Sequences and series investigation By Neil

from 1 onwards and addingthem up (according to the sequence). We then take the summation(Å) of these odd numbers and multiply them by two. After doing thiswe add on the next consecutive odd number to the doubled total.I have also noticied something through the drawings I have madeof the patterns. If we look at the symetrical sides of the patternand 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 6No.of squar. (c) 1 5 13 25 41 611st differ. (a+b) 0 4 8 12 16 202nd differ. (2a) 4 4 4 4 4We can now use the equationan2 + bn + c'n' indicating the position in the sequence.If a = 2 then c = 1 and a + b = 0 If2 is equal to b- then b = -2I will now work out the equation using the information I have obtainedthrough 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 Numberof Squares in Higher SequencesI will now prove my equation by applying it to a number of sequencesand 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 besuccessful.Sequence 5:1. 2(52) - 10 + 12. 2(25) - 10 + 13. 50 - 10 + 14. 50 - 95. = 41SuccessfulSequence 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. = 113SuccessfulThe formula I found seems to be successful as I have shown on theprevious 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 findthe 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 methodI used at the start of this piece of coursework. The method in whichIused to look for any patterns in the sequences. I will use thisto 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 Establishthe Number of Squares in a 3D Version of the PatternPos.in seq. 0 1 2 3 4 5No.of squar. -1 1 7 25 63 1291st differ. 2 6 18 38 662nd differ. 4 12 20 28 363rd differ. 8 8 8 8So therefore we get the equation;anƒ + bn2 + cn + dWe already know the values of 'n' (position in sequence) in theequation 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 findthe 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 +d7a + 3b + cD will always be added to each side of the equation.2nd8a + 4b + 2c = 8 = 4a + 2b + c = 42So then n = 2 8a + 4b + 2c = 8 = 4a + 2b + c = 4n = 3 27a + 9b + 3c = 26n = 4 64a + 16b +4c = 64 = 16a + 4b + c = 164To 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 latestequation 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 intothe 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 - 6Yb= -2We 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 = 4by 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 sequuencenot 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 eachsequence 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 patternwas a quadratic. This can be proven through the fact that the 2nddifference was a constant, a necessary element of any quadraticand also the fact that the first value has to be squared. This canalso be proved by illustrating the equation on the graph, creating a curve.I have also established that the top triangular half of the 2D patternalways turns out to be a square number.If we now look at the 3D pattern, the equation I achieved for ithas turned out to a cubic equation. This can be proven through theconstant, again a necessary characteristic of any cubic equationand also the fact that its 1st value must be cubed and its secondsquared. If we drew a graph we would get a ccurved graph in whichthe line falls steeply, levels off and then falls again.The Differentiation Method developed by Jean Holderness played avery important role in this investigation. It helped us to gainknowledge of any pattern and anything that would help in the invetigation,giving us our constant, but most importantly it gave us the equationon which to base our solutions.It was:an2 + bn + cThis proved very helpful.To find our equation we then substituted in different values whichwe 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