#### Maths Investigation - Pile 'em High

Maths Investigation - Pile 'em High "Jo has started in a local supermarket and her first job is to build displays of soup tins. To make them stable they are stacked using a brick bond so that each tins stands on two others. The tins are stacked against the wall. Each stack is complete with one tin in the top row." This piece of Maths coursework is about investigating sequences from a practical situation. In this investigation, tins are used to build stacks using a brick bond so that each new tin stands on two others. The tins are stacked flat against a structure and each stack is complete with one tin in the top row. An Example: First I will investigate a two row stack. In this two row stack there are: Two tins on the base (row 2) One tin on the top (row 1) Three tins altogether In a three row stack there are: Three tins on the base (row 3) Two tins in the middle (row 2) One tin on the top (row 1) Six tins altogether An early pattern that I can see is that whatever the row number (counting down from row 1 at the top) there are that many tins in the row e.g. row 3 - 3 tins. It appears to me that there is some kind of pattern forming with the total number of tins in each built up stacks. (See table) Number of Stacks The Amount of Tins 2 3 3 6 4 0 5 ? I predict that for five stacks, the amount of tins needed will be fifteen based on other stacks e.g.

#### To see if three horizontal rectangle numbers e.g. “12,13,14” – have the same result when you multiply the middle number by 2 and add the 1st and last number together.

Diagram.1 Case.1: To see if three horizontal rectangle numbers e.g. "12,13,14" - have the same result when you multiply the middle number by 2 and add the 1st and last number together. Testing: Conclusion: This happens because the mean of the 1st and last number is equal to the middle number i.e.: Mean of 12 and 14 = 13 Middle number = 13 The next part of the scenario was to multiply the middle number by 2 and plus the 1st and last number together; this is the same result - it can be proven when we write as it's shortest form i.e.: ( + ( = (2 ( ( 2= (2 Formula: Middle number multiplied by 2 = (First number) + (Last number) Case.2: To see if "Case.1" applies for horizontal and diagonal rectangles. Testing: Conclusion: The original formula (Case.1) also works for Variable.2. Case.3: Investigate what happens when Mary draws a rectangle around five numbers. Testing: I first tried to see if the formula for Case.1 applied for Case.3: Conclusion: The formula for Case.3 also works for rectangles with 5 squares whether it is diagonal, horizontal or vertical. Case.4 Investigate if the formula also works for any amount of rectangle squares: Testing: Conclusion: After doing the previous scenarios I have realised that by adding the first and last numbers together and then dividing by 2 makes the middle number (see formula in Case.4.) This works for any

#### Matrix Powers

Table of contents: Questions: .) Consider the Matrix M= Calculate Mn for n= 2, 3, 4, 5, 10, 20, 50. Describe in words any pattern you observe. Use this pattern to find a general expression for the matrix Mn in terms of n. 2.) Consider the matrices P= and S= P2= 2 = =; S2= 2 = = Calculate Pn and Sn for other values of n and describe any pattern you observe. 3.) Now consider matrices of the form steps 1 and 2 contain examples of these matrices for K=1 2 and 3. Consider other values of k, and describe any pattern(s) you observe. Generalize these results in terms of K and N 4.) Use technology to investigate what happens with further values of k and n. State the scope or limitations of k and n. 5.) Explain why your results holds true in general. SL type 1: Matrix Powers ) a) To calculate the value for matrix 'M' when n=2, the matrix must be multiplied by an exponent of 2. This would be shown and calculated as, x Therefore the value of matrix M2 = b) To calculate the value for matrix 'M' when n=3, the matrix must be multiplied by an exponent of 3. Therefore the value of M3 = c) To calculate the value for matrix 'M' when n= 4, the matrix x must be multiplied by an exponent of 4. Therefore the value of M4= d) To calculate the value for matrix 'M' when n=5, the matrix must be multiplied by an exponent of 5. Therefore the value of M5= e) To calculate the

#### Investigate calendars, and look for any patterns.

Maths Coursework Introduction: I was given a task to investigate calendars, and look for any patterns. I noticed several patterns, the first of which was the relationship between the starting days of different months, also I noticed the relationship between numbers in columns of the calendar, the relationship between numbers in the rows, Studying diagonal relationships, and Studying relationships between adjacent numbers. . Days on which months start First, I explore the days on which different months start. Which months are the same/ have a pattern? Ex 1.1 Study Sample Calendar: Month Starting day Friday 2 Monday 3 Monday 4 Thursday 5 Saturday 6 Tuesday 7 Thursday 8 Sunday 9 Wednesday 0 Friday 1 Monday 2 Wednesday From the above table, I can see that some of the months start on the same day, which means there may be a pattern when compared with other years. If so, then that means there is a pattern of which months start on the same day each year. The results of this first test are as follows: , 10 = same 2, 3, 11 = same 4, 7 = same 5 6 8 9, 12 = same Now I must investigate to find out if the pattern is the same in other years. In order to do this, I check a calendar of the year 2004. Ex 1.2: Results for 2004 Months ( n ) Start day Thursday 2 Sunday 3 Monday 4 Thursday 5 Saturday 6 Tuesday 7 Thursday 8 Sunday 9

#### I am to conduct an investigation involving a number grid.

Maths Coursework Part 1 Introduction I am to conduct an investigation involving a number grid. The first part of the investigation will involve me: ü Drawing a box around various random numbers on a number grid i.e. 2 x 2 numbers, 3 x 3 numbers, 4 x 4 numbers. ü I will then find the product of the top left number and the bottom right number in the boxes by multiplying them. ü I will then do the same with the top right and bottom left numbers. ü I am then going to calculate the difference between these products 10x10 Grid ü 2x2 Boxes Box 1 52 53 X X+1 62 63 X+10 X+11 [image001.gif] [image002.gif] 52 x 63 = 3276 x (x + 11) x2 + 11x [image003.gif] 62 x 53 = 3286 (x + 1) (x + 10) x2 + 11x + 10 = (x2 + 11x + 10) - (x2 + 11x) = 10 3286 - 3276 = 10 The difference between the two numbers is 10 Box 2 81 82 91 92 [image004.gif] 81 x 92 = 7452 91 x 82 = 7462 7462 - 7452 = 10 The difference between the two numbers is 10 Box 3 69 70 79 80 [image005.gif] 69 x 80 = 5520 79 x 70 = 5530 5530 - 5520 = 10 The difference between the two numbers is 10 ü 3x3 Boxes Box 1 8 9 10 X X+1 X+2 18 19 20 X+10 X+11 X+12 28 29 30 X+20 X+21 X+22 [image006.gif] [image007.gif] 8 x 30 = 240 x (x + 22) x2 +

#### Explaining the Principle of mathematical induction

Explaining the Principle of mathematical induction: Formally the principle of a proof by induction can be stated as follows: A proposition P (n) involving a positive integer n, is true for all positive integral values of n if, P (1), and P (k) ? P (k +1) is true. This can be explained using a staircase as a simple analogy. Image the proposition that a man can climb a given uniform staircase, to prove this statement we need to show two things. These are that the man can get onto the first step and that he is able to climb from one step to an other. Now relating this to the formal principle of induction, the staircase can be considered the general proposition P (n). The first step of the staircase is P (1), the second P (2), the third P (3), and so on. If we can show that the man can get onto the first step P (1) then we have ironically finished the first step of proving the proposition. The second step would be to prove that he can get from one step to an other formally put P (k) ? P (k +1). If we can show this than it follows that man can get from the first to the second step, second to the third,... n steps. Thus it can be said that P (n) is true for all positive integral values of n. 2 Definition of the derivative function f (x) The derivative function is a general expression for the gradient of a curve at any given point. It is based on the principle of limiting

#### About Triangular Square Numbers

About Triangular Square Numbers By August Pieres January 18th, 2003 I believe I have discovered an algorithm which generates an infinity of triangular squares. "Triangular squares" are triangular numbers which are also perfect squares. These are triangular numbers: 1,3,6,10,15,21,28,36,45,55,66,78,91,105,120,136,153,171,190,210,... Notice that 120=5! (and 6=3!) and that 1,3,21, and 55 are also Fibonacci numbers; one might call them "Fibonacci triangles." Are there any more Fibonacci triangles? These are the perfect squares: 1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400,... The only triangular squares listed so far are 1 and 36. Earlier today I thought that these were the only existing triangular squares, but I found out that there are more and quite possibly an infinity of them. I made a program on my programmable Texas Instruments TI-86 calculator. Here it is: PROGRAM:TRISQUAR -->N Lbl A N*(N+1)/2-->M If ?M==iPart?M Then Disp M End +N-->N Goto A I ran the simple program above and it found the following additional triangular squares: 1225, 41616, 1413721. Then I "played" with these new numbers -- with the help of the calculator -- trying to find patterns. To each triangular square corresponds a pair of parameters: s and t, such that a triangular square N is the sth perfect square and the tth triangular number, i.e. N=s2=Tt. So

#### Study the topic of trios and work on from that, to discover patterns and links.

Trios and Extensions of Trios Aim: My aim is to study the topic of trios and work on from that, to discover patterns and links. I will study trios using permutations as I believe these to have the most regularity and a much easier sequence to work with. I am going to investigate how many trios there are for the number 5. Once I have worked this out, I will work out how many trios there are for the number 6; then for the number 7 and so on. I will then try to make a connection between the above numbers and come up with a formula which can be used to work out a trio. I will then explore further into this question, by investigating 'quartets', 'quintets', 'sextet' etc. What is a trio? A trio is a set of 3 numbers greater than zero, that when added together make another number. For example, 1 + 2 + 2 = 5, also 2 + 1 + 2 = 5. TRIOS: I will, first of all, investigate the number of trios there are for the number 5: Trios for 5: 2 2 2 1 2 2 2 1 1 3 3 1 3 1 1 Trios for 6: 2 2 2 3 2 1 3 1 2 2 1 3 2 3 1 4 1 1 4 1 1 4 2 3 3 2 Trios for 7: 3 2 2 5 1 1 2 3 2 1 5 1 2 2 3 1 1 5 2 4 1 3 3 4 2 3 1 3 2 4 1 3 3 1 2 1 4 4 2 1 4 1 2 Trios for 8: 2 5 5 2 2 5 1 2 1 5 5 1 2 5 2 1 3 4 4 3 3 4 1 3 1 4 4 3 1 4 1 3 6 1 1 6 1 1 6 2 2 4 2 4 2 4 2 4 2 3 3 3 2 3 3 3 2 Numbers of trios: 5 --> 6 6 --> 10 7 --> 15 8 --> 21 I

#### Find out how many squares would be needed to make up a certain pattern according to its sequence.

In this experiment I am going to require the following: A calculator A pencil A pen Variety of sources of information Paper Ruler In this investigation I have been asked to find out how many squares would be needed to make up a certain pattern according to its sequence. The pattern is shown on the front page. In this investigation I hope to find a formula which could be used to find out the number of squares needed to build the pattern at any sequential position. Firstly I will break the problem down into simple steps to begin with and go into more detail to explain my solutions. I will illustrate fully any methods I should use and explain how I applied them to this certain problem. I will firstly carry out this experiment on a 2D pattern and then extend my investigation to 3D. The Number of Squares in Each Sequence I have achieved the following information by drawing out the pattern and extending upon it. Seq. no. 1 2 3 4 5 6 7 8 No. Of cubes 1 5 13 25 41 61 85 113 I am going to use this next method to see if I can work out some sort of pattern: Sequence Calculations Answer =1 1 2 2(1)+3 5 3 2(1+3)+5 13 4 2(1+3+5)+7 25 5 2(1+3+5+7)+9 41 6 2(1+3+5+7+9)+11 61 7 2(1+3+5+7+9+11)+13 85 8 2(1+3+5+7+9+11+13)+15 113 9 2(1+3+5+7+9+11+13+15) +17 145 What I am doing above is shown with the aid of a diagram below; If we take sequence 3: 2(1+3)+5=13 2(1

#### Borders - a 2 Dimensional Investigation.

BORDERS The shape on the left is a dark cross shape that has been surrounded by white squares to create a bigger cross shape. The shape consists of 25 small squares in total. To make the next shape in the sequence, white squares are added to the existing shape, creating a bigger cross shape made up in the same way. For my investigation, I am going to work out how many squares would be needed to make up any cross shape built up in this way by working out a general formula. Then I am going to extend my investigation to 3 dimensions. I intend to use structure to work out a general formula. 2 Dimensional Investigation I am going to start by building a 1x1 cross-shape and adding on the borders. The complete shape will be known as the cross-shape (this includes the border squares). I have decided to define this shape as a 1x1 shape. x1 shape = 1 square 2x2 shape = 5 squares 3x3 shape = 13 squares 4x4 shape = 25 squares 5x5 shape = 41 squares 6x6 shape = 61 squares I now have enough data to analyse my results: Number in sequence (n) Total number of squares (tn) 2 5 3 3 4 25 5 41 6 61 To work out which formula to use I will now put my results in a table showing the differences between the numbers: n tn 1st difference 2nd difference 1 4 2 5 4 8 3 13 4 12 4 25 4 16 5 41 4 20 6 61 I can clearly see that the 2nd difference is the