Given an equilateral triangle as shown in figure (a), if we draw one line parallel to each sides such that the 3 lines divide each side into 2 segments of equal length, then we will get the figure as shown in figure (b). Let us call the triangle in (a) So, the figure in (b) Si. Similarly, we can draw k lines parallel to each side such that the 3k lines divide each side into (k+1) segments of equal length. Let us call the figure Sk. As an example, figure (c) shows the figure S2. side into (k+1) segments of equal length. Let us call the figure St. As an example, figure (c) shows the figure S2. 3 (a) So (b) Si (c) S2 Now, we wish to count how many upright triangles are contained in Sk. A triangle is called upright if the base is at the lower side. In figure (b), the triangle in Si is not a upright triangle because the line (a, b) is at the upper side of node c. Thus, S contains 4 upright triangles, including 3 smaller ones plus It is not difficult to see that S2 contains 10 upright triangles. Let an be the number of upright triangles that Sn contains. Find a recurrence relation for an and solve it