The first tablecloth was a 5x5 grid, with 1 line of symetry. the different coloured squares in the grid were in a 5x3 rectangle, a total of 15. 5 was 'n' so my nth term formula would start with an n. I called the 3 'd' because I knew it would change along with n, so I first had to work out the formula for 'd'. My formula would be 'nd', and when I found that 'd', in this case 3, was 1/2 of n,and then add 0.5. I replaced 'd' in my formula with this, so my first formula would be: n(n/2+0.5)
The second tablecloth had a 3x2 set of different colours, and one more colourin the middle. I could see that 3 was 'd' so my formula would begin: n/2+0.5. Then there was the 2. this was 1 less than 'd', and it turned out that this occured in every odd-numbered grid, so this was 'd-1'. The formula for 'd-1' was n/2+0.5+1, then simplified it became n/2-0.5. then I added the extra 1 on the end, and I had my formula: (n/2+0.5)(n/2-0.5)+1
The third tablecloth had 9 different colours in a 3x3. When I looked at tables of different sizes I noticed my number 'd' again. In the 3x3 there wer 4 different colours, and in the 7x7 there were 16 colours. I could see that the number of different colours in each grid was always a square number, or rather, my special number 'd' squared. Since d was n/2+0.5, my formula was (n/2+0.5)².
The fourth table cloth appeared to be the triangle of 'd', added to the triangle of 'd-1'. However, when I counted the different colours in different sized grids, I once again had the result d². So my formula was once again (n/2+0.5)².
The last grid had 6 different colours, but when I looked at the pattern they were in, I noticed a pattern. When I checked the other grids I noticed that the number of different colours was the triangular number of 'd'. I used the formula for triangle numbers: d(d+1)/2, and replaced the 'd's with their value in comparison to 'n', giving me: (n/2+0.5)((n/2+0.5)+1)/2