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...

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...

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.]