The first term of a an infinite sequence is an equilateral triangle inscribed in a circle. The triangle has side n. Each new polygon in the sequence is formed by replacing the middle third of each side by two segments that have the same length as each third. As n, increases the area of polygon P increases but remains less than the area of the circle. The area of polygon 1 is A. Explain why the limit of the area of polygon p = A + A (1/3 + 4/27 + 16/243 + ...) Let Sp be the area of polygon p. Find the limit of Sp.
but remains less than the area of a circle...
area pologon <A given
So how can area ever be this?
area pologon=A+A(1/3+..._ ?
To explain why the limit of the area of polygon p is equal to A + A(1/3 + 4/27 + 16/243 + ...), we first need to understand the pattern in the sequence and how the areas of the polygons are related.
Let's call the equilateral triangle in the first polygon Triangle 1. The side length of Triangle 1 is denoted as n. Each subsequent polygon in the sequence is formed by replacing the middle third of each side of the previous polygon with two segments of the same length as each third.
As n increases, the area of each polygon P increases but remains less than the area of the circle.
Now, let's denote the area of polygon p as Sp.
To find the limit of Sp, we can consider the process of constructing the polygons and analyze the areas involved.
1. Polygon 1: The area of Polygon 1 is denoted as A.
2. Polygon 2: To construct Polygon 2, we replace the middle third of each side of Triangle 1 with two segments of length n/3, while maintaining the remaining segments of length 2n/3. This process reduces the area of Polygon 1 by (1/3)^2 = 1/9. Therefore, the area of Polygon 2 is A - A/9 = A(1 - 1/9).
3. Polygon 3: Similarly, to construct Polygon 3, we replace the middle third of each side of Polygon 2 with two segments of length (n/3)^2, which decreases the area further. The area of Polygon 3 is A(1 - 1/9) - (A/9) * (1/9) = A(1 - 1/9 - 1/81) = A(1 - 1/9 - 1/9^2).
4. Polygon 4: Continuing this pattern, for Polygon 4, we replace the middle third of each side of Polygon 3 with two segments of length (n/3)^3, decreasing the area even more. The area of Polygon 4 is A(1 - 1/9 - 1/9^2) - (A/9^2) * (1/9) = A(1 - 1/9 - 1/9^2 - 1/9^3).
By observing this pattern, we can see that the area of Polygon p can be represented as A multiplied by the series (1 - 1/3^2 - 1/3^3 - 1/3^4 + ...). This is a geometric series with the common ratio r = 1/3.
Using the formula for the sum of an infinite geometric series, which is S = a / (1 - r), where a is the first term and r is the common ratio, we can find the sum of the series in the parentheses.
In this case, a = 1 and r = 1/3. Plugging these values into the formula, we get:
S = 1 / (1 - 1/3) = 1 / (2/3) = 3/2.
Therefore, the area of Polygon p is Sp = A * (3/2).
To find the limit of Sp as p approaches infinity, we multiply A by the sum of the series (3/2). As the series converges to a value of 3/2, the limit of Sp is A * (3/2) = 3A/2.
Thus, the limit of the area of Polygon p is A + A * (1/3 + 4/27 + 16/243 + ...), where the series in the parentheses converges to 3/2.