1. Bellwork 1) Find the fifth term of the sequence: a n = 3n + 2 a n = 3n + 2 2) Find the next three terms of the sequence 3, 5, 7, 9, 11, … Explain your answer. 3)Find the next three terms of the sequence 1, 2, 4, 8, 16, … Explain your answer.

2. Solution 1) Find the fifth term of the sequence: a n = 3n + 2; a n = 3n + 2; a 5 = 3(5) + 2 = 17 2) Find the next three terms of the sequence 3, 5, 7, 9, 11, … Explain your answer. 13, 15, 17 Each term is 2 more than the previous term. 3) Find the next three terms of the sequence 1, 2, 4, 8, 16, … Explain your answer. 32, 64, 128 Each term is 2 times the previous term.

3. A quick review… We know a sequence is a series of numbers that have a specific order. The numbers in the sequence are called terms. We refer to each term as a n. For example, the fifth term is a 5 and the eighth term is a 8 (pronounced “a sub 5” and “a sub 8”).

4. Let’s look at an example Consider the sequence 4, 8, 12, 16, 20, 24, … The third term in the sequence, a 3, is 12. What is the first term? What is the sixth term? What is the seventh term?

5. In the previous example, we came up with the seventh term by looking at the pattern and applying it to the next term in the sequence. But what if we are asked to find the 150 th term? Today, we are going to look at 2 ways to write a rule (an equation) for finding the n th term in a sequence: closed (explicit) formula and recursive formula.

6. Explicit Formula You’ve already been using “explicit formula” sequences for the past few days. With an explicit formula, we can easily find any term in the series. Let’s Review! a n = a 1 + (n-1) * d for Arithmetic Write the explicit formula for the following: 3, 6, 9, 12, … 4, 2, 0, -2, ….

7. With an explicit formula, we do not need to know what the previous terms are in order to calculate the next term. Let’s practice: a) Find the 5 th term: a n = n + 8 b) Find the 8 th term: a n = n 2 – 4 c) Find the 150 th term: a n = 2n + 5 d) Find the 50 th term: a n = 3n

8. Recursive Formula What does recursive mean? The dictionary defines recursive as pertaining to or using a rule or procedure that can be applied repeatedly. So, to simplify things, when you see recursive, I want you to think repeat.

9. Example of a recursive formula a 1 = 4 a n = a n-1 + 4 Notice that the formula has 2 parts: 1. It defines the first term. In the above example, a 1 = 4. 2. It defines the remaining terms. In our example, a n =a n-1 + 4. What does a n-1 mean???

10. Find the first six terms of the recursive sequence a 1 = 3 a n = a n-1 + 5 a 1 = 3 a 2 = 3 + 5 = 8 a 3 = 8 + 5 = 13 a 4 = 13 + 5 = 18 a 5 = 18 + 5 = 23 a 6 = 23 + 5 = 28 Repeat Repeat

11. If you wanted to find the 50 th term, you would have to repeat your calculations 50 times. Can you see why we associate recursive with repeat?

12. Just to summarize An explicit formula has only one equation. You can easily find any term in the sequence. ***** A recursive formula has 2 parts: the first term, and a rule, or equation, for finding the remaining terms based on knowing the previous term. In order to find the 50 th term, you will first have to find the previous 49 terms.

13. Writing formulas We are going to look at how to write a recursive formula and a closed formula when we are given a sequence. First we’ll write the recursive formula, and from that, we will write the explicit formula. We will look at two kinds of sequences: one where you add to get the next term and one where you multiply to get the next term.

14. Sequences involving addition

15. 1, 3, 5, 7, 9, … In the series above, we notice that each term in the series is just 2 more than the term before it. We can say that a 1 = 1 a 2 = a 1 + 2 a 3 = a 2 + 2 a 4 = a 3 + 2 Instead of writing a 2, a 3, a 4, etc, we can shorten it to: a n = a n-1 + 2

16. 1, 3, 5, 7, 9, … So our recursive formula is: a 1 = 1 a n = a n-1 + 2 Notice our formula has both parts: a 1 and a n

17. Let’s look at another example: 4, 8, 12, 16, 20, … First we define a 1 : a 1 = 4 Next we write the rule for finding the nth term. We notice that each term is 4 more than the previous term. So, a n = a n-1 + 4

18. You try… Write the recursive formula for the following sequences: 1) 0, 5, 10, 15, 20, … 2) 3, 10, 17, 24, … 3) 1, 11, 21, 31, … 4)2, 4, 6, 8, …

19. Let’s see the results… 1) 0, 5, 10, 15, 20, … Each term is 5 more than the previous term. a 1 = 0 a n = a n-1 + 5 2)3, 10, 17, 24, … Each term is 7 more than the previous term. a 1 = 3 a n = a n-1 + 7

20. Let’s see the results… 3)1, 11, 21, 31, …Each term is 10 more than the previous term. a 1 = 1 a n = a n-1 + 10 4)2, 4, 6, 8, …Each term is 2 more than the previous term. a 1 = 2 a n = a n-1 + 2

21. Let’s write the explicit formula for the four sequences we just defined. Pay attention to the pattern… Problem 1: Recursive: a 1 = 0 a n = a n-1 + 5 Explicit: a n = 0 + (n-1)5 Problem 2: Recursive: a 1 = 3 a n = a n-1 + 7 Explicit: a n = 3 + (n-1) 7

22. Can you see the pattern??? Problem 3: Recursive: a 1 = 1 a n = a n-1 + 10 Explicit: a n = 1 + (n-1) 10 Problem 4: Recursive: a 1 = 2 a n = a n-1 + 2 Explicit: a n = 2 + (n-1) 2

23. Sequences involving multiplication

24. 1, 3, 9, 27, 81, … In the sequence above, we notice that each term in the series is 3 times the term before it (Note that this is a geometric sequence) We can say that a 1 = 1 a 2 = 3a 1 a 3 = 3a 2 a 4 = 3a 3 Instead of writing a 2, a 3, a 4, etc, we can shorten it to: a n = 3a n-1

25. 1, 3, 9, 27, 81, … So our recursive formula is: a 1 = 1 a n = 3a n-1 Notice our formula has both parts: a 1 and a n

26. Let’s look at another example: 4, 8, 16, 32, 64, … First we define a 1 : a 1 = 4 Next we write the rule for finding the nth term. We notice that each term is 4 more than the previous term. So, a n = 4a n-1

27. You try… Write the recursive formula for the following sequences: 1) 1, 5, 25, 125, … 2) 3, 12, 48, 192, … 3) 4, 24, 144, 864, … 4)2, 20, 200, 2000, …

28. Let’s see the results… 1) 1, 5, 25, 125, … Each term is 5 times the previous term. a 1 = 1 a n = 5a n-1 2)3, 12, 48, 192, … Each term is 4 more than the previous term. a 1 = 3 a n = 4a n-1

29. Let’s see the results… 3)4, 24, 144, 864, … Each term is 6 times the previous term. a 1 = 4 a n = 6a n-1 4)2, 20, 200, 2000, … Each term is 10 more than the previous term. a 1 = 2 a n = 10a n-1

30. Let’s write the explicit formula for the four sequences we just defined. Pay attention to the pattern… Problem 1: Recursive: a 1 = 1 a n = 5a n-1 Explicit: a n = 5 (n-1) Check: a 4 = 5 3 = 125 Problem 2: Recursive: a 1 = 3 a n = 4a n-1 Explicit: a n = 3*4 n-1 Check: a 4 = 3(4) 3 = 192

31. Let’s write the explicit formula for the four sequences we just defined. Pay attention to the pattern… Problem 3: Recursive: a 1 = 4 a n = 6a n-1 Explicit: a n = 4*6 (n-1) Check: a 3 = 4(6) 2 = 144 Problem 4: Recursive: a 1 = 2 a n = 10a n-1 Explicit: a n = 2*10 n-1 Check: a 4 = 2(10) 3 = 2000