Question 1168952: The side of the square is 20 inches . A second square is formed by joining in the proper order , the midpoints of the sides of the first square. A third square is formed by joining the midpoints of the second square , and so on . Find the sum of the perimeters and areas up to the 7th square.
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! Let s = the side length of the 1st square.
Then the side length of the 2nd square = s/sqrt(2), s/2 for the 3rd square and so on.
Thus the side length for the nth square is s_n = s/sqrt(2^(n-1))
The perimeter of the nth triangle = 4s/sqrt(2^(n-1))
The perimeters represent a geometric sequence, with common ratio = 1/sqrt(2)
So the nth perimeter in the sequence is given by p_n = 4s*(1/sqrt(2))^(n-1)
The sum of n terms of a geometric sequence is a1*(1-r^n)/(1-r), where a1 is the 1st term.
Thus the sum of the perimeters up to the 7th square is 4s*(1-(1/sqrt(2))^7)/(1-1/sqrt(2)) = 124.497
Similarly, the nth area is given by A_n = s_n^2 = s^2*(1/2)^(n-1)
The sum of the 1st seven areas is 198.438
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