Using the Difference Method to Find an Equation to Establish the Number of Squares in a 3D Version of the Pattern

= 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 invetigation, 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