Infinite Geometric Series

Hello, , F19!

The midpoints of the sides of an equilateral triangle form new triangles.
If AB = 2 m, find the total area of the triangles given by the sequence.
Code: 
                  * C
                 / \
                /   \
               /     \
              /       \
             /    I    \
          E * - - * - - * D
           / \   / \   / \
          /   \ /   \ /   \
         /   C *- - -* H   \
        /       \   /       \
       /         \ /         \
    A * - - - - - * - - - - - * B
                  F
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\(\displaystyle \text{There area of an equilateral triangle with side }x\text{ is: }\:A \:=\:\frac{\sqrt{3}}{4}x^2\)

\(\displaystyle \Delta ABC\text{ has side 2, so its area is: }\:A_1 \:=\:\frac{\sqrt{3}}{4}(2^2) \:=\:\sqrt{3}\)

\(\displaystyle \text{The area of }\Delta DEF\text{ is }\tfrac{1}{4}\text{ of }\Delta ABC\!:\;\;A_2 \:=\:\frac{\sqrt{3}}{4}\)

\(\displaystyle \text{The area of }\Delta GHI\text{ is }\tfrac{1}{4}\text{ of }\Delta DEF\!:\;\;A_3 \:=\:\frac{\sqrt{3}}{4^2}\)

\(\displaystyle \text{The next triangle is }\tfrac{1}{4}\text{ of }\Delta GHI\!:\;\;A_4 \:=\:\frac{\sqrt{3}}{4^3} \quad\hdots \text{ and so on.}\)


\(\displaystyle \text{The sum of the areas is: }\;S \;=\;\sqrt{3} + \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4^2} + \frac{\sqrt{3}}{4^3} + \hdots\)

. . . . . . . . . . . . . . . . . . . \(\displaystyle = \;\sqrt{3}\underbrace{\left(1 + \frac{1}{4} + \frac{1}{4^2} + \frac{1}{4^3} + \hdots \right)}_{\text{a geometric series}}\)


\(\displaystyle \text{The geometric series has first term }a = 1\text{ and common ratio }r = \tfrac{1}{4}\)
. . \(\displaystyle \text{Its sum is: }\:\frac{1}{1-\frac{1}{4}} \:=\:\frac{1}{\frac{3}{4}} \:=\:\tfrac{4}{3}\)


\(\displaystyle \text{Therefore: }\:S \;=\;\sqrt{3}\left(\frac{4}{3}\right) \;=\;\frac{4\sqrt{3}}{3}\text{ m}^2\)

 
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