{4}\text{,}{e}^{5}\text{,}{e}^{6}\text{,}{e}^{7}\text{,}{e}^{8}\text{,}\ldots \}\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
Look for the pattern in each sequence.
- ⓐThe terms alternate between positive and negative. We can use \({(-1)}^{n}\) to make the terms alternate. The numerator can be represented by \(n+1.\) The denominator can be represented by \(2n+9.\)
\({a}_{n}=\frac{{(-1)}^{n}(n+1)}{2n+9}\)
- ⓑ
The terms are all negative.
So we know that the fraction is negative, the numerator is 2, and the denominator can be represented by \({5}^{n+1}.\)
\[{a}_{n}=-\frac{2}{{5}^{n+1}}\] - ⓒ
The terms are powers of \(e.\) For \(n=1,\) the first term is \({e}^{4}\) so the exponent must be \(n+3.\)
\[{a}_{n}={e}^{n+3}\]
-
Write an explicit formula for the \(n\text{th}\) term of the sequence.
\[\text{\{9,}\ -\text{81,}\ \text{729,}\ -\text{6,561,}\ \text{59,049,}\ \text{\ldots \}}\]ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\({a}_{n}={(-1)}^{n+1}{9}^{n}\)
-
Write an explicit formula for the \(n\text{th}\) term of the sequence.
\[\{-\frac{3}{4},-\frac{9}{8},-\frac{27}{12},-\frac{81}{16},-\frac{243}{20},...\}\]ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\({a}_{n}=-\frac{{3}^{n}}{4n}\)
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Write an explicit formula for the \(n\text{th}\) term of the sequence.
\[\{\frac{1}{{e}^{2}},\frac{1}{e},1,e,{e}^{2},...\}\]ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\({a}_{n}={e}^{n-3}\)
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Write the first five terms of the sequence defined by the recursive formula.
\[\begin{array}{l}\begin{array}{l} \\ {a}_{1}=9\end{array} \\ {a}_{n}=3{a}_{n-1}-20\text{, for }n\ge 2\end{array}\]ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
The first term is given in the formula. For each subsequent term, we replace \({a}_{n-1}\) with the value of the preceding term.
\[\begin{array}{llllll}n=1\begin{array}{lllll} & & & & \end{array} & {a}_{1}=9 \\ n=2 & {a}_{2}=3{a}_{1}-20=3(9)-20=27-20=7 \\ n=3 & {a}_{3}=3{a}_{2}-20=3(7)-20=21-20=1 \\ n=4 & {a}_{4}=3{a}_{3}-20=3(1)-20=3-20=-17 \\ n=5 & {a}_{5}=3{a}_{4}-20=3(-17)-20=-51-20=-71\end{array}\]The first five terms are \(\{9,\ 7,\ 1,\ -17,\ -71\}.\) See .
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Write the first five terms of the sequence defined by the recursive formula.
\[\begin{array}{l}{a}_{1}=2 \\ {a}_{n}=2{a}_{n-1}+1\text{, for }n\ge 2\end{array}\]ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\(\{2,5,11,23,47\}\)
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Write the first six terms of the sequence defined by the recursive formula.
\[\begin{array}{l}{a}_{1}=1 \\ {a}_{2}=2 \\ {a}_{n}=3{a}_{n-1}+4{a}_{n-2}\text{, for }n\ge 3\end{array}\]ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
The first two terms are given. For each subsequent term, we replace \({a}_{n-1}\) and \({a}_{n-2}\) with the values of the two preceding terms.
\[\begin{array}{lllllll}n=3 & & & & & & {a}_{3}=3{a}_{2}+4{a}_{1}=3(2)+4(1)=10 \\ n=4 & & & & & & {a}_{4}=3{a}_{3}+4{a}_{2}=3(10)+4(2)=38 \\ n=5 & & & & & & {a}_{5}=3{a}_{4}+4{a}_{3}=3(38)+4(10)=154 \\ n=6 & & & & & & {a}_{6}=3{a}_{5}+4{a}_{4}=3(154)+4(38)=614\end{array}\]The first six terms are \(\text{\{1,2,10,38,154,614\}}\text{.}\) See .
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Write the first 8 terms of the sequence defined by the recursive formula.
\[\begin{array}{l}\begin{array}{l} \\ {a}_{1}=0\end{array} \\ {a}_{2}=1 \\ {a}_{3}=1 \\ {a}_{n}=\frac{{a}_{n-1}}{{a}_{n-2}}+{a}_{n-3}\text{, for }n\ge 4\end{array}\]ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\(\{0,1,1,1,2,3,\frac{5}{2},\ \frac{17}{6}\}.\)
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Write the first five terms of the sequence defined by the explicit formula \({a}_{n}=\frac{5n}{(n+2)!}.\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
Substitute \(n=1,n=2,\) and so on in the formula.
\[\begin{array}{lllll}n=1 & & & & {a}_{1}=\frac{5(1)}{(1+2)!}=\frac{5}{3!}=\frac{5}{3\cdot 2\cdot 1}=\frac{5}{6} \\ n=2 & & & & {a}_{2}=\frac{5(2)}{(2+2)!}=\frac{10}{4!}=\frac{10}{4\cdot 3\cdot 2\cdot 1}=\frac{5}{12} \\ n=3 & & & & {a}_{3}=\frac{5(3)}{(3+2)!}=\frac{15}{5!}=\frac{15}{5\cdot 4\cdot 3\cdot 2\cdot 1}=\frac{1}{8} \\ n=4 & & & & {a}_{4}=\frac{5(4)}{(4+2)!}=\frac{20}{6!}=\frac{20}{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}=\frac{1}{36} \\ n=5 & & & & {a}_{5}=\frac{5(5)}{(5+2)!}=\frac{25}{7!}=\frac{25}{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}=\frac{5}{1\text{,}008}\end{array}\]The first five terms are \(\{\frac{5}{6},\frac{5}{12},\frac{1}{8},\frac{1}{36},\frac{5}{1,008}\}.\)
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Write the first five terms of the sequence defined by the explicit formula \({a}_{n}=\frac{(n+1)!}{2n}.\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
The first five terms are \(\{1,\frac{3}{2},4,\ 15,\ 72\}.\)
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Discuss the meaning of a sequence. If a finite sequence is defined by a formula, what is its domain? What about an infinite sequence?
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
A sequence is an ordered list of numbers that can be either finite or infinite in number. When a finite sequence is defined by a formula, its domain is a subset of the non-negative integers. When an infinite sequence is defined by a formula, its domain is all positive or all non-negative integers.
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Describe three ways that a sequence can be defined.
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Is the ordered set of even numbers an infinite sequence? What about the ordered set of odd numbers? Explain why or why not.
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
Yes, both sets go on indefinitely, so they are both infinite sequences.
-
What happens to the terms \({a}_{n}\) of a sequence when there is a negative factor in the formula that is raised to a power that includes \(n?\) What is the term used to describe this phenomenon?
-
What is a factorial, and how is it denoted? Use an example to illustrate how factorial notation can be beneficial.
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
A factorial is the product of a positive integer and all the positive integers below it. An exclamation point is used to indicate the operation. Answers may vary. An example of the benefit of using factorial notation is when indicating the product It is much easier to write than it is to write out \(\text{13}⋅\text{12}⋅\text{11}⋅\text{10}⋅\text{9}⋅\text{8}⋅\text{7}⋅\text{6}⋅\text{5}⋅\text{4}⋅\text{3}⋅\text{2}⋅\text{1}\text{.}\)
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\({a}_{n}={2}^{n}-2\)
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\({a}_{n}=-\frac{16}{n+1}\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
First four terms: \(-8,\ -\frac{16}{3},\ -4,\ -\frac{16}{5}\)
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\({a}_{n}=-{(-5)}^{n-1}\)
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\({a}_{n}=\frac{{2}^{n}}{{n}^{3}}\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
First four terms: \(2,\ \frac{1}{2},\ \frac{8}{27},\ \frac{1}{4}\) .
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\({a}_{n}=\frac{2n+1}{{n}^{3}}\)
-
\({a}_{n}=1.25⋅{(-4)}^{n-1}\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
First four terms: \(1.25,\ -5,\ 20,\ -80\) .
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\({a}_{n}=-4⋅{(-6)}^{n-1}\)
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\({a}_{n}=\frac{{n}^{2}}{2n+1}\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
First four terms: \(\frac{1}{3},\ \frac{4}{5},\ \frac{9}{7},\ \frac{16}{9}\) .
-
\({a}_{n}={(-10)}^{n}+1\)
-
\({a}_{n}=-(\frac{4⋅{(-5)}^{n-1}}{5})\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
First four terms: \(-\frac{4}{5},\ 4,\ -20,\ 100\)
-
\({a}_{n}=\{\begin{array}{ll}{(-2)}^{n}-2 & \text{if }n\ \text{is even} \\ {(3)}^{n-1} & \text{if }n\ \text{is odd}\end{array}\)
-
\({a}_{n}=\{\begin{array}{ll}\frac{{n}^{2}}{2n+1} & \text{if }n\ \le \text{5} \\ {n}^{2}-5 & \text{if }n\ \text{>5}\end{array}\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\(\frac{1}{3},\ \frac{4}{5},\ \frac{9}{7},\ \frac{16}{9},\ \frac{25}{11},\ 31,\ 44,\ 59\)
-
\({a}_{n}=\{\begin{array}{ll}{(2n+1)}^{2} & \text{if }n\ \text{is divisible by 4} \\ \frac{2}{n} & \text{if }n\ \text{is not divisible by 4}\end{array}\)
Symbols used here
n × (n−1) × … × 1; the number of orderings of n things. 0! = 1.
Inequalities that allow equality; < and > exclude it.
Distance from zero: |−3| = 3. For a complex number, distance from the origin.
Number of k-element subsets of n things: n!/(k!(n−k)!).
Ratio of a circle's circumference to its diameter, 3.14159…
2.71828…, the base whose exponential is its own derivative.
i² = −1.
The usual name for an angle.
The exponent b must be raised to for x; ln uses base e.
A quantity with magnitude and direction; a column of numbers.
How to: Sequences and Their Notations
- Write the terms of a sequence defined by an explicit formula.
- Write the terms of a sequence defined by a recursive formula.
- Use factorial notation.
- Write the first few terms of a sequence (IA 12.1.1)
- Find a formula for the general term (nth term) of a sequence (IA 12.1.2)
- Substitute each value of
- To find the second term,
- Continue in the same manner until you have identified all
Questions people ask
What is a function, really?
A rule that assigns exactly one output to each input. The vertical-line test on a graph is the same idea: no input may have two outputs.
Why do we need complex numbers?
Because x² + 1 = 0 has no real solution, and allowing one new number i with i² = −1 makes every polynomial equation solvable. They then turn out to describe rotation, waves and alternating current more naturally than real numbers do.
ನಿಮ್ಮದೇ ಆದದ್ದನ್ನು ಪ್ರಯತ್ನಿಸಿ
Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0), OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.
ಇನ್ನಷ್ಟು Precalculus
Complex numbersPolynomial functionsRational functionsSequences and seriesThe binomial theoremConic sectionsVectorsExponential and logarithmic functionsPolynomial division and the remainder theoremParametric equations and polar coordinates