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I understand the formula for infinite geometric series as

$$S = \frac{a_{1}}{1-r}$$ if $0<r<1$

However I'm having trouble applying it to these images enter image description here

It seems to me that in the first image, the first square represents 1/4 of the entire square

For the second and third images, the respective rectangle and triangle make up 1/2 of the entire square.

Not sure what to do with this. Does it mean that for the first image for example, the image is $$\sum_{n=0}^\infty \left(\frac{1}{4}\right)^n$$

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In your first example, with the squares, you color in $(\frac{1}{4})^n$ with each new square. These squares add to $\sum_{n=1}^\infty (\frac{1}{4})^n = \frac{\frac14}{1-\frac14}=\frac{1}{3}.$

In your second example, with the rectangles, the first rectangle is $\frac{1}{2}$ of the square, but your second rectangle is $\frac{1}{4}\cdot\frac{1}{2}$, so this sum is $\frac{1}{2} \sum_{n=0}^\infty (\frac{1}{4})^n = \frac12\cdot\frac{1}{1-\frac14}=\frac12\cdot\frac43=\frac{2}{3}.$

Your triangles also sum in the same way to $\frac{2}{3}.$

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  • $\begingroup$ im having trouble understanding the difference between the first image and the second and third $\endgroup$ – user477465 Apr 7 '19 at 4:37
  • $\begingroup$ i dont understand how you get the formula for the second image as different from the first one $\endgroup$ – user477465 Apr 7 '19 at 4:39
  • $\begingroup$ In Image 1, the largest square is $\frac{1}{4}$ the total area. In Image 2, the largest rectangle is $\frac{1}{2}$ the total area. $\endgroup$ – mjw Apr 7 '19 at 4:40
  • $\begingroup$ Before doing the calculation: Clearly, the total gray area in Image 1 is less than $\frac{1}{2}$ and the gray areas in Images 2 & 3 are each greater than $\frac{1}{2}$. $\endgroup$ – mjw Apr 7 '19 at 4:43
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You are correct, in the first image the largest gray square is $\frac 12 \times \frac 12$ of the original square area. Now what is the area of the next gray square? It might help to continue the sides of the first gray square to the other side of the original square. The idea is that the sides of the gray squares form a geometric progression, as does the area of the gray squares. What is the sum of the progression of the sides? What is the ratio of that progression? What is the ratio of the progression of the areas?

The same ideas will solve the others.

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You can solve these without geometric series.

Say the proportion of coverage of the first image is $s$. Split the first image up into four squares. The topleft and bottomright are not covered at all. The bottomleft square is fully covered. And the topright square? Well it's covered in exactly the same way as $s$ itself. So we find:

\[s = \frac{1}{4}(0 + s + 1 + 0)\] \[4s = 1 + s\] \[s = \frac{1}{3}\]

Similarly for the other images we find equations $s = \frac{1}{4}(1 + s + 1 + 0)$ and $s = \frac{1}{4}(\frac12 + s + 1 + \frac12)$.

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The first terms are $1/4; 1/2; 1/2$, respectively.

The second and following terms are $1/4$ times the previous. To see this: take next smaller figure and place it inside previous figure to check four of them can fill it.

Knowing the first term $a_1$ and the ratio $r$, you can use the sum formula $S=\frac{a_1}{1-r}$.

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1) The area of the first shaded square is a fourth part of the original square: $\frac{1}{4}$. The area of the second shaded square is a fourth of a fourth of the original area: $\frac{1}{4}\cdot\frac{1}{4}$. The area of the third square would be a fourth of that: $\frac{1}{4}\cdot\frac{1}{4}\cdot\frac{1}{4}$. Do you see the pattern?

$$ \frac{1}{4}+\left(\frac{1}{4}\cdot\frac{1}{4}\right)+\left(\frac{1}{4}\cdot\frac{1}{4}\cdot\frac{1}{4}\right)+...=\\ \left(\frac{1}{4}\right)^1+\left(\frac{1}{4}\right)^2+\left(\frac{1}{4}\right)^3+...=\\ \sum_{n=1}^{\infty}\left(\frac{1}{4}\right)^n= \sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^n-1= \frac{1}{1-\frac{1}{4}}-1=\frac{4}{3}-1=\frac{1}{3}. $$

2) The first rectangle is area $\frac{1}{2}$. The second is a half of the original area divided by four $\frac{1}{2}\cdot\frac{1}{4}$. The third part is one fourt of that $\frac{1}{2}\cdot\frac{1}{4}\cdot\frac{1}{4}$:

$$ \frac{1}{2}+\left(\frac{1}{2}\cdot\frac{1}{4}\right)+\left(\frac{1}{2}\cdot\frac{1}{4}\cdot\frac{1}{4}\right)+...=\\ \frac{1}{2}\left(\frac{1}{4}\right)^0+\frac{1}{2}\left(\frac{1}{4}\right)^1+ \frac{1}{2}\left(\frac{1}{4}\right)^2+...=\\ \sum_{n=0}^{\infty}\frac{1}{2}\left(\frac{1}{4}\right)^n=\frac{1}{2}\cdot\frac{1}{1-\frac{1}{4}}=\frac{2}{3}. $$

3) The first triangle is area $\frac{1}{2}$. The second triangle is area $\frac{1}{2}\cdot\frac{1}{4}$ (a fourth part of a half of the original triangle). The third triangle is going to have area $\frac{1}{2}\cdot\frac{1}{4}\cdot\frac{1}{4}$. I think you see that the pattern is the same as in the previous case:

$$ \frac{1}{2}+\left(\frac{1}{2}\cdot\frac{1}{4}\right)+\left(\frac{1}{2}\cdot\frac{1}{4}\cdot\frac{1}{4}\right)+...=\\ \frac{1}{2}\left(\frac{1}{4}\right)^0+\frac{1}{2}\left(\frac{1}{4}\right)^1+ \frac{1}{2}\left(\frac{1}{4}\right)^2+...=\\ \sum_{n=0}^{\infty}\frac{1}{2}\left(\frac{1}{4}\right)^n=\frac{1}{2}\cdot\frac{1}{1-\frac{1}{4}}=\frac{2}{3}. $$

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