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Firstly let's count the number of coins (spots, discs etc..) needed to make the The sequence is 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66 ... The nth term is given by 6n. This is easy to see if you start at a corner and count the spots on a side up to but excluding the next corner. There are six sides like this. Each number in the sequence is 6 more than the previous number. The ancient Greeks studied 'polygonal' numbers. Here is how the triangular and square numbers are formed: The The sequence is 1, 6, 15, 28, 45, 66, 91, 120, 153, 190, 231 ... The nth term is given by n.(2n-1) Every hexagonal number is also a triangular number The The sequence is 1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331 ... We can find the formula for the nth term by partitioning the hexagon into 3 parallelograms, plus the central spot. The nth term is therefore 3n.(n-1) + 1. The The sequence is 1, 13, 37, 73, 121, 181, 253, 337, 433, 541, 661 ... Again we can partition them, this time into 6 parallelograms plus the central spot which gives us the formula for the nth term: 6n.(n-1) + 1. The The sequence goes 1, 7, 22, 50, 95, 161, 252, 372, 525, 715, 946 ...: The nth term of the sequence is given by n.(n+1).(4n-1)/6 We can form The sequence goes: 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331 ... The nth term of the sequence is given by n^3 - very simple and surprising! Proof:
if the nth term is n^3, then the (n+1)th term will be the same plus the number of spots in the new base layer of the pyramid, ie
Now the first (and second) terms are indeed cubes, and so therefore are all others. QED Note that, from this, the hexagonal centred numbers themselves can be represented as the difference between two adjacent cubes. | ||||

page date: 23Oct05. I enjoy correspondence stimulated by this site. You can contact me here. | ||||