In this investigation I have been asked to find out how many squares would be needed to make up a certain pattern accorting to its sequence.

onwards and adding them up (according to the sequence). We then take the summation (Å) of these odd numbers and multiply them by two. After doing this we add on the next consecutive odd number to the doubled total.I have also noticied something through the drawings I have made of the patterns. If we look at the symetrical sides of the pattern and add up the number of squares we achieve a square number.Attempting to Obtain a Formula Through the Use of the Difference MethodI will now apply Jean Holderness' difference method to try and find a formula.Pos.in seq. 1 2 3 4 5 6 No.of squar. (c) 1 5 13 25 41 61 1st differ. (a+b) 0 4 8 12 16 20 2nd differ. (2a) 4 4 4 4 4 We can now use the equation an2 + bn + c'n' indicating the position in the sequence.If a = 2 then c = 1 and a + b = 0 If 2 is equal to b- then b = -2I will now work out the equation using the information I have obtained through using the difference method:1) 2(n -1) (n - 1) + 2n - 12) 2(n2 - 2n + 1) + 2n - 13) 2n2 - 4n + 2 + 2n - 14) 2n2 - 2n + 1Therefore my final equation is: 2n2 - 2n + 1Proving My Equation and Using it to Find the Number of Squares in Higher SequencesI will now prove my equation by applying it to a number of sequences and higher sequences I have not yet explored.Sequence 3:1. 2(32) - 6 + 12. 2(9) - 6 + 13. 18 -54. = 13The formula when applied to sequence 3 appears to be successful.Sequence 5:1. 2(52) - 10 + 12. 2(25) - 10 + 13. 50 - 10 + 14. 50 - 95. = 41 SuccessfulSequence 6:1. 2(62) - 12 + 12. 2(36) - 12 +13. 72 - 12 + 14. 72 - 115. = 61SuccessfulSequence 8:1. 2(82) - 16 + 12. 2(64) - 16 + 13. 128 - 16 + 14. 128 - 15 5. = 113 SuccessfulThe formula I found seems to be successful as I have shown on the previous page. I will now use the formula to find the number of squares in a higher sequence.So now I wil use the formula 2n2 - 2n + 1 to try and find the number of squares contained in sequence 20.Sequence 20: 2 (202) - 40 + 12(400) - 40 + 1800 - 40 + 1800 - 49= 761Instead of illustrating the pattern I am going to use the method I used at the start of this piece of coursework. The method in which Iused to look for any patterns in the sequences. 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 = 761I 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 PatternPos.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;anƒ + bn2 + cn + dWe 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 = 1I 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 + da + b + c +d 7a + 3b + c D will always be added to each side of the equation.2nd8a + 4b + 2c = 8 = 4a + 2b + c = 42 So then n = 2 8a + 4b + 2c = 8 = 4a + 2b + c = 4n = 3 27a + 9b + 3c = 26n = 4 64a + 16b +4c = 64 = 16a + 4b + c = 16 4To get rid of 'c' I will use this calculation;_16a + 4b + c = 164a + 2b + c = 412a + 2b = 12We can simplify this equation to: 6a + b = 6My next calculation is below:N =3 _27a + 9b + 3c = 2612a + 6b + 3c = 12 15a + 3b =14 (15a + 3b = 14) ÷ 3 = 5a + b = 4YIf I use the equation above 6a + b = 6. I can take my latest equation and subtract it from it to find 'a'.So, _6a + b = 65a + b = 4Ya = 15Now that we nave obtained 'a' we can now substitute its value into the equation to find the other values.We can now find 'b' by substituting in 15 into the equation as follows:5(_) + b = 4Y5(_) = 6Yb = 4Y - 6Y b= -2 We have now found values a & b.I will now attempt to find values c and d by substituting in the two values I now possess.So we will now sub. in the values to the equationIf we take sequence 2 for our value of n with our present values we will get:_(8) - 2(4) + c = 8this can then be simplified to, _(2) - 2(2) + c = 4 by dividing the equation by two.We will continue with the calculation using the simplified version of the equation;To find 'c' we will use the following calculations:1) _(4) - 2(2) + c = 42) 55 - 4 + c = 43) 55 + c = 84) c = 8 - 555) c = 2YTherefore my four values are:a = 15b = -2c = 2Yd = -1My equation for the working out of the number of cubes in a 3D version of the pattern is:_(nƒ) - 2n2 + 2Y(n) - 1Testing out the New EquationI will take sequence 10 to try and test this new equation.Sequence 5 :N = 5_(5ƒ) - 2(52) + 2Y(5) - 1 =166Y - 50 + 135 - 1 =129The formula on this sequence seems to be successful.I will now apply it to another sequence to be 100% correct:N = 4 _(4ƒ) - 2(42) + 2Y(4) - 1 =855 - 32 + 10Y - 1 = 63The formula again proves to be successful.Using the formula to find the number of squares in a higher sequuence not yet explored in this investigation.Sequence 10:N = 10_(10ƒ) - 2(102) + 2Y(10) - 1 =13335 - 200 + 26Y - 1 =1159Sequence 15:N = 15_(15ƒ) - 2(152) + 2Y(15) - 1 =4500 - 450 + 40 - 1 =4089This equation has correctly given me the number of squares in each sequence which again proves it can be applied to any of the 3D sequences to give the correct answer.My ConclusionsI have made a number of conclusions from the investigation I have carried out.Firstly I have deciferred that the equation used in the 2D pattern was a quadratic. 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 investigation, giving us our constant, but most importantly it gave us the equation on which to base our solutions. It was:an2 + bn + cThis 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 + 1For the 3D pattern the equation is;_(nƒ) - 2n2 + 2Y(n) - 1 essaybank.co.uk