Sum of infinite squares.

Area of first square=12X12=144 square units. Side of second square = 6 $\sqrt{2}$ so area of second square=72 square units. And so on. the area of third square will be 36 square units.

This forms a infinite G.P. and sum of terms of infinite G.P. is $\frac{a}{1-r}$ .

a=144 and r= $\frac{1}{2}$ so Sum= $\frac{144}{1/2}$ = $144 \times 2$ = 288

The side of the biggest square is the diagonal of the second square and the side of the second square is the diagonal of the next smaller square and so on.A right triangle is formed by the diagonals and the 2 congruent sides.So by application of Pythagorean Theorem,we get sides as $\color{#D61F06}{12,6\sqrt2,,6,3\sqrt2,3\dotsm}$ .So the sum of areas is $\color{#3D99F6}{12^2+(6\sqrt2)^2+6^2+\dotsm=144+72+36+18+\dotsm}$ . $\color{#20A900}{Let\;S=144+72+36+\dotsm}\\ \color{#69047E}{Then\;\frac{S}{2}=72+36+18+\dotsm=S-144}\\ \color{#EC7300}{\frac{S}{2}=S-144\\144=S-\frac{S}{2}\\144=\frac{S}{2}\\ S=2\times144=\boxed{288}}$

Let A = Sum of areas, starting at 0. i a number of iterations, set to 100. and L the area of the current square, starting at 144, and halfing on each iteration (as each square will have 1/2 the area of the parent of the parent), this is easy to demonstrate using pitagoras, as the sides of the first sub-square will be $6 * \sqrt{2}$ , and the square of that is 72 = 144/2...

we can make the calculations with the following algorithm

i=100

A=0

while (i > 0) {

A=A+L

L=L/2.0

i=i-1

}

A will converge to 288.0