... 1) ² 2n +1 2n 4n² 2n +1 2n 1 = 4n² + 4n + 1 Middle² = (2n² + 2n)² 2n² +2n 2n² 4n4 4n³ +2n 4n³ 4n² = 4n4 + 8n³ + 4n² Longest² = (2n² + 2n + 1)² 2n² +2n +1 2n² 4n4 4n³ 2n² +2n 4n³ 2n² 2n +1 2n² 2n 1 = 4n4 + 8n³ + 8n² +4n + 1 A + B = C 4n² + 4n + 1 + 4n4 + 8n³ + 4n² = 4n4+ 8n³ + 8n² +4n + 1 My results show that shortest² + middle² = longest². I have Pythagoras' theorem. For this table, I am using the term a, b, b + 2 Triangle Number (n) Length of shortest side Length of middle side Length of longest side Perimeter Area 1 6 8 10 24 48 2 8 15 17 40 120 3 10 24 26 60 240 4 12 35 37 84 420 5 14 48 50 112 672 6 16 63 65 144 1008 7 18 80 82 180 1440 8 20 99 101 220 1980 Formulas Shortest side = 2n + 4. This is because: 2 x 1 + 4 = 6 2 x 2 + 4 = 8 2 x 3 + 4 = 10 etc Middle side = n² + 4n + 3 Longest side = n² + 4n + 5 Box Methods Shortest² = (2n + 4)² 2n +4 2n 4n² 8n +4 8n 16 = 4n² + 16n + 16 Middle² = (n² + 4n + 3)² n² +4n +3 n² n4 4n³ 3n² +4n 4n³ 16n² 12n +3 3n² 12n 9 = n4 + 8n³ + ...

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