To find the pattern within the shapes we were given, we first counted up how many squares there were in each figure. To find out the nth term, we used the values we found to create a sequence, and from there, solved for the nth term. For the first sequence of figures, the first four values were 8,16,24 and 32. We quickly recognized that they were all multiples of 8, and from there, found that Tn=8n. From there, it was simple to find the next few values in the sequence(40, 48, and 56), and the values of any term in said sequence(e.g., when n=20, Tn= 160, when n= 50, Tn=400).
Taking the next sequence, we quickly realized that these were the same shapes, but filled in, rather than hollow. Using the same approach, we counted each number of squares, and used a numerical sequence to find a pattern. When finding the differences to solve the pattern, we saw that the first differences were not equal, and but the second differences were, making quadratic. The first values were 8,24,48 and 80. The difference between the first two numbers was 16, between the second and the third it was 24. These are not equal differences so we have ti fund the second difference which is 8. Since to find the multiplier of the squared number, you must divide it by two, we found 4n^2, and from those values, found that you also must add 4n to find the term. So therefore Tn=4n^2+4n. From there we found the next 3 terms (120,168,224), and then found the 20th term(1680) and the 50th(10200).
As we looked at the next two, we found that the same pattern occurred: the first(linear) sequence was simply the outside 'layer' of the other(quadratic) sequence. Solving them the exact same way as the other two, we found the first of the two was Tn=6n+2, and the second was Tn=2n^2+4n+2. Again, the first was linear and the second was quadratic. As we looked at them, we realized the reason why was because one was simply finding the perimeter, and the other was finding the area, like if they had taken a square, with sides of x length. The area would be x^2. But the perimeter would be 4x, without the second dimension. This is the same as the figures in all patterns given, where the hollow one(the perimeter) is linear, and the filled one(the area) is a quadratic.