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1 solution
Let the side of the first square be a, so its area a1=a^2; Next square will have the diagonal equal to a, so its area a2=a^2/2; And so on.
So the areas of the squares will form infinite geometric progression (GP): a^2, a^2/2, a^2/4, a^2/8, a^2/16 with common ratio (r) equal to 1/2.
Sides of the squares will form another GP: a, a/√2, a/2, a/2√2, a/4 with r = 1/√2
& perimeters of the squares will also form GP: 4a, 2√2a, 2a, √2a, a with r = 1/√2
Sum of perimeters = 4a+2√2a + 2a + √2a + a = (7+3√2)a
OR using the sum of the progression formula,
For GP with r, |r|<1, the sum of the progression is sum=b(1-r^n)/(1−r), where b is the first term and n is number of terms. b=4a, n=5
So the sum of the perimeters will be sum of GP= 4a*(1-(1/√2)^5)/(1-(1/√2)) = (7+3√2)a