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A Sierpinski triangle, starts with a white equilateral triangle with sides of length $1$. First, the middle triangle is colored green. At the second step, 3 triangles are colored blue. At the third step, 9 triangles are colored red, and so on. At the $n$-th step, $3^{ n-1 }$ triangles are colored. Each step reduces the amount of the original white triangle that is still visible.

The original white triangle has area $\frac { \sqrt { 3 } }{ 4 } $ ​square units, and the area remaining white after $n$ steps is given by the formula $\large\frac { \sqrt { 3 } }{ 4 } \left(1-\frac { 1 }{ 4 } -\frac { 3^{ 1 } }{ 4^{ 2 } } -...-\frac { 3^{ n-1 } }{ 4^{ n } } \right)$ for $n≥1$.

How many square units of the original area remain white after 10 steps?

I feel completely lost and can't comprehend where to even start with this question. I am aware of the formulas that I can use to find the sum of a series but this seems to be different. I would like this broken down step by step for a layman like me. A direct answer will not help me. Thank you.

Question taken from Khan Academy's precalculus section

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  • $\begingroup$ it's easier to calculate how much of the original triangle is colored after 10 iterations, and then subtract that amount from the area of the original white triangle. $\endgroup$ – John Joy Dec 27 '14 at 18:12
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$-\frac { 1 }{ 4 } -\frac { 3^{ 1 } }{ 4^{ 2 } } -...-\frac { 3^{ n-1 } }{ 4^{ n } }$ is a geometric series which you may be able to solve.

Alternatively see that after one step you have $\frac { \sqrt { 3 } }{ 4 } \times \frac { 3 }{ 4 }$, after two steps $\frac { \sqrt { 3 } }{ 4 } \times \left(\frac { 3 }{ 4 }\right)^2$, and so on. Each step removes a quarter of the remaining area.

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  • $\begingroup$ Am I headed in the right direction? $a=\frac { \sqrt { 3 } }{ 4 } ,\quad r=(-\frac { 1 }{ 4 } ),\quad n = 10$ $\endgroup$ – Cherry_Developer Dec 27 '14 at 9:36
  • $\begingroup$ Try instead $a=-\frac { 1 }{ 4 } ,\quad r=\frac { 3 }{ 4 } ,\quad n = 10$, and then add that result to $1$ and then multiply the whole thing by $\frac { \sqrt { 3 } }{ 4 }$ $\endgroup$ – Henry Dec 27 '14 at 9:41
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You obviously suspect that the problem has to do with geometric series, probably because it's filed under "Geometric series" chapter. In this case you should look for a geometric series, but do not expect it to be right in front of your nose.

You correctly got to the sum $$\frac{\sqrt{3}}{4} \cdot \left( 1 - \frac{1}{4} - \frac{3^1}{4^2} - \frac{3^2}{4^3} - \cdots - \frac{3^9}{4^{10}} \right).$$ But where is the geometric series? You have to look at various parts of your formula to find it. First of all, a geometric series is going to have terms which are all positive, all negative, or alternating in sign. So it can't be $1, -1/4, -3^1/4^2, \ldots$ because the first term is positive and the rest negative. We might try taking every other term (and get two interleaves series), or take all but the first 27 terms – there is no recipe in general. In our case the most obvious candidate is $$- \frac{1}{4} - \frac{3^1}{4^2} - \frac{3^2}{4^3} - \cdots - \frac{3^9}{4^{10}},$$ whose $k$-th term is $-\frac{3^k}{4^{k+1}}$ (if we start counting $k$ from $0$). But is it geometric? Remember the definition of a geometric series: a series $x_0 + x_1 + x_2 + \cdots$ is geometric when the ratio of two consecutive terms $x_{k+1}/x_k$ is always the same constant, i.e., it does not depend on $k$. In our case: $$\left(\frac{3^k}{4^{k+1}}\right) \Big/ \left(\frac{3^{k+1}}{4^{k+2}}\right) = \frac{4}{3}.$$ It is geometric with ratio $r = \frac{4}{3}$? The first term is $a = -\frac{1}{4}$, so from here it is easy using the formula: $$- \frac{1}{4} - \frac{3^1}{4^2} - \frac{3^2}{4^3} - \cdots - \frac{3^9}{4^{10}} = -\frac{989527}{1048576},$$ if I didn't make a calculating error (the trickiest thing is to get $n$ right; pay attention to how you were taught the formula – does a series begin with $0$-th term or $1$-st term?). The final result seems to be $$\frac{2038103 \sqrt{3}}{4194304}.$$

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  • $\begingroup$ "The trickiest part is to get $n$ right". Please elaborate on this. I have noticed this with other problems. Can you elaborate on that? $\endgroup$ – Cherry_Developer Dec 27 '14 at 9:50
  • $\begingroup$ It's about an off-by-one error: to add the first ten terms of the series, should you plug in $n = 9$, $n = 10$, or $n = 11$? The answer depends on how you start indexing the series, i.e., if the first term corresponds to $n = 0$ then you should plug in $n = 9$ to get ten terms. However, if you start counting terms with $n = 1$ then you should plug in $n = 10$. This error is so prevalent in programming that it has a name: "Obi-Wan error". $\endgroup$ – Andrej Bauer Dec 27 '14 at 9:53
  • $\begingroup$ How about in the case of this problem? $\endgroup$ – Cherry_Developer Dec 27 '14 at 9:55
  • $\begingroup$ You won't learn anything by getting all the answers. Figure it out by yourself. Once you are convinced that you know what is right, come back to check. (By "convinced" I mean that you won't just immediately assume you're wrong if my answer is different from yours.) $\endgroup$ – Andrej Bauer Dec 27 '14 at 10:05

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