help geometry

The entire triangle itself will have an area of 16, because the side length of the entire triangle is 4, so 4x4=16. 

Then, we'll have to find the area of the space that is NOT the stacked squares.

If you look closely there are two half-triangles to accompany each square so we can put them together and if you look even MORE closely, all of the side triangles equal HALF of all of the stacked squares combined.

I hope this gave you a clue, but I am warning you that I am not exactly great at math and let alone geometry.

It appears   that  the   side of the  bottom  square, S,  can  be  computed as

sin 60    =   S/2

2sin 60 =  S

2 sqrt (3) /2   = S

sqrt (3)  =  S

And it appears  that  each  square after  this one has side lengths 1/2 as long as the previous square  =   sqrt (3) / 2 , sqrt (3)/4, sqrt (3) /8

So  the combined areas are

(sqrt (3))^2  +  ( sqrt (3)/2)^2  + (sqrt (3)/4)^2  + (sqrt (3)/ 8)^2   =

3  + 3/4  +   3/16  +  3/64   =

3 (  1+  1/4 + 1/16 + 1/64)   =

3  (64 + 16 + 4 + 1)  / 64   =

255/ 64  units^2

If the given equilateral triangle has a side measure of 4.
What is the sum of the areas of the stacked squares inside it?

My attempt:

$\text{Let}~ \tan(60^\circ) = \sqrt{3} \\ \text{Let the side of the equilateral triangle }~ s_0 = 4 \\ s_1~\text{side of stacked square 1} \\ s_2~\text{side of stacked square 2} \\ s_3~\text{side of stacked square 3} \\ s_4~\text{side of stacked square 4} \\ \ldots \\ s_n~\text{side of stacked square n} \\$

$\begin{array}{|rcll|} \hline s_0 &=& s_1 + 2x_1 \quad | \quad \tan(60^\circ) = \dfrac{s_1}{x_1}~ \text{or}~ s_1 = x_1\tan(60^\circ)=x_1\sqrt{3} \\ s_0 &=& x_1\sqrt{3} + 2x_1 \\ s_0 &=& x_1(2+\sqrt{3}) \\ x_1 &=& \dfrac{s_0}{(2+\sqrt{3})}\times \dfrac{(2-\sqrt{3})} {(2-\sqrt{3}) } \\ x_1 &=& \dfrac{s_0(2-\sqrt{3})} { (2+\sqrt{3})(2-\sqrt{3}) } \\ x_1 &=& \dfrac{s_0(2-\sqrt{3})} { 4-3 } \\ \mathbf{x_1} &=& \mathbf{s_0(2-\sqrt{3})} \\ \hline s_0 &=& s_1 + 2x_1 \quad | \quad \mathbf{x_1=s_0(2-\sqrt{3})} \\ s_0 &=& s_1 + 2s_0(2-\sqrt{3}) \\ s_1 &=& s_0 - 2s_0(2-\sqrt{3}) \\ s_1 &=& s_0\Big(1-2(2-\sqrt{3})\Big) \\ s_1 &=& s_0(1-4+2\sqrt{3}) \\ \mathbf{s_1} &=& \mathbf{s_0(2\sqrt{3}-3)} \\ \hline \end{array}$

In the same way...

$\begin{array}{|rcll|} \hline s_2 &=& s_1(2\sqrt{3}-3) \quad | \quad \mathbf{s_1=s_0(2\sqrt{3}-3)} \\ s_2 &=& s_0(2\sqrt{3}-3)(2\sqrt{3}-3) \\ s_2 &=& s_0(2\sqrt{3}-3)^2 \\ \hline \ldots \\ s_3 &=& s_0(2\sqrt{3}-3)^3 \\ s_4 &=& s_0(2\sqrt{3}-3)^4 \\ \ldots \\ \mathbf{s_n} &=& \mathbf{s_0(2\sqrt{3}-3)^n} \\ \hline \end{array}$

sum

$\begin{array}{|rcll|} \hline \mathbf{sum_4} &=& \mathbf{s_1^2+s_2^2+s_3^2+s_4^2} \\ sum_4 &=& s_0^2(2\sqrt{3}-3)^2+s_0^2(2\sqrt{3}-3)^4+s_0^2(2\sqrt{3}-3)^6+s_0^2(2\sqrt{3}-3)^8 \\ sum_4 &=& s_0^2\Big( \underbrace{ (2\sqrt{3}-3)^2+(2\sqrt{3}-3)^4+(2\sqrt{3}-3)^6+(2\sqrt{3}-3)^8\Big) }_{Geometric~Sequence~q=(2\sqrt{3}-3)^2} \\ sum_4 &=& s_0^2(q+q^2+q^3+q^4) \\ sum_4 &=& s_0^2\dfrac{q}{1-q} (1-q^4) \quad | \quad q = 0.2153903092 \\ sum_4 &=& 4^2\times 0.2745190528 \times 0.9978476909 \\ sum_4 &=& 4.3923048454 \times 0.9978476909 \\ \mathbf{sum_4} &=& \mathbf{4.3828512478} \\ \hline \end{array}$

Infinite Geometric Sequence

$\begin{array}{|rcll|} \hline sum_{\infty} &=& s_0^2\dfrac{q}{1-q} \quad | \quad q = 0.2153903092 \\ sum_{\infty} &=& 4^2\times 0.2745190528 \\ \mathbf{sum_{\infty}} &=& \mathbf{4.3923048454} \\ \hline \end{array}$