Hello!
1. When constructing the Koch snowflake, we start with an equilateral triangle (stage ). On each subsequent stage the figure remains a closed polygon, all segments of this polygon have the same length. Denote the perimeter at the
-th stage as
it is given that
After -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 ). On each subsequent stage the figure remains a closed polygon, all segments of this polygon have the same length. Denote the perimeter at the
-th stage as
it is given that
After -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 is replaced with
segments of the length
each, the new length becomes
The same ratio applies to the perimeters because they are the sums of the lengths. This way we see that
A sequence whose next term is fixed times more than the previous is called geometric progression, and its -th term is
For first 's the perimeters are
[The second part, about the Sierpinski triangle, is somewhat similar but different. I can answer it as a separate question.]
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