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. 

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 \(-3\). 

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!