Beyond Pythagros
...y +4 each term ? ('a' x 'n') + n = 'b' From these observations I have worked out the next two terms. I will now put the first five terms in a table format. Term Number 'n' Shortest Side 'a' Middle Side 'b' Longest Side 'c' Perimeter Area 1 3 4 5 12 6 2 5 12 13 30 30 3 7 24 25 56 84 4 9 40 41 90 180 5 11 60 61 132 330 I have worked out formulas for How to get 'a' from 'n' How to get 'b' from 'n' How to get 'c' from 'n' How to get the perimeter from 'n' How to get the area from 'n' My formulas are 2n + 1 2n2 + 2n 2n2 + 2n + 1 4n2 + 6n + 2 2n3 + 3n2 + n To get these formulas I did the following Take side 'a' for the first five terms 3, 5, 7, 9, 11. From these numbers you can see that the formula is 2n + 1 because these are consecutive odd numbers (2n + 1 is the general formula for consecutive odd numbers) You may be able to see the formula if you draw a graph From looking at my table of results, I noticed that 'an + n = b'. So I took my formula for 'a' (2n + 1) multiplied it by 'n' to get '2n2 + n'. I then added my other 'n' to get '2n2 + 2n'. This is a parabola as you can see from the equation and also the graph Side 'c' is just the formula for side 'b' +1 The perimeter = a + b + c. Therefore I took my formula for 'a' (2n + 1), my formula for 'b' (2n2 + 2n) and my formula for 'c' (2n2 + 2n + 1). I then did the following: - 2n + 1 + 2n2 + 2n + 2n2 + 2n + 1 = perimeter Rearranges to equal 4n2 + 6n + 2 = perimeter The area = (a x b) divided by 2. Therefore I took my formula for 'a' (2n + 1) and my formula for 'b' (2n2 + 2n). I then did the following: - (2n + 1)(2n2 + 2n) = area 2 Multiply this out to get 4n3 + 6n2 + 2n = area 2 Then divide 4n3 + 6n2 + 2n by 2 to get 2n3 + 3n2 + n To prove my formulas for 'a', 'b' and 'c' are correct. I decided incorporate my formulas into a2 + b2 = c2: - a2 + b2= c2 (2n + 1)2+ (2n2 + 2n)2= (2n2 + 2n + 1)2 (2n + 1)(2n + 1) + (2n2 + 2n)(2n2 + 2n) = (2n2 + 2n + 1)(2n2 + 2n + 1) 4n2 + 2n + 2n + 1 + 4n4 + 4n2 + 4n3 + 4n3= 4n4 + 8n3 + 8n2 + 4n + 1 4n2 + 4n + 1 + 4n4 + 8n3 + 4n2= 4n4 + 8n3 + 8n2 + 4n + 1 4n4 + 8n3 + 8n2 + 4n + 1 = 4n4 + 8n3 + 8n2 + 4n + 1 This proves that my 'a', 'b' and 'c' formulas are correct Here are the formulas for the perimeter and the area Perimeter = 4n2 + 6n + 2...