1. The Koch snowflake at stage 0 is an equilateral triangle with side length 1 unit. a) Find the perimeter for stages 1-5 if the perimeter for stage 0 is 3. b) Write an expression for the perimeter at stage n. 2. (Refer to image)

Hello!
1. When constructing the Koch snowflake, we start with an equilateral triangle (stage 0). On each subsequent stage the figure remains a closed polygon, all segments of this polygon have the same length. Denote the perimeter at the n-th stage as P_n, it is given that P_0 = 3.
After n-th stage we take each segment of the polygon and broke it onto three equal sub-segments. Two sub-segments at the ends remain at their places, while the middle sub-segment is replaced by two segments of the same length looking outwards of the center.
Thus, each segment of length x is replaced with 4 segments of the length x/3 each, the new length becomes 4/3 x. The same ratio applies to the perimeters because they are the sums of the lengths. This way we see that  P_(n+1) = 4/3 P_n.
A sequence whose next term is fixed times more than the previous is called geometric progression, and its n-th term is
P_n = P_0 * (4/3)^n = 3* (4/3)^n.
For first n's the perimeters are
P_1 = 4, P_2 = 16/3, P_3 = 64/9, P_4 = 256/27, P_5 = 1024/81.
 
[The second part, about the Sierpinski triangle, is somewhat similar but different. I can answer it as a separate question.]