Sequences Defined by a Recursive Formula

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

$a_{10}=a_9+a_8=34+21=55$

Recursive Formula

A recursive formula is a formula that defines each term of a sequence using preceding term(s). Recursive formulas must always state the initial term, or terms, of the sequence.

Q&A

Must the first two terms always be given in a recursive formula?

No. The Fibonacci sequence defines each term using the two preceding terms, but many recursive formulas define each term using only one preceding term. These sequences need only the first term to be defined.

How To

Given a recursive formula with only the first term provided, write the first $n$ terms of a sequence.

1. Identify the initial term, $a_1$, which is given as part of the formula. This is the first term.

2. To find the second term, $a_2$, substitute the initial term into the formula for $a_{n−1}$. Solve.

3. To find the third term, $a_3$, substitute the second term into the formula. Solve.

4. Repeat until you have solved for the $nth$ term.

Example 5

Writing the Terms of a Sequence Defined by a Recursive Formula

Write the first five terms of the sequence defined by the recursive formula.

$\begin{array}{lll}a_1 = 9 \\a_n = 3a_{n-1} - 20, \text { for } n \geq 2\end{array}$

Solution

The first term is given in the formula. For each subsequent term, we replace $a_{n−1}$ with the value of the preceding term.

$\begin{array}{ll}n=1 & a_{1}=9 \\n=2 & a_{2}=3 a_{1}-20=3(9)-20=27-20=7 \\n=3 & a_{3}=3 a_{2}-20=3(7)-20=21-20=1 \\n=4 & a_{4}=3 a_{3}-20=3(1)-20=3-20=-17 \\n=5 & a_{5}=3 a_{4}-20=3(-17)-20=-51-20=-71\end{array}$

The first five terms are $\{9,7,1,–17,–71\}$. See Figure 5.

Figure 5

Try It #7

Write the first five terms of the sequence defined by the recursive formula.

$\begin{array}{ll}a_1 = 2 \\a_n = 2a_{n-1}+1, \text { for } n \geq 2\end{array}$

How To

Given a recursive formula with two initial terms, write the first $n$ terms of a sequence.

1. Identify the initial term, $a_1$, which is given as part of the formula.

2. Identify the second term, $a_2$, which is given as part of the formula.

3. To find the third term, substitute the initial term and the second term into the formula. Evaluate.

4. Repeat until you have evaluated the $nth$ term.

Example 6

Writing the Terms of a Sequence Defined by a Recursive Formula

Write the first six terms of the sequence defined by the recursive formula.

$\begin{array}{ll}a_1 = 2 \\a_2 = 2 \\a_n = 3a_{n-1} + 4a_{n-2}, \text { for } n \geq 3\end{array}$

Solution

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.

$\begin{array}{ll}n=3 & a_{3}=3 a_{2}+4 a_{1}=3(2)+4(1)=10 \\n=4 & a_{4}=3 a_{3}+4 a_{2}=3(10)+4(2)=38 \\n=5 & a_{5}=3 a_{4}+4 a_{3}=3(38)+4(10)=154 \\n=6 & a_{6}=3 a_{5}+4 a_{4}=3(154)+4(38)=614\end{array}$

The first six terms are $\{1,2,10,38,154,614\}$. See Figure 6.

Figure 6

Try It #8

Write the first 8 terms of the sequence defined by the recursive formula.

$\begin{array}{ll}a_1 = 0 \\a_2 = 1 \\a_3 = 1 \\a_2 = \frac{a_{n-1}}{a_{n-2}}+a_{n-3}, \text { for } n \geq 4\end{array}$

$n$ Factorial

$n$ factorial is a mathematical operation that can be defined using a recursive formula. The factorial of $n$, denoted $n!$, is defined for a positive integer $n$ as:

$\begin{array}{ll}0! = 1 \\1! = 1 \\n!=n(n-1)(n-2) \cdot\cdot\cdot(2)(1), \text { for } n \geq 2\end{array}$

Q&A

Can factorials always be found using a calculator?

No. Factorials get large very quickly – faster than even exponential functions! When the output gets too large for the calculator, it will not be able to calculate the factorial.

Example 7

Writing the Terms of a Sequence Using Factorials

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.

$\begin{array}{ll}n=1 & a_{1}=\frac{5(1)}{(1+2) !}=\frac{5}{3 !}=\frac{5}{3 \cdot 2 \cdot 1}=\frac{5}{6} \\n=2 & a_{2}=\frac{5(2)}{(2+2) !}=\frac{10}{4 !}=\frac{10}{4 \cdot 3 \cdot 2 \cdot 1}=\frac{5}{12} \\n=3 & a_{3}=\frac{5(3)}{(3+2) !}=\frac{15}{5 !}=\frac{15}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}=\frac{1}{8} \\n=4 & a_{4}=\frac{5(4)}{(4+2) !}=\frac{20}{6 !}=\frac{20}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}=\frac{1}{36} \\n=5 & a_{5}=\frac{5(5)}{(5+2) !}=\frac{25}{7 !}=\frac{25}{7\cdot 6 \cdot 5 \cdot 4 \cdot \cdot \cdot \cdot\cdot \cdot 1}=\frac{5}{1,008}\end{array}$

The first five terms are $\{\frac{5}{6},\frac{5}{12},\frac{1}{8},\frac{1}{36},\frac{5}{1,008}\}$.

Analysis

Figure 7 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 $n$ increases. This means the quotient gets smaller and, as the plot of the terms shows, the terms are decreasing and nearing zero.

Figure 7

Try It #9

Write the first five terms of the sequence defined by the explicit formula $a_n=\frac{(n+1)!}{2n}$.