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 [latex]{a}_{n}[/latex] in terms of preceding terms. For example, suppose we know the following:

Solution

The first term is given in the formula. For each subsequent term, we replace [latex]{a}_{n - 1}[/latex] with the value of the preceding term.

[latex]\begin{array}{ll}n=1\begin{array}{lllll}\hfill & \hfill & \hfill & \hfill & \hfill \end{array}\hfill & {a}_{1}=9\hfill \\ n=2\hfill & {a}_{2}=3{a}_{1}-20=3\left(9\right)-20=27 - 20=7\hfill \\ n=3\hfill & {a}_{3}=3{a}_{2}-20=3\left(7\right)-20=21 - 20=1\hfill \\ n=4\hfill & {a}_{4}=3{a}_{3}-20=3\left(1\right)-20=3 - 20=-17\hfill \\ n=5\hfill & {a}_{5}=3{a}_{4}-20=3\left(-17\right)-20=-51 - 20=-71\hfill \end{array}[/latex]

The first five terms are [latex]\left\{9,\text{ }7,\text{ }1,\text{ }-17,\text{ }-71\right\}[/latex].

Figure 6

Solution

The first two terms are given. For each subsequent term, we replace [latex]{a}_{n - 1}[/latex] and [latex]{a}_{n - 2}[/latex] with the values of the two preceding terms.

[latex]\begin{array}{lllllll}n=3\hfill & \hfill & \hfill & \hfill & \hfill & \hfill & {a}_{3}=3{a}_{2}+4{a}_{1}=3\left(2\right)+4\left(1\right)=10\hfill \\ n=4\hfill & \hfill & \hfill & \hfill & \hfill & \hfill & {a}_{4}=3{a}_{3}+4{a}_{2}=3\left(10\right)+4\left(2\right)=38\hfill \\ n=5\hfill & \hfill & \hfill & \hfill & \hfill & \hfill & {a}_{5}=3{a}_{4}+4{a}_{3}=3\left(38\right)+4\left(10\right)=154\hfill \\ n=6\hfill & \hfill & \hfill & \hfill & \hfill & \hfill & {a}_{6}=3{a}_{5}+4{a}_{4}=3\left(154\right)+4\left(38\right)=614\hfill \end{array}[/latex]

The first six terms are [latex]\text{{1,2,10,38,154,614}}\text{.}[/latex]

Figure 7

Using Factorial Notation

Analysis of the Solution

Figure 8 shows the graph of the sequence. Notice that, since factorials grow very quickly, the presence of the factorial term in the denominator results in the denominator becoming much larger than the numerator as [latex]n[/latex] increases. This means the quotient gets smaller and, as the plot of the terms shows, the terms are decreasing and nearing zero.

Figure 8

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