Alternative Format — Lesson 1: Introducing Sequences

Let's Start Thinking

Making Predictions

Much of the study of mathematics involves looking for patterns. Finding patterns allows us to explain relationships and make predictions about unknown numbers.

For example, here are the first few numbers in some lists. Can you predict what number comes next? 

• \(9,~7,~5,~3,\dots\)
• \(1,~3,~4,~7,~11,\dots\)
• \(2,~10,~50,~250,\dots\)

There appear to be some patterns in each of these lists.

• In \(9,~7,~5,~3,\dots\), the numbers consistently decrease by \(2\).
• In \(1,~3,~4,~7,~11,\dots\), a number is obtained by adding the \(2\) previous numbers.
• In \(2,~10,~50,~250,\dots\), each number is \(5\) times bigger than the number preceding it.

But can we be absolutely sure that these patterns continue?

Here's an even trickier question: What numbers should follow \(1,~3,~2,~2,\dots\)? Do we even have enough information to answer this last question?

In the next few lessons, we will be looking for patterns and making predictions for sequences of numbers like these, and representing a sequence in many different ways.

Try This

A sequence of numbers is defined as follows:

• The first term is \(5\).
• The second term is \(-7\).
• Each new term, after the first two, is obtained by taking the preceding term and subtracting from it the term before it. For example:
  • The third term is equal to the second term minus the first term (i.e., the third term is \(-7-5\), or \(-12\)).
  • The fourth term is equal to the third term minus the second term. 

What is the \(100\)^th term in this sequence? 

Introduction to Sequences

Here are some examples:

• These figures represent a sequence of three images.
  
  

  

  Expand/Collapse Alternative Format for Figures

  In Figure 1, circles from a 3 by 3 grid where circles are missing from the following positions:

  • Row 2 has a circle in column 2 missing.
  • Row 3 has a circle in column 3 missing.

  In Figure 2, circles form a 5 by 5 grid where circles are missing from the following positions:

  • Rows 2 and 3 have circles in columns 2 and 3 missing.
  • Rows 4 and 5 have circles in columns 4 and 5 missing.

  In Figure 3, circles form a 7 by 7 grid where circles are missing from the following positions:

  • Rows 2, 3, and 4 have circles in columns 2, 3, and 4 missing.
  • Rows 5, 6, and 7 have circles in columns 5, 6, and 7 missing.
• \(A,~S,~L,~E,~E,~P\) and \(P,~L,~E,~A,~S,~E\) represent two different sequences of six terms. Although they contain the same six letters, they represent two different sequences since order matters. 
• \(1,~1,~2,~3,~5,~8,~13,\dots\) is the beginning of a famous sequence of numbers called the Fibonacci sequence. New terms are added to this sequence based on the previous terms. 

There are two types of sequences to consider:

Examples

• The sequence \(A,~S,~L,~E,~E,~P\) is a finite sequence with six terms. 
• The Fibonacci sequence \(1,~1,~2,~3,~5,~8,~13,\dots\) is an infinite sequence. The sequence continues beyond the numbers listed here.

Representing Sequences

The natural numbers are often referred to when discussing sequences. This is because the term number, or index, of any term in a sequence is typically a natural number. 

For the Fibonacci sequence \(1,~1,~2,~3,~5,~8,~13,\dots\), we have that \(t_2=1\) (since the second term is \(1\)) and \(t_7=13\) (since the seventh term is \(13\)). 

Thus, a sequence of numbers can be considered as a special type of function. As with other functions, it is possible to represent a sequence of numbers with a table or with a graph. 

Example 1

Solution

To create the table, let the first column represent the term number, \(n\), and let the second column represent each term in the sequence, \(t_n\). 

The term number, \(n\), is the independent variable, while the corresponding term in the sequence, \(t_n\), is the dependent variable. So the graph will consist of seven coordinate pairs: \((1, 1) \), \((2, -2)\), and so on. Plot these coordinates to produce the graph. 

Notice that the points have not been joined and that the graph does not extend past \(n=7\). This is because there are only seven terms in the sequence, and so the only possible values of \(n\) are \(1,~2,~3,\dots,~7\). This set of possible values of \(n\) is the domain of the function, which means that no other values of \(n\) should be included in the graph (for example, it is not possible to have a term number somewhere between \(1\) and \(2\)). 

Check Your Understanding 1

Question — Version 1

The graph below represents the first five terms in a sequence. What is \(t_2\)?

Answer — Version 1

\(t_2=8\)

Feedback — Version 1

\(t_{2}\) is represented by the point on the graph where \(n=2\).

Therefore, \(t_2=8\).

Question — Version 2

The graph below represents the first five terms in a sequence. What is \(t_1\)?

Answer — Version 2

\(t_1=2\)

Feedback — Version 2

\(t_{1}\) is represented by the point on the graph where \(n=1\).

Therefore, \(t_1=2\).

Recursive Sequences

We are often interested in predicting future terms in a sequence (beyond those listed), or in developing a rule that describes all the terms in the sequence. It is only possible to do this if there is an observable pattern. This pattern is often described in terms of how each term relates to the previous terms in the sequence. 

Let's begin by identifying a pattern in a given sequence and, assuming that pattern continues, use this pattern to write more terms. 

Example 2

Solution — Part A

Since the first four terms have been listed, we need to find \(t_5\), \(t_6\), and \(t_7\).

We notice that the first four terms have a pattern — each term is \(5\) less than the previous term in the sequence.

If this pattern continues, we need to subtract \(5\) from the fourth term to find the fifth term.

\(\begin{align*}t_5&=t_4-5 \\ &= -3-5 \\ &= -8\end{align*}\)

Similarly, \(t_6=-13\), and \(t_7=-18\). 

So, the next three terms in the sequence are \(-8 \), \(-13\), and \(-18\).

Solution — Part B

Here, the first five terms have been listed. We need to find \(t_6\), \(t_7\), and \(t_8\).

We notice another pattern — this time, each term is obtained by dividing the previous term by \(-2\).

For example, \(-192 = \dfrac{384}{-2}\), \(96=\dfrac{-192}{-2}\), and so on.

So, if this pattern continues, we need to divide the fifth term by \(-2\) to find the sixth term.

 \(\begin{align*} t_6&=\dfrac{t_5}{-2} \\ &= \dfrac{24}{-2} \\ &= -12 \end{align*}\)

Similarly, \(t_7=6\) and \(t_8=-3\).

Thus, the next three terms in the sequence are \(-12\), \(6\), and \(-3\). \(\)

An Unpredictable Sequence

The number that faces up when a six-sided die is rolled repeatedly can be represented by a sequence:

\(3,~4,~6,~3,~5,~4,~1,~5,~1,~3,~5,~2,~3,~1,\dots\)

There is no discernible pattern in these numbers (and in fact, there shouldn't be, since the results of rolling a die should be random). This makes it impossible to predict any future terms in the sequence, or any future results of the die.

For the rest of this lesson, we will only be examining sequences that have predictable patterns. 

Recursive Sequences

In a previous example, you may have seen the following sequences.

Each of these sequences have patterns. And these patterns can be described in terms of how each term relates to the term immediately before it.

Because of this, both of these sequences can be described as recursive sequences.

Sometimes more than one term is provided with the formula in the case of more complicated recursive definitions. We'll see cases like this later. For now, we'll discuss how to express each of the sequences using a recursion formula.

Example 3

Solution — Part A

We noted earlier that each term, which we'll denote as \(t_n\), is five less than the previous term. If we are considering the term \(t_n\), the previous term has a term number that is \(1\) lower. So the previous term is actually ​\(t_{n-1}\).

This means that we can write the following formula indicating that term \(n\) is \(5\) less than the previous term.

\(t_n=t_{n-1}-5\)

This is an important part of the recursion formula. However, on its own, it's not enough to define the sequence. We need a starting point or a known term in the sequence.

So as part of our definition of the sequence, we'll include the initial term, term 1, which is \(12\), or \(t_1=12\).

All together, the sequence is defined by the following recursion formula

\(t_1=12\), \(t_n=t_{n-1}-5\) for \(n \ge 2\).

This means that we can use the recursion formula to calculate any term except for the first term. We can check to see that this formula works for the values we already know.

For example, according this this recursion formula,

 \(\begin{align*} t_2 &= t_1 - 5 \\ &= 12 -5 \\ &= 7\end{align*}\)

\(7\) is in fact the second term in the sequence.

Solution — Part B

Recall: Represent \(384,-192,~96,-48,~24,\dots\) using a recursion formula.

Earlier we saw that each term, \(t_n\), is obtained by dividing the previous term,  ​​​​​​\(t_{n-1}\), by \(-2\). From this information, we can write the formula

\(t_n=\dfrac{t_{n-1}}{-2}\) or  \(t_n=-\dfrac{t_{n-1}}2\)

Remember that we also need to define a known term in the sequence. So again, we'll include the initial term, term 1, which in this case is \(384\). That is, \(t_1=384\).

Altogether the sequence is defined by the following recursion formula:

\(t_1=384\), \(t_n=-\dfrac{t_{n-1}}2\) for \(n \ge 2\)

Check Your Understanding 2

Question — Version 1

Represent the following sequence using a recursion formula.

\(-6, -2, ~2,~6, \ldots\)

1. Determine the initial term, \(t_1\).
2. State the recursion formula, \(t_n\), for \(n\geq 2\).

Answer — Version 1

1. The initial term is \(t_1=-6\)
2. The recursion formula is \(t_n=t_{n-1}+4\) for \(n \geq 2\).

Feedback — Version 1

In the sequence \(-6,-2,~2,~6, \ldots\) each term, \(t_{n},\) is obtained by adding \(4\) to the previous term, \(t_{n-1}\).

Therefore, the initial term is \(t_{1}=-6\) and the recursion formula is \(t_{n}=t_{n-1}+4\) for \(n \geq 2\).

Question — Version 2

Represent the following sequence using a recursion formula.

\(16~384,~4096,~1024,~256, \ldots\)

1. Determine the initial term, \(t_1\).
2. State the recursion formula, \(t_n\), for \(n\geq 2\).

Answer — Version 2

1. The initial term is \(t_1=16~384\)
2. The recursion formula is \(t_{n}=\dfrac{t_{n-1}}{4}\) for \(n \geq 2\).

Feedback — Version 2

In the sequence \(16~384,~4096,~1024,~256, \ldots\) each term, \(t_{n},\) is obtained by dividing the previous term, \(t_{n-1}\) by \(4 .\)
Therefore, the initial term is \(t_{1}=16~384\) and the recursion formula is \(t_{n}=\dfrac{t_{n-1}}{4}\) for \(n \geq 2\)

Now let's look at a recursive sequence that is a little bit more complicated to define.

Example 4

Solution — Option A

It is fairly straightforward to write a recursion formula if there is a constant difference between each of the terms, that is, a number that is added to or subtracted from each term to obtain the next term. Unfortunately, that's not the case here, as the difference between the first two terms, \(t_2-t_1\), is \(1\), while \(t_3-t_2\) equals \(2\).

• The difference between terms is not constant. Terms in the sequence are not obtained by adding a constant to the previous term. 

We can also check if it's possible to multiply or divide each term by the same number to obtain the next term. 

However, \(t_2=1.5t_1\), but \(t_3 \neq 1.5 t_2\). So this doesn't work either. 

• Terms in this sequence are not obtained by multiplying the previous term by a constant. 

This means that the recursion formula for the sequence is a little bit more complicated than others we've seen.

Even though the differences between the terms aren't constant, they may still help us to write a recursion formula.

So let's calculate each of them. The difference between the first and second terms is \(1\). Similarly, the difference between the second and third terms is \(2\). And the rest of the differences are \(4\), \(8\), and \(16\).

We summarize these findings in a table:

It turns out that there is a pattern in these differences that we can use. Each of them is a power of \(2\).

• \(1\) is \(2^0\).
• \(2\) is \(2^1\).
• \(4\) is \(2^2\), and so on. 

Let's consider how to write some of the terms in the sequence in terms of the previous term.

The second term is \(1\) larger than the first term, or \(t_2=t_1+1\). We can rewrite this using a power of \(2\), where \( t_2= t_1 + 2^0\).

Similarly, term 3 is \(2\) larger than the second term. Or \(t_3=t_2+2\). We can rewrite this as \( t_3= t_2 + 2^1\).

We also know that \(t_4=t_3+4\), which can be written as \( t_4= t_3 + 2^2\).

It turns out that each term after the first can be written as a sum of the previous term and the power of \(2\).

• And notice that for term 2, the exponent on the \(2\) is \(0\).
• For term 3, the exponent on the \(2\) is \(1\).
• And for term 4, the exponent on the \(2\) is \(2\).

The exponent on the \(2\) is always \(2\) less than the term number.

Each term after the first is the previous term, plus \(2\) raised to the exponent two less than the term number ​​\((n-2) \):

 \(t_n=t_{n-1}+2^{n-2}\)

Remember that since each term depends on the preceding term, the formula needs to include a known term.

The sequence can be defined as:

\(t_1=2\), \(t_n=t_{n-1}+2^{n-2}\) for \(n \ge 2\)

You can check this by ensuring that each term produces the correct following term. For example, substitute \(t_4\) in \(n=5\) to ensure that the formula gives the correct value for \(t_5\).

Solution — Option B

Recall the sequence: \(2,~3,~5,~9,~17,~33,\dots\)

Checking differences is often an effective strategy. But it turns out that there is a quicker way to define the sequence if you can see it. 

Notice:

\(2t_1=4\), so \(2t_1-1=3\), or \(t_2\)

\(2t_2=6\), \(\)so \(2t_2-1=5\), or \(t_3\)

\(2t_3=10\), so \(2t_3-1=9\), or \(t_4\)

Each term is one less than two times the previous term:

 \(t_n=2t_{n-1}-1\)

Including the initial term, the sequence can be defined as \(t_1=2\), \(t_n=2t_{n-1}-1\) for \(n \ge 2\).

Again, you may wish to check this by ensuring that each term produces the correct following term when substituted into the formula.

Example 4 Summary

It seems that we have found two recursion formulas that are both valid representations of the sequence \(2,~3,~5,~9,~17,~33,\dots\). They both produced the first six terms of the sequence we were given. It turns out they also both produced the same value for the next term, term 7. And, in fact, both formulas represent the exact same sequence. 

However, in general, it is important to remember that even if two sequences begin with the same six terms, the rest of the terms in the sequence may not be the same. For example, it would be possible to find another recursion formula that satisfies these six terms and produces a different term 7, although the pattern may be much more difficult to find.

Example 5

Solution — Part A

\(t_1=75\), so after the first cappuccino, there is \(75\) mg of caffeine in Barney's system. This means that each cappuccino must contain \(75\) mg of caffeine. 

In the formula \(t_n=75+0.2t_{n-1}\), the \(75\) represents the \(75\) mg of caffeine that is in his most recent cappuccino.

• \(t_{n-1}\) represents the amount of caffeine in Barney's system after the previous cappuccino.
• Since \(0.2t_{n-1}\) represents \(20 \%\) of \(t_{n-1}\), it means that \(20\%\) of the previous caffeine remains in Barney's body after one hour.

Solution — Part B

We can use the recursion formula to generate the six terms in the sequence. 

We are given that \(t_1=75\).

\(t_2=75+0.2t_1\), so we can use \(t_1=75\) to calculate \(t_2\):

Similarly, use \(t_2\) to calculate \(t_3\):

Use \(t_3\) to calculate \(t_4\):

Use \(t_4\) to calculate \(t_5\):

Finally, use \(t_5\) to calculate \(t_6\):

The sequence is \(75,~90,~93,~93.6,~93.72,~93.744 \).

These numbers represent the amount of caffeine (in mg) in Barney's system immediately after each of his six cappuccinos.

Solution — Part C

Graph \(n\) (the term number, or number of cappuccinos) as the independent variable and \(t_n\) as the dependent variable. Since the only possible values of \(n\) are \(n=1,~2,~3,~4,~5,~6\), we do not join the points on the graph. 

We can see on the graph that the amount of caffeine in Barney's system hits a plateau (where the amount of caffeine remains close to constant). From the numbers in the sequence, we know that this plateau occurs somewhere between \(93.7\) and \(93.8\) mg.

In the previous example, we were able to write the terms of the sequence from the recursion formula. Let's practise this some more. 

Example 6

Solution — Part A

We are given that \(\)\(t_1=30\). 

To calculate \(t_2\), use \(t_1\) and \(n=2\):

To calculate \(t_3\), use \(t_2\) and \(n=3\):

To calculate \(t_4\), use \(t_3\) and \(n=4\):

To calculate \(t_5\), use \(t_4\) and \(n=5\):

The first five terms are \(30\), \(22\), \(10\), \(-6\), and \(-26\). 

Solution — Part B

Notice that for this sequence, we are given two known terms. This is because \(t_n\) depends on two of the previous terms. \(t_{n-1}\) is the term that precedes \(t_n\), and \(t_{n-2}\) is the term that precedes \(t_{n-1}\). 

We are given both \(t_1=9\) and \(t_2=-\dfrac13\).

To calculate \(t_3\), use \(t_1\) and \(t_2\):

To calculate \(t_4\), use \(t_2\) and \(t_3\):

To calculate \(t_5\), use \(t_3\) and \(t_4\)

The first five terms are \(9\), \(-\dfrac13\), \(-3\), \(1\), and \(-3\). 

Check Your Understanding 3

Question — Version 1

Given \(t_{1}=2\) and \(t_{n}=\left(t_{n-1}\right)^{2}\) for \(n \geq 2\), find \(t_2\), \(t_3\), and \(t_{4}\)

Answer — Version 1

• \(t_2 = 4\)
• \(t_3 = 16\)
• \(t_4 = 256\)

Feedback — Version 1

Question — Version 2

Given \(t_{1}=30\) and \(t_{n}=t_{n-1}-5 n\) for \(n \geq 2\), find \(t_2\), \(t_3\), and \(t_{4}\)

Answer — Version 2

• \(t_2 = 20\)
• \(t_3 = 5\)
• \(t_4 = -15\)

Feedback — Version 2

Question — Version 3

Given \(t_{1}=-4\) and \(t_{n}=t_{n-1}+3^{n-2}\) for \(n \geq 2\), find \(t_2\), \(t_3\), and \(t_{4}\)

Answer — Version 3

• \(t_2 = -3\)
• \(t_3 = 0\)
• \(t_4 = 9\)

Feedback — Version 3

Question — Version 4

Given \(t_{1}=8\), \(t_{2}=7\), and \(t_{n}=t_{n-1}-t_{n-2}\) for \(n \geq 3\), find \(t_{3}\), \(t_{4}\), and \(t_{5}\)

Answer — Version 4

• \(t_3 = -1\)
• \(t_4 = -8\)
• \(t_5 = -7\)

Feedback — Version 4

Next, let's look at some more sequences that depend on two preceding terms, including the famous Fibonacci sequence.

Example 7

Solution — Part A

We are given the first seven terms. So we need to find \(t_8\), \(t_9\), and \(t_{10}\).

You may have noticed that each term is the sum of the two preceding terms. So term \(8\) can be calculated by adding term \(6\) and term \(7\), and so on. 

Therefore, the next three terms in the sequence are \(21\), \(34\), and \(55\).

Solution — Part B

The two terms that precede \(t_n\) are \(t_{n-1}\) and \(t_{n-2}\).

So in the Fibonacci sequence, \(t_n=t_{n-1}+t_{n-2}\).

Since each term depends on two preceding terms, the recursive formula needs to include two known terms (\(t_1\) and \(t_2\)).

The Fibonacci sequence has a lot of interesting applications, and its numbers show up in many different contexts. 

One of the flowers below has \(5\) petals, while the other has \(21\). Both of these numbers are found in the Fibonacci sequence.

It is so difficult to find a four-leaf clover since most clovers have three leaves — also a number in the Fibonacci sequence.

Source: Clovers - Scacciamosche/iStock/Getty Images

In one direction, this pinecone has eight spirals, while in the other there are thirteen spirals — more Fibonacci numbers!

Let's return to the Try This question from the beginning of the lesson.

Try This Revisited

Solution

Each of the terms and the sequence depend on the two previous terms. So it's a recursive sequence. Thus, we can define a recursion formula. Because each term depends on two preceding terms, we need to have defined two known terms. Fortunately, we have that information.

We know that the first term is \(5\) and that the second term is \(-7\). That is, \(t_1=5\) and \(t_2=-7\).

And from the third term on, term \(n\) is obtained by subtracting the two terms that preceded, which are term \(n -1\) and term \(n -2\).

Recursion formula: \(t_1=5\), \(t_2=-7\), \(t_n=t_{n-1}-t_{n-2}\) for \(n \ge 3\).

Together with the known terms, this completes the recursion formula.

We're asked to find the \(100\)^th term in the sequence. Because of the way the sequence is defined, it appears that to answer this question we need to find all of the terms up to the \(100\)^th term. This is a lot of terms to calculate by hand. So hopefully we'll be able to identify some other pattern in the sequence.

Let's start by finding some more terms.

This last term, \(t_7\), has the same value as \(t_1\), which could mean that the sequence is repeating itself! We'll know for sure if this is true after checking the next term. 

So \(t_9\) will be the same as \(t_3\), and so on. The rest of the sequence will repeat. This makes it significantly easier for us to find the \(100\)^th term. 

We've now identified the sequence is \(5,-7,-12,-5,~7,~12,~5,-7,-12,-5,~7,~12,\dots\)

There are only six different numbers in this sequence, so the \(100\)^th term must be one of them. Organizing our thinking in a table may help us determine which of the six numbers is found at term 100. The headings of this table are each of the six numbers that appear in the sequence in the order that they appear. These numbers represent terms 1 to 6. They also make up the next six terms, or terms 7 to 12. And we know that they will repeat again from terms 13 to 18. The terms are always in groups of six.

We can keep filling in this table until we reach term 100, or we can note that the term numbers in the last column of the table are always multiples of \(6\). Let's try to skip ahead and think of the multiples of \(6\) that are closest to \(100\). Turns out that \(96\) is a multiple of \(6\), and so is \(102\). So term \(96\) and term \(102\) are both \(12\). We can now work backwards from term 102 to identify term 101 and term 100. 

The \(100\)^th term in the sequence is \(-5\). 

If the sequence had had no pattern, finding the \(100\)^th term would have been a very long and repetitive process, unless we set up a computer program or spreadsheet to do it. For a question like this, it's better if we can directly relate each term in the sequence to its term number without having to know any of the previous terms. We'll examine this in the next section.

The General Term of a Sequence

The general term allows us to calculate any term in a sequence, without knowing any of the preceding terms. 

Example 8

Solution

\(t_n=\dfrac{n}{n+7}\) is the general term of this sequence.

To find \(t_2\), substitute \(n=2\):

To find \(t_7\), substitute \(n=7\):

To find \(t_{15}\), substitute \(n=15\):

Check Your Understanding 4

Question — Version 1

For the sequence \(t_{n}=-3 n^{2}-6 n+47\), determine \(t_{3}\).

Answer — Version 1

\(t_3 = 2\)

Feedback — Version 1

\(\begin{align*} t_{n} &=-3 n^{2}-6 n+47 \\ t_{3} &=-3(3)^{2}-6(3)+47 \\ &=-27-18+47 \\ &=2 \end{align*}\)

Question — Version 2

For the sequence \(t_{n}=-\dfrac{840}{n}\), determine \(t_{7}\).

Answer — Version 2

\(t_7 = -120\)

Feedback — Version 2

\(\begin{align*} t_{n} &=-\frac{840}{n} \\ t_{7} &=-\frac{840}{7} \\ &=-120 \end{align*}\)

Question — Version 3

For the sequence \(t_{n}=(-6)^{n-4}\), determine \(t_{6}\).

Answer — Version 3

\(t_6=36\)

Feedback — Version 3

\(\begin{align*} t_{n} &=(-6)^{n-4} \\ t_{6} &=(-6)^{6-4} \\ &=(-6)^{2} \\ &=36 \end{align*}\)

Let's do another example involving the general term of a sequence.

Example 9

Solution

We can determine the first five terms of the sequence by substituting the term numbers 1 to 5 into the general term, \(t_n=7-3n\).

These five terms produce a graph consisting of five points:

Note that if we were to graph the entire sequence, the graph would consist of infinitely many more points.

Comparing \(t_n\) and \(f(x)\)

Here is a more extended look at the graph of the sequence \(t_n=7-3n\). Let's compare it to the graph of the linear function \(f(x)=7-3x\).

Both algebraic representations are almost identical. And the graphs are very similar, as well. You may notice that all of the points that lie on the graph of the sequence also lie on the graph of the line. In fact, the graph of the sequence is just a subset of the graph of the linear function, where the values of \(x\) can only be positive whole numbers or natural numbers. That is, \(x \in \mathbb{N}\). So, in fact, we can represent the sequence \(t_n=7-3n\) with the function \(f(x)=7-3x\), as long as we restrict the domain to the natural numbers.

The graph of \(f(x)=7-3x\) appears different because, for the linear function, the domain has not been restricted. It consists of all the real numbers. The other function, \(f(x)=7-3x\), \(x \in \mathbb{N}\), is an example of a discrete function.

Discrete Functions

Another example of a discrete function is the function \(f(x)=2(x-1)(x-3)\), where \(x \in \mathbb{N}\). Here we can see the points on its graph, which are limited to points with positive whole number values of \(x\). Compare this to the graph of a function that is not discrete, the function \(f(x)=2(x-1)(x-3)\), where \(x \in \mathbb{R}\).

All of the points on the discrete function are part of the graph of \(f(x)=2(x-1)(x-3)\) as well, but a parabola is drawn through them since there are no restrictions on \(x\).

Sequences as Discrete Functions

This gives us another way to represent a sequence algebraically, in addition to a recursion formula or a general term. Although the discrete function is directly related to the general term, in fact, it's quite straightforward to represent a sequence in function notation, if we already know or are given the general term, \(t_n\).

To represent a sequence in function notation, given the general term, \(t_n\):

• Replace \(n\) with \(x\) and replace \(t_n\) with \(f(x)\).
• Remember to restrict the domain of \(f(x)\) to the natural numbers. \(\)

For example, the sequence represented by \(t_n=\dfrac n{n+7}\) can also be represented by \(f(x)=\dfrac x{x+7}\), \(x \in \mathbb{N}\).

Example 10

Solution — Part A

To write a general term, we need to be able to relate each term directly to the term number. When looking at the numbers in the sequence, we notice that some of the terms include square roots. It would be nice to write all of the terms with the square root for consistency. And in fact, we can do this.

The first term, \(1\), is equal to \(\sqrt{1}\). And the fourth term, \(2\), is equal to \(\sqrt{4}\). 

The sequence can be rewritten with these square roots included:

\(\sqrt1,~\sqrt 2,~\sqrt 3,~\sqrt4,~\sqrt 5,\dots\)

Each term is obtained by taking the square root of the term number. This means that the general term is

\(t_n=\sqrt n\)

We can also write this as the discrete function

\(f(x)=\sqrt x\), where \(x \in \mathbb{N}\)

Solution — Part B

Again, we need to find a relationship between each term and the term number. In this sequence, many of the terms are fractions with a \(5\) in the numerator. It turns out that we can write the other terms in this form as well.

The first term, \(5\), is equal to \(\dfrac{5}{1}\), and the fifth term, \(1\), is equal to \(\dfrac{5}{5}\).

So we can rewrite the sequence so that every term is a fraction, and we notice that the sequence can be re-written as 

\(\dfrac51,~\dfrac52,~\dfrac53,~\dfrac54,~\dfrac55,~\dfrac56,\dots\)

Each term is \(5\) divided by the term number.

The general term is

\(t_n=\dfrac5n\)

Using function notation, we represent this as

\(f(x)=\dfrac5x\), where \(x \in \mathbb{N}\)

Multiple Representations of Sequences

So far in this lesson, we have seen how to represent the terms of a sequence in a table or graph. We have also seen three different algebraic representations of a sequence: a recursion formula, a general term, or a discrete function. A recursion formula relates each term to at least one of the preceding terms, while the general term and discrete function relate each term directly to the term number. 

In this part of the lesson, we will look at finding multiple algebraic representations for the same sequence, and explore how to represent the sequence in one form, when given another. 

Example 11

Solution — Part A

Let's examine how each term in the sequence relates to the preceding term. 

\(9=3 \times 3\), so \(t_2=3t_1\), \(27=3 \times 9\), so \(t_3=3t_2\), and so on.

Each term in the sequence is three times the preceding term.

Since each term depends on the preceding term, the recursion formula must provide one known term. This is most often the first term, \(t_1=3\).

The recursion formula representing this sequence is: \(t_1=3\), \(t_n=3t_{n-1}\) for \(n \ge 2\). 

Solution — Part B

We can also note that each term is a power of \(3\).

\(t_1=3^1\), \(t_2=3^2\), \(t_3=3^3\), and so on.

Each term is obtained by evaluating a power of \(3\), where the exponent is the term number, \(n\).

So the general term is \(t_n=3^n\).

Solution — Part C

The discrete function is almost identical to the general term: \(f(x)=3^x\), where \(x \in \mathbb{N}\).

Here's an example of a sequence whose algebraic representations are a little bit more difficult to identify. 

Example 12

Solution — Recursion Formula

Giving the terms of the sequence a quick glance, we can see that there is no common difference or ratio. That is, the terms don't increase or decrease by the same number each time. And the terms can't be generated by multiplying or dividing the previous term by the same number every time. The relationship between a term and the preceding term is a little more complicated.

Even though the differences between the terms are not the same, there still might be a pattern in them that might help us define the sequence. So let's calculate the first differences. We can organize our thinking by entering the terms into a table where the first column is for the term number, the second is for the terms themselves, and the last is for the first differences.

Each of these differences is a multiple of \(-2\). 

This is enough for us to identify a consistent pattern in the terms of the sequence.

Each term consists of the previous term, subtracted by a multiple of \(2\).

• We can also see that for term two, the \(2\) is multiplied by \(0\).
• For term three, the \(2\) is multiplied by \(1\).
• For term four, the \(2\) is multiplied by \(2\).

The \(2\) is multiplied by a number that is two less than the term number ​​(​\(n-2\)), which is represented by the expression \(n -2\). 

Therefore, the recursion formula is \(t_1=0\), \(t_n=t_{n-1}-2(n-2)\) for \(n \ge 2\).

Solution — General Term and Function Notation

Let's move on to the general term in the discrete function. These will both relate the term directly to its term number. So far, the connection between these two things is not necessarily obvious. However, in developing the recursion formula, we found the first differences, and found that there was a pattern in these differences.

In fact, if we calculate the second differences, we find that they're all constant at \(-2\).

So the terms in the sequence can be represented by a discrete quadratic function, \(f(x)\), where \(x\) is the term number. 

Since \(f (x)\) is \(0\) when \(x =1\) and when \(x =2\), the zeros of the function are ‌\(1\) and ‌\(2\). So, in factored form, the function is written as

\(f(x)=a(x-1)(x-2)\)

We can use another term to calculate to find the value of \(a\). Since term 3 is \(-2\), \(f(3)=-2\). Substitute these values into the function and solve:

\(\begin{align*} -2 &=a(3-1)(3-2) \\ -2 &=a(2)(1) \\ -2 &=2 a \\ a &=-1 \end{align*}\)

Therefore, the function representing the sequence is \(f(x)=-(x-1)(x-2)\), where \(x \in \mathbb{N}\). 

We can write the general term based on the discrete function. The general term is \(t_n=-(n-1)(n-2)\). 

Example 13

Solution

From the general term, we can see that the sequence can be represented by a discrete quadratic function. 

There will be no common difference between the terms, but there will be a pattern in the first differences (since the second differences will be constant).
To find this pattern, evaluate some of the terms in the sequence: 

\(\begin{align*} t_1&=\dfrac32(1)^2+\dfrac32(1)-13 \\ &= -10 \end{align*}\)

\(\begin{align*} t_2&=\dfrac32(2)^2+\dfrac32(2)-13 \\ &= -4 \end{align*}\)

\(\begin{align*} t_3&=\dfrac32(3)^2+\dfrac32(3)-13 \\ &= 5 \end{align*}\)

\(\begin{align*} t_4&=\dfrac32(4)^2+\dfrac32(4)-13 \\ &= 17 \end{align*}\)

The first four terms of the sequence are \(-10\), \(-4\), \(5\), and \(17\). 

To find a relationship between each term and the preceding term, we can use a table to examine the first differences: 

Each of the first differences is a multiple of \(3\). So each term is obtained by adding a multiple of \(3\) to the previous term. 

More specifically: 

• \(t_2\) is obtained by adding \(6\) to \(t_1\), so \(t_2=t_1+6\), or \(t_2=t_1+3(2)\).
• \(t_3\) is obtained by adding \(9\) to \(t_2\), so \(t_3=t_2+9\), or \(t_3=t_2+3(3)\). 
• \(t_4\) is obtained by adding \(12\) to \(t_3\), so \(t_4=t_3+12\), or \(t_4=t_3+3(4)\). 

Notice that the \(3\) is always multiplied by the term number!

This means that the \(n\)^th term (\(t_n\)) will be the previous term (\(t_{n-1}\)) plus \(3n\).

Since each term depends on one previous term, we also need to provide one known term in the recursion formula. This can be the first term: \(t_1=-10\). 

So altogether, the recursion formula is: \(t_1=-10\), \(t_n=t_{n-1}+3n\) for \(n \ge 2\).

Example 14

Solution

To find \(t_{50}\) using the recursion formula, we would need to find each of the \(49\) previous terms. It would be best if we could find the general term for this sequence so that we can calculate the \(50\)^th term directly. 

To do this, evaluate the first few terms of the sequence to identify any patterns:

We are given that \(t_1=\dfrac13\).

Use \(t_1=\dfrac13\) and \(n=2\) to evaluate \(t_2\):

 \(\begin{align*} t_2&=t_1-\dfrac1{(2+1)(2+2)}\\ &= \dfrac13 - \dfrac1{12} \\ &= \dfrac14 \end{align*}\)

Use \(t_2=\dfrac14\) and \(n=3\) to evaluate \(t_3\):

 \(\begin{align*} t_3&=t_2-\dfrac1{(3+1)(3+2)}\\ &= \dfrac14 - \dfrac1{20} \\ &= \dfrac15 \end{align*}\)

Use \(t_3=\dfrac15\) and \(n=4\) to evaluate \(t_4\):

 \(\begin{align*} t_4&=t_3-\dfrac1{(4+1)(4+2)}\\ &= \dfrac15 - \dfrac1{30} \\ &= \dfrac16 \end{align*}\)

We have found the first four terms of the sequence to be \(\dfrac13\), \(\dfrac14\), \(\dfrac15\), and \(\dfrac16\).

Each term is a fraction with \(1\) in the numerator, while the denominator changes depending on the term.

In fact, the denominator is always \(2\) greater than the term number, or \(n+2\). 

So the general term for this sequence is \(t_n=\dfrac1{n+2}\).

(Note: It is possible to show that this formula satisfies \(t_n=t_{n-1}-\dfrac1{(n+1)(n+2)}\).)  

Use the general term to evaluate \(t_{50}\): 

\(\begin{align*} t_{50}&= \dfrac1{50+2} \\ &=\dfrac1{52} \end{align*}\)

Example 15

Solution

Equating the two formulas produces the equation \(t_{n-1}+2^{n-2}=2t_{n-1}-1\). 

Subtract \(t_{n-1}\) from both sides of this equation: 

 \(2^{n-2}=t_{n-1}-1 \) 

Rearrange the result:

 \(t_{n-1}=2^{n-2}+1\) 

This formula can be used to calculate any term in the sequence. 

For example,

 \(\begin{align*} t_{2-1}&=t_1 \\&=2^{2-2}+1 \\ &= 2^0+1 \\ &= 1+1 \\ &= 2 \end{align*}\)

This is the correct first term. 
We can write a formula for \(t_n\), instead of \(t_{n-1}\), by noticing that the exponent on the 2 is always one less than the term number.

• For \(t_{n-1}\), the exponent on the \(2\) is \(n-2\). 

• So for \(t_n\), the exponent on the \(2\) should be \(n-1\). 

Therefore, the general term for the sequence is \(t_n=2^{n-1}+1\). 

(It is possible to check that this formula produces all of the correct terms in the sequence.) 

Take a moment to consider why the strategy we used in this example worked.  Is it always possible to find the general term this way if you have two different recursion formulas?  Why or why not?

Summary of Algebraic Representations of a Sequence

In this lesson, we discussed three different types of algebraic representations for a sequence: a recursion formula, a general term, and a discrete function. For some sequences, it is equally easy to write all three of these types of representations. For other sequences, a recursion formula is much simpler to write than a general term or discrete function, and vice-versa.

It is important to keep in mind that each type of representation has its own benefits and provides different information about the sequence. Here is a summary of some of those features: 

Check Your Understanding 5

Question

What are some different ways a sequence can be represented?

Match each cell in the table with the appropriate option from the options provided.

Options

1. \(t_{1}=160\), \(t_{n}=t_{n-1}-4 n\) for \(n \geq 2\)
2. \(f(x)=\dfrac{160}{x}\), where \(x \in \mathbb{N}\)
3. \(f(x)=-4 x+164\), where \(x \in \mathbb{N}\)
4. \(t_{1}=160\), \(t_{n}=t_{n-1}-\dfrac{160}{n(n-1)}\) for \(n \geq 2\)
5. \(t_n = \dfrac{160}{n}\)
6. \(t_n = 164 - 4n\)
7. \(f(x)=-2 x^{2}-2 x+164\), where \(x \in \mathbb{N}\)
8. \(t_{1}=160\),  \(t_{n}=t_{n-1}-4\) for \(n \geq 2\)
9. \(t_{n}=-2 n^{2}-2 n+164\)

Hints:

• Recursion Formula: Consider how each term is related to the preceding term. Try evaluating \(t_{2}\) using one of the recursion formulas to help you differentiate between the sequences.
• General Term: Consider how each term is related to the term number. Try evaluating \(t_{2}\) using one of the general terms to help you differentiate between the sequences.
• Discrete Function: The discrete function is almost identical to the general term. Replace \(n\) with \(x\) and replace \(t_{n}\) with \(f(x)\) . Remember also to restrict the domain.

Answer

• Cell 1: H. \(t_{1}=160\),  \(t_{n}=t_{n-1}-4\) for \(n \geq 2\)
• Cell 2: F. \(t_n = 164 - 4n\)
• Cell 3: C. \(f(x)=-4 x+164\), where \(x \in \mathbb{N}\)
• Cell 4: A. \(t_{1}=160\), \(t_{n}=t_{n-1}-4 n\) for \(n \geq 2\)
• Cell 5: I. \(t_{n}=-2 n^{2}-2 n+164\)
• Cell 6: G. \(f(x)=-2 x^{2}-2 x+164\), where \(x \in \mathbb{N}\)
• Cell 7: D. \(t_{1}=160\), \(t_{n}=t_{n-1}-\dfrac{160}{n(n-1)}\) for \(n \geq 2\)
• Cell 8: E. \(t_n = \dfrac{160}{n}\)
• Cell 9: B. \(f(x)=\dfrac{160}{x}\), where \(x \in \mathbb{N}\)

Interactive Version

Representations of Sequences

Wrap-Up

Take It With You

You may have noticed that many sequences are defined recursively as \(t_n=t_{n-1}+f(n)\), where \(f(n)\) is some function involving the term number. An example of this is \(t_n=t_{n-1}+2^{n-2}\) for \(n \ge 2\). In this sequence, we would define \(f(n)=2^{n-2}\). 

If the pattern in the sequence \(t_n=t_{n-1}+f(n)\) for \(n \ge 2\) is quadratic, what type of function is \(f(n)\)?