1. Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing this problem.

2. Designed by David Jay Hebert, PhD A pattern of numbers in a particular order is called a number sequence, and the individual numbers in the sequence are called terms.

3. Designed by David Jay Hebert, PhD Each number or term in the sequence is associated with a position, which is also a number.

4. Designed by David Jay Hebert, PhD Examples of Sequences 1,2,3,4,5,…. 1,1,2,3,5,8,… 1,2,4,8,16,32,… 2,1,7,3,9,3,… The dots at the end of the sequence indicate that the sequence continues without end. Note not all sequences have a pattern.

5. Designed by David Jay Hebert, PhD Arithmetic Sequences The sequence 2, 5, 8, 11, 14,… Has first differences of 3 all the time, this makes this sequence arithmetic. If a sequence is arithmetic the first differences must be the same.

6. Designed by David Jay Hebert, PhD First Differences First differences are the differences found by subtracting two consecutive numbers in a sequence. Ex. 2, 5, 8, 11, 14,….

7. Designed by David Jay Hebert, PhD First Differences First differences are the differences found by subtracting two consecutive numbers in a sequence. Ex. 2, 5, 8, 11, 14,…. 5 – 2 = 3 3

8. Designed by David Jay Hebert, PhD First Differences First differences are the differences found by subtracting two consecutive numbers in a sequence. Ex. 2, 5, 8, 11, 14,…. 8 – 5 = 3 33

9. Designed by David Jay Hebert, PhD First Differences First differences are the differences found by subtracting two consecutive numbers in a sequence. Ex. 2, 5, 8, 11, 14,…. 11 - 8 = 3 333

10. Designed by David Jay Hebert, PhD First Differences First differences are the differences found by subtracting two consecutive numbers in a sequence. Ex. 2, 5, 8, 11, 14,…. Since all the first differences are the same this must be an arithmetic sequence 3333

11. Designed by David Jay Hebert, PhD Position Numbers Each term in the sequence is paired with a position number 2, 5, 8, 11, 14,…

12. Designed by David Jay Hebert, PhD Arithmetic Sequences 2, 5, 8, 11, 14,… (recall the first difference is 3) 5 = 2 + 3

13. Designed by David Jay Hebert, PhD Arithmetic Sequences 2, 5, 8, 11, 14,… (recall the first difference is 3) 5 = 2 + 3 8 = 5 + 3 = 2 + 3 + 3 (the underlined part is the previous term)

14. Designed by David Jay Hebert, PhD Arithmetic Sequences 2, 5, 8, 11, 14,… (recall the first difference is 3) 5 = 2 + 3 8 = 5 + 3 = 2 + 3 + 3 11 = 8 + 3 = 2 + 3 + 3 + 3 (the underlined part is the previous term)

15. Designed by David Jay Hebert, PhD Arithmetic Sequences 2, 5, 8, 11, 14,… (recall the first difference is 3) 5 = 2 + 3 8 = 5 + 3 = 2 + 3 + 3 11 = 8 + 3 = 2 + 3 + 3 + 3 14 = 11 + 3 = 2 + 3 + 3 + 3 + 3 (the underlined part is the previous term)

16. Designed by David Jay Hebert, PhD Arithmetic Sequence: Predictability The question is how many times does one move from the first term in a sequence. Ex. 2, 5, 8, 11, 14,…. Take one step from 2 to get to 5 Take two steps from 2 to get to 8 Take three steps from 2 to get to 11

17. Designed by David Jay Hebert, PhD Arithmetic Sequences Have a constant first difference Are predictable

18. Designed by David Jay Hebert, PhD Predictability If the sequence in question is arithmetic the position numbers will begin with a 0 th position. i.e., 2, 5, 8, 11, 14, …

19. Designed by David Jay Hebert, PhD Predictability If the sequence in question is arithmetic the position numbers will begin with a 0 th position. i.e., 2, 5, 8, 11, 14, … 2 is in the 0 th position. Zero 3’s were added to 2 to get to 2. But 2 is the 1 st term in the sequence.

20. Designed by David Jay Hebert, PhD Predictability If the sequence in question is arithmetic the position numbers will begin with a 0 th position. i.e., 2, 5, 8, 11, 14, … 5 is in the 1st position. One 3 was added to 2 to get to 5. But 5 is the 2 nd term in the sequence.

21. Designed by David Jay Hebert, PhD Predictability If the sequence in question is arithmetic the position numbers will begin with a 0 th position. i.e., 2, 5, 8, 11, 14, … 8 is in the 2 nd position. Two 3’s were added to 2 to get to 8. But 8 is the 3 rd term in the sequence.

22. Designed by David Jay Hebert, PhD Predictability If the sequence in question is arithmetic the position numbers will begin with a 0 th position. i.e., 2, 5, 8, 11, 14, … 11 is in the 3 rd position. Three 3’s were added to 2 to get to 11. But 11is the 4 th term in the sequence.

23. Designed by David Jay Hebert, PhD Predictability If we wanted to know the term in the 22 nd position, we would add 22 threes to 2 to get the result, i.e., 22(3) +2 = 66 + 2 = 68. 68 is in the 22 nd position, but it is the 23 rd term in the sequence.

24. Designed by David Jay Hebert, PhD Predictability Mathematics is the study of patterns, and therefore we must look for a pattern. The position number is the same as the number of first differences that must be added to the first term. This leads me to believe that the following formula might work. Term = (Position Number)(Step Size) + Initial Step

25. Designed by David Jay Hebert, PhD Arithmetic Sequences Start with a sequence -2, 3, 8, 13, 18, 23,… Find the step size (first difference). 3 – (-2) = 5 Find the initial step (term in 0 th position) -2

26. Designed by David Jay Hebert, PhD Arithmetic Sequences Term = (position)(step size) + initial step From above Term= (position)(5) + -2 This is a relationship between the position of a term and the term in a position.

27. Designed by David Jay Hebert, PhD Arithmetic Sequences Term = (position)(step size) + initial step From above Term= (position)(5) + -2 This is a relationship between the position of a term and the term in a position. What is the 45 th term?

28. Designed by David Jay Hebert, PhD Arithmetic Sequence Term = (position)(5) – 2 Term = (45)(5) – 2 Term = 210 – 2 Term = 208 By knowing the position we are able to determine the term, what about the other way around.

29. Designed by David Jay Hebert, PhD Arithmetic Sequence What if we know the term but not the position, can the position be determined? Suppose that 183 is a number in the sequence –2, 3, 8, 13, 18,…

30. Designed by David Jay Hebert, PhD Arithmetic Sequence The relationship Term = (Position)(Step Size) + Initial Step Replace with known values 183 = (Position)(5) – 2 37 = Position

31. Designed by David Jay Hebert, PhD New Problem: Find the sum of the sequence: -2, 3, 8, 13, …,603, 608, 613

32. Designed by David Jay Hebert, PhD New Problem: S = -2 + 3 + 8 + … + 603 + 608 + 613 Is the same as S = 613 + 608 + 603 + … + 8 + 3 + (-2)

33. Designed by David Jay Hebert, PhD New Problem: S = -2 + 3 + 8 + … + 603 + 608 + 613 Notice the following S = 613 + 608 + 603 + … + 8 + 3 + (-2) Add down in columns defined by the plus signs.

34. Designed by David Jay Hebert, PhD New Problem: S = -2 + 3 + 8 + … + 603 + 608 + 613 Notice the following S = 613 + 608 + 603 + … + 8 + 3 + (-2) 2S = 611 + 611+ 611 + … + 611 + 611 + 611 If we knew the number of terms in the original sequence we could answer the question 611+ 611 + … + 611 + 611.

35. Designed by David Jay Hebert, PhD New Problem: S = -2 + 3 + 8 + … + 603 + 608 + 613 Notice the following S = 613 + 608 + 603 + … + 8 + 3 + (-2) 2S = 611 + 611+ 611 + … + 611 + 611 + 611 The sequence –2, 3, 8, 13,… Is arithmetic therefore we have predictability.

36. Designed by David Jay Hebert, PhD New Problem: S = -2 + 3 + 8 + … + 603 + 608 + 613 Notice the following S = 613 + 608 + 603 + … + 8 + 3 + (-2) 2S = 611 + 611+ 611 + … + 611 + 611 + 611 613 = (Position)(5) – 2 123 = Position, which means there are 124 terms in the sequence.

37. Designed by David Jay Hebert, PhD New Problem: S = -2 + 3 + 8 + … + 603 + 608 + 613 Notice the following S = 613 + 608 + 603 + … + 8 + 3 + (-2) 2S = 611 + 611+ 611 + … + 611 + 611 + 611 There are 124 - 611’s in the sequence

38. Designed by David Jay Hebert, PhD New Problem: S = -2 + 3 + 8 + … + 603 + 608 + 613 Notice the following S = 613 + 608 + 603 + … + 8 + 3 + (-2) 2S = 611 + 611+ 611 + … + 611 + 611 + 611 2S = (611)(124)

39. Designed by David Jay Hebert, PhD New Problem: S = -2 + 3 + 8 + … + 603 + 608 + 613 Notice the following S = 613 + 608 + 603 + … + 8 + 3 + (-2) 2S = 611 + 611+ 611 + … + 611 + 611 + 611 2S = (611)(124) 2S = 75764

40. Designed by David Jay Hebert, PhD New Problem: S = -2 + 3 + 8 + … + 603 + 608 + 613 Notice the following S = 613 + 608 + 603 + … + 8 + 3 + (-2) 2S = 611 + 611+ 611 + … + 611 + 611 + 611 2S = (611)(124) 2S = 75764 S = 37882

41. Designed by David Jay Hebert, PhD New Problem: Let us find the sum of the sequence 5, 9, 13, …, 365, 369, 373 First we must find out how many terms are in the sequence. In order to do this we must determine the step size and initial step of the sequence.

42. Designed by David Jay Hebert, PhD New Problem: Let us find the sum of the sequence 5, 9, 13, …, 365, 369, 373 Step Size = 4 Initial Step = 5 Hence Term = 4(Position) + 5

43. Designed by David Jay Hebert, PhD New Problem: Let us find the sum of the sequence 5, 9, 13, …, 365, 369, 373 Term = 4(Position) + 5 373 = 4(Position) + 5 92 = Position We will need this information later.

44. Designed by David Jay Hebert, PhD Step 1: Write down the sum you wish to determine S = 5 + 9 + 13 + … + 365 + 369 + 373

45. Designed by David Jay Hebert, PhD Step 2: reverse the order of the sum and write it under the first sum. S = 5 + 9 + 13 + … + 365 + 369 + 373 S = 373 + 369 + 365 + …+ 13 + 9 + 5

46. Designed by David Jay Hebert, PhD Step 3: add down in columns determined by the plus signs in the problem. S = 5 + 9 + 13 + … + 365 + 369 + 373 S = 373 + 369 + 365 + …+ 13 + 9 + 5 2S = 378 + 378 + 378+ … + 378 + 378 + 378 Notice that all the columns have the same sum, if only we knew how many they were!

47. Designed by David Jay Hebert, PhD Step 4: recall we had predictability with arithmetic sequences and we already determined that 373 was in position 92 of the original sequence. Therefore 93 terms in the sequence. S = 5 + 9 + 13 + … + 365 + 369 + 373 S = 373 + 369 + 365 + …+ 13 + 9 + 5 2S = 378 + 378 + 378+ …+ 378 + 378 + 378

48. Designed by David Jay Hebert, PhD Step 5: replace the long sum with the associated multiplication problem. S = 5 + 9 + 13 + … + 365 + 369 + 373 S = 373 + 369 + 365 + …+ 13 + 9 + 5 2S = 378 + 378 + 378+ …+ 378 + 378 + 378 2S = 378(93)

49. Designed by David Jay Hebert, PhD Step 6: solve for S and be done with the problem. S = 5 + 9 + 13 + … + 365 + 369 + 373 S = 373 + 369 + 365 + …+ 13 + 9 + 5 2S = 378 + 378 + 378+ …+ 378 + 378 + 378 2S = 378(93) 2S = 35154 S = 17577