Lesson 5
Sequences are Functions
 Let's learn how to define a sequence recursively.
5.1: Bowling for Triangles (Part 1)
Describe how to produce one step of the pattern from the previous step.
5.2: Bowling for Triangles (Part 2)
Here is a visual pattern of dots. The number of dots \(D(n)\) is a function of the step number \(n\).
 What values make sense for \(n\) in this situation? What values don't make sense for \(n\)?
 Complete the table for Steps 1 to 5.
\(n\) \(D(n)\) 1 1 2 \(D(1)+2=3\) 3 \(D(2)+3=6\) 4 5  Following the pattern in the table, write an equation for \(D(n)\) in terms of the previous step. Be prepared to explain your reasoning.
Consider the same triangular pattern.

Is the sequence defined by the number of dots in each step arithmetic, geometric, or neither? Explain how you know.

Can you write an expression for the number of dots in Step \(n\) without using the value of \(D\) from a previous step?
5.3: Let's Define Some Sequences
Use the first 5 terms of each sequence to state if the sequence is arithmetic, geometric, or neither. Next, define the sequence recursively using function notation.
 \(A\): 30, 40, 50, 60, 70, . . .
 \(B\): 80, 40, 20, 10, 5, 2.5, . . .
 \(C\): 1, 2, 4, 8, 16, 32, . . .
 \(D\): \(1, \frac12, \frac14, \frac18, \frac{1}{16},\) . . .
 \(E\): 20, 13, 6, 1, 8, . . .
 \(F\): 1, 3, 7, 15, 31, . . .
Summary
Sometimes we can define a sequence recursively. That is, we can describe how to calculate the next term in a sequence if we know the previous term.
Here’s a sequence: 6, 10, 14, 18, 22, . . . This is an arithmetic sequence, where each term is 4 more than the previous term. Since sequences are functions, let's call this sequence \(f\) and then we can use function notation to write \(f(n) = f(n1) + 4\). Here, \(f(n)\) is the term, \(f(n1)\) is the previous term, and + 4 represents the rate of change since \(f\) is an arithmetic sequence.
When we define a function recursively, we also must say what the first term is. Without that, there would be no way of knowing if the sequence defined by \(f(n) = f(n1) + 4\) started with 6 or 81 or any other number. Here, one possible initial condition is \(f(1) = 6\). (It could also make sense to number the terms starting with 0, using \(f(0) = 6,\) and we'll talk more about this later.)
Combining this information gives the recursive definition: \(f(1) = 6\) and \(f(n) = f(n1) + 4\) for \(n \ge2\), where \(n\) is an integer. We include the \(n\ge2\) at the end since the value of \(f\) at 1 is already given and the other terms in the sequence are generated by inputting integers larger than 1 into the definition.
Glossary Entries

arithmetic sequence
A sequence in which each term is the previous term plus a constant.

geometric sequence
A sequence in which each term is a constant times the previous term.