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enter image description here Let $C_0$ an equilateral triangle of side 1 and $A_{n}$ the area of the figure $C_{n+1}, n\in\mathbb{N}\cup {0}$

Write $A_{n+1}$ based on $A_n$ and the area of $C_0$

I noted that $C_1$ has been built by adding a smaller equilateral triangle on each side of $C_0$. Then $C_2$ is built by adding an even smaller equilateral triangle on each side of $C_1$. Figure $C_3$ is constructed by adding a smaller equilateral triangle on each side of $C_2$. So each new figure is formed by adding small equilateral triangles on the sides of the previous figure.

The area of one smaller triangle added in each figure can be calculated using the formula: $$a=\frac{s^2\sqrt{3}}{4}$$ s is the lengt of the side, and knowing that the side of $C_0$ is $1$, I noted that the lenght of each segment of $C_{n+1}$ is givig by $\left(\frac{1}{3}\right)^{n+1}$ so

$$a=\frac{\left(\left(\frac{1}{3}\right)^{n+1}\right)^2\sqrt{3}}{4}$$

I also noted that the number of sides in figure $C_{n+1}$ is $3 \times 4 ^ n$ so: $$A_{n+1}=A_n+\frac{\left(\left(\frac{1}{3}\right)^{n+1}\right)^2\sqrt{3}}{4}\times 3\times 4^n=A_n+\left(\frac{1}{9}\right)^{n+1}\times \frac{\sqrt{3}}{4}\times3\times4^n$$ $$\boldsymbol{A_{n+1}=A_n+\frac{3\sqrt{3}}{16}\left(\frac{4}{9}\right)^{n+1}}$$


Is there another way or reasoning to get the same recursive sequence?

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  • $\begingroup$ There is a small mistake: the number of sides in figure $C_n$ is $3 \times 4^n$ (not $C_{n + 1}$). $\endgroup$ – Claude Apr 30 '18 at 17:57
  • $\begingroup$ @Claude Thanks for corrected me, I did not notice that mistake. Do you know if is there another way to get the same recursive sequence? $\endgroup$ – B. David Apr 30 '18 at 18:09
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One alternative reasoning might follow from:

$$A_{n+2} - A_{n+1} = \frac{4}{9} \left(A_{n + 1} - A_{n}\right) \quad;\quad A_1 = \frac{4}{3} A_0 \quad;\quad A_0 = \frac{\sqrt{3}}{4}\\ A_{n+2} - A_0 = \left(A_{n+2} - A_{n+1}\right) + \left(A_{n+1} - A_n\right) + \cdots + \left(A_1 - A_0\right)$$

I'll leave you to flesh out the details.

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  • $\begingroup$ Good idea! Thank your for your help $\endgroup$ – B. David Apr 30 '18 at 18:19

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