Step-by-step explanation:
I guess, the silent request here is that there will be no wire left after the 10th square.
and I guess, the following square does not reuse one side of the previous square. all squares are "standalone".
T stands for area ? well, ok.
when the side length doubles, the area quadruples (the area is side×side, and factor 2 × factor 2 = factor 4). but the perimeter still only doubles.
T1 = x × x = x²
T2 = x×2 × x×2 = x² × 4 = T1 × 4
Tn = Tn-1 × 4 = T1 × 4^(n-1) = x² × 4^(n-1)
so, Tn/Tn-1 = 4
but the geometric series for the perimeter is
p1 = 4x
pn = pn-1 × 2 = x × 2^(n+1)
and pn/pn-1 = 2
the sum of a finite geometric series with n elements is
s = p1 × (1 - r^(n))/(1 - r)
r is the common factor. in our case it is 2.
and we have 10 elements.
so, we know
8184 = 4x × (1 - 2^(10))/(1 ‐ 2) = 4x × (1-2¹⁰)/-1 =
= 4x × -1023 / - 1 = 4x × 1023
2046 = x × 1023
x = 2046/1023 = 2 cm
so, now finally for iii)
T10 (the last, biggest square) = x² × 4⁹ = 4×4⁹ = 4¹⁰ =
= 1,048,576 cm²
