1. Chapter 1 Mathematical Reasoning Section 1.4 Patterns

2. Patterns are common to every area of study and are found in both man made objects and in nature. Many patterns develop in the study mathematics in arithmetic and geometry. The problem is that in mathematics patterns are explained or described with great detail. Discovering and extending patterns requires both inductive and deductive reasoning. In this section we will talk about certain number patterns and show how they can explain some other patterns in geometry for example. Number Sequence Patterns Find the next three terms in each of the following number sequences. Sequence 1:2, 5, 8, 11, 14, 17, 20,,,,…,77, (Keep adding 3 “Arithmetic Sequence”) Sequence 2:25, 250, 2500,,,…, 2500000,… (Keep Multiplying by 10 “Geometric Sequence”) Sequence 3:100, 88, 76, 64,,,,…,4,,… (Keep subtracting 12 “Arithmetic Sequence”) 23262980 25000250000 524028-8

3. The arithmetic sequence and geometric sequence are two of the most well-known number sequence patterns in mathematics. We will give some more details about each one. Arithmetic Sequences The next number in an arithmetic sequence can be found by adding (or subtracting) the same number from the current number in the sequence. The number you start with this book calls the INITIAL value. You can get your CURRENT value by adding a certain number to the PREVIOUS value. For Example In Sequence 1 (previous slide):2, 5, 8, 11, 14, 17, 20, 23,… Initial=2and CURRENT = PREVIOUS + 3 Using the INITIAL value and the amount you increase or decrease by we can get a formula for the sequence. n1234567 2+3(n-1) 25811141720

4. Let try this for Sequence 3:100, 88, 76, 64, 52,… INITIAL = CURRENT = PREVIOUS The formula can be seen in the following table: 100 - 12 n1234567 100-12(n-1) 100887664524028 For an arithmetic sequence with INITIAL = a and CURRENT=PREVIOUS + d we get the following formula: n1234567 a  d(n-1) a adada  2da  3da  4da  5da  6d Use a + when the same number is added and a – when it is subtracted.

5. Geometric Sequences Sequence 2:25, 250, 2500, 25000, 250000,… This can also be written using the INITIAL, CURRENT and PREVIOUS values in the following way: INITIAL = 25andCURRENT = PREVIOUS  10 n12345 25250250025000250000 In general with INITIAL = a and CURRENT = PREVIOUS  r we get the following: n12345

6. AREAPERIMETER The is no one set method that works for finding patterns. If there where that would be what we would teach you to do. The challenge is how can you apply the techniques of inductive and deductive reasoning to begin to find a pattern. Example In the example below the squares are of length 1 on each side. I want to find the area (number of squares) and perimeter (distance around the outside) of each pattern. 1 2 3 4 4 6 8 10 AREA:INITIAL = 1andCURRENT = PREVIOUS + 1 PERIMETER: INITIAL = 4andCURRENT = PREVIOUS + 2

7. This will give the following pattern for the n th shape. AREA of nth shape = 1+1  (n-1)=1+n-1=n PERIMETER of the nth shape = 4+2(n-1)=4+2n-2=2+2n If the area is 7 what is the perimeter? If the perimeter is 42 what is the area? 16 20