Lesson 5

Describe how to produce one step of the pattern from the previous step.

 

Here is a visual pattern of dots. The number of dots \(D(n)\) is a function of the step number \(n\).

1. What values make sense for \(n\) in this situation? What values don't make sense for \(n\)?
2. 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

3. Following the pattern in the table, write an equation for \(D(n)\) in terms of the previous step. Be prepared to explain your reasoning.

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.

1. \(A\): 30, 40, 50, 60, 70, . . .
2. \(B\): 80, 40, 20, 10, 5, 2.5, . . .
3. \(C\): 1, 2, 4, 8, 16, 32, . . .
4. \(D\): \(1, \frac12, \frac14, \frac18, \frac{1}{16},\) . . .
5. \(E\): 20, 13, 6, -1, -8, . . .
6. \(F\): 1, 3, 7, 15, 31, . . .

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 \(f(n) = f(n-1) + 4\). Here, \(f(n)\) is the term, \(f(n-1)\) 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 \(f(n) = f(n-1) + 4\) 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 \(f(n) = f(n-1) + 4\) 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.