- Level: AS and A Level
- Subject: Maths
- Document length: 1877 words
Sequences and series investigation
Extracts from this essay...
Introduction
Sequences and series investigation By Neil 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 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 squares)
Middle
I will use this to prove the number of squares given by the equation is correct. As shown below: 2(1+3+5+7+9+11+13+15+17+19+21+23+25+27+29+31+33+35+37) + 39 = 761 I feel this proves the equation fully. Using the Difference Method to Find an Equation to Establish the Number of Squares in a 3D Version of the Pattern Pos.in seq. 0 1 2 3 4 5 No.of squar. -1 1 7 25 63 129 1st differ. 2 6 18 38 66 2nd differ. 4 12 20 28 36 3rd differ. 8 8 8 8 So therefore we get the equation; an + bn2 + cn + d We already know the values of 'n' (position in sequence) in the equation so now we have to find out the values of a, b, c, and d. If n = 0 then d = -1 and if n = 1 then d = 1 I can now get rid of d from the equation to make it easier to find the rest of the values. I will will take n = 2 to do this in the following way: 1st calculation _ 8a + 4b + 2c + d a + b + c +d 7a + 3b + c D will always be added to each side of the equation. 2nd 8a + 4b + 2c = 8 = 4a + 2b + c = 4 2 So then n = 2 8a + 4b + 2c = 8 = 4a + 2b + c = 4 n = 3 27a + 9b +
Conclusion
This can be proven through the fact that the 2nd difference was a constant, a necessary element of any quadratic and also the fact that the first value has to be squared. This can also be proved by illustrating the equation on the graph, creating a curve. I have also established that the top triangular half of the 2D pattern always turns out to be a square number. If we now look at the 3D pattern, the equation I achieved for it has turned out to a cubic equation. This can be proven through the constant, again a necessary characteristic of any cubic equation and also the fact that its 1st value must be cubed and its second squared. If we drew a graph we would get a ccurved graph in which the line falls steeply, levels off and then falls again. The Differentiation Method developed by Jean Holderness played a very important role in this investigation. It helped us to gain knowledge of any pattern and anything that would help in the invetigation, giving us our constant, but most importantly it gave us the equation on which to base our solutions. It was: an2 + bn + c This proved very helpful. To find our equation we then substituted in different values which we could find in our differentiation table. I have concluded that both the equations proved to be very successful. Therefore the equations are: For the 2D pattern the equation is; 2n2 - 2n + 1 For the 3D pattern the equation is; _(n) - 2n2 + 2Y(n) - 1
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