Interpreting images representing geometric seriesFind the common ratio of the geometric series with the sum...
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Interpreting images representing geometric series
Find the common ratio of the geometric series with the sum and the first termGeometric Series, finding the r valueGeometric Series of $a = 2$, $r = 2$Unexpected result while calculation geometric series sumWhen is a Power Series a Geometric Series?Name for Finite Geometric Series Summing to 1Geometric Series Question Variable answerGeometric series : Find common ration 'r'In what sense is a geometric series geometric?Substitution in Infinite Geometric Series
$begingroup$
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
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 $$sum_{n=0}^infty left(frac{1}{4}right)^n$$
sequences-and-series algebra-precalculus geometric-series
$endgroup$
add a comment |
$begingroup$
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
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 $$sum_{n=0}^infty left(frac{1}{4}right)^n$$
sequences-and-series algebra-precalculus geometric-series
$endgroup$
add a comment |
$begingroup$
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
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 $$sum_{n=0}^infty left(frac{1}{4}right)^n$$
sequences-and-series algebra-precalculus geometric-series
$endgroup$
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
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 $$sum_{n=0}^infty left(frac{1}{4}right)^n$$
sequences-and-series algebra-precalculus geometric-series
sequences-and-series algebra-precalculus geometric-series
edited 12 hours ago
Blue
49.5k870157
49.5k870157
asked 14 hours ago
user477465user477465
130113
130113
add a comment |
add a comment |
5 Answers
5
active
oldest
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$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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 $s = frac{1}{4}(frac12 + s + 1 + frac12)$.
$endgroup$
add a comment |
$begingroup$
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 $S=frac{a_1}{1-r}$.
$endgroup$
add a comment |
$begingroup$
In your first example, with the squares, you color in $(frac{1}{4})^n$ with each new square. These squares add to $sum_{n=1}^infty (frac{1}{4})^n = frac{1}{3}.$
In your second example, with the rectangles, the first rectangle is $frac{1}{2}$ of the square, but your second rectangle is $frac{1}{4}cdotfrac{1}{2}$, so this sum is $frac{1}{2} sum_{n=0}^infty (frac{1}{4})^n = frac{2}{3}.$
Your triangles also sum in the same way to $frac{2}{3}.$
$endgroup$
$begingroup$
im having trouble understanding the difference between the first image and the second and third
$endgroup$
– user477465
14 hours ago
$begingroup$
i dont understand how you get the formula for the second image as different from the first one
$endgroup$
– user477465
14 hours ago
$begingroup$
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
14 hours ago
$begingroup$
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
14 hours ago
$begingroup$
The first sum is $frac{frac{1}{4}}{1 - frac{1}{4}}=frac{1}{3}$.
$endgroup$
– mjw
13 hours ago
|
show 1 more comment
$begingroup$
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: $frac{1}{4}cdotfrac{1}{4}$. The area of the third square would be a fourth of that: $frac{1}{4}cdotfrac{1}{4}cdotfrac{1}{4}$. Do you see the pattern?
$$
frac{1}{4}+left(frac{1}{4}cdotfrac{1}{4}right)+left(frac{1}{4}cdotfrac{1}{4}cdotfrac{1}{4}right)+...=\
left(frac{1}{4}right)^1+left(frac{1}{4}right)^2+left(frac{1}{4}right)^3+...=\
sum_{n=1}^{infty}left(frac{1}{4}right)^n=
sum_{n=0}^{infty}left(frac{1}{4}right)^n-1=
frac{1}{1-frac{1}{4}}-1=frac{4}{3}-1=frac{1}{3}.
$$
2) The first rectangle is area $frac{1}{2}$. The second is a half of the original area divided by four $frac{1}{2}cdotfrac{1}{4}$. The third part is one fourt of that $frac{1}{2}cdotfrac{1}{4}cdotfrac{1}{4}$:
$$
frac{1}{2}+left(frac{1}{2}cdotfrac{1}{4}right)+left(frac{1}{2}cdotfrac{1}{4}cdotfrac{1}{4}right)+...=\
frac{1}{2}left(frac{1}{4}right)^0+frac{1}{2}left(frac{1}{4}right)^1+
frac{1}{2}left(frac{1}{4}right)^2+...=\
sum_{n=0}^{infty}frac{1}{2}left(frac{1}{4}right)^n=frac{1}{2}cdotfrac{1}{1-frac{1}{4}}=frac{2}{3}.
$$
3) The first triangle is area $frac{1}{2}$. The second triangle is area $frac{1}{2}cdotfrac{1}{4}$ (a fourth part of a half of the original triangle). The third triangle is going to have area $frac{1}{2}cdotfrac{1}{4}cdotfrac{1}{4}$. I think you see that the pattern is the same as in the previous case:
$$
frac{1}{2}+left(frac{1}{2}cdotfrac{1}{4}right)+left(frac{1}{2}cdotfrac{1}{4}cdotfrac{1}{4}right)+...=\
frac{1}{2}left(frac{1}{4}right)^0+frac{1}{2}left(frac{1}{4}right)^1+
frac{1}{2}left(frac{1}{4}right)^2+...=\
sum_{n=0}^{infty}frac{1}{2}left(frac{1}{4}right)^n=frac{1}{2}cdotfrac{1}{1-frac{1}{4}}=frac{2}{3}.
$$
$endgroup$
add a comment |
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5 Answers
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active
oldest
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5 Answers
5
active
oldest
votes
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$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered 14 hours ago
Ross MillikanRoss Millikan
301k24200375
301k24200375
add a comment |
add a comment |
$begingroup$
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 $s = frac{1}{4}(frac12 + s + 1 + frac12)$.
$endgroup$
add a comment |
$begingroup$
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 $s = frac{1}{4}(frac12 + s + 1 + frac12)$.
$endgroup$
add a comment |
$begingroup$
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 $s = frac{1}{4}(frac12 + s + 1 + frac12)$.
$endgroup$
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 $s = frac{1}{4}(frac12 + s + 1 + frac12)$.
answered 7 hours ago
orlporlp
7,6491433
7,6491433
add a comment |
add a comment |
$begingroup$
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 $S=frac{a_1}{1-r}$.
$endgroup$
add a comment |
$begingroup$
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 $S=frac{a_1}{1-r}$.
$endgroup$
add a comment |
$begingroup$
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 $S=frac{a_1}{1-r}$.
$endgroup$
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 $S=frac{a_1}{1-r}$.
answered 13 hours ago
farruhotafarruhota
21.8k2842
21.8k2842
add a comment |
add a comment |
$begingroup$
In your first example, with the squares, you color in $(frac{1}{4})^n$ with each new square. These squares add to $sum_{n=1}^infty (frac{1}{4})^n = frac{1}{3}.$
In your second example, with the rectangles, the first rectangle is $frac{1}{2}$ of the square, but your second rectangle is $frac{1}{4}cdotfrac{1}{2}$, so this sum is $frac{1}{2} sum_{n=0}^infty (frac{1}{4})^n = frac{2}{3}.$
Your triangles also sum in the same way to $frac{2}{3}.$
$endgroup$
$begingroup$
im having trouble understanding the difference between the first image and the second and third
$endgroup$
– user477465
14 hours ago
$begingroup$
i dont understand how you get the formula for the second image as different from the first one
$endgroup$
– user477465
14 hours ago
$begingroup$
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
14 hours ago
$begingroup$
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
14 hours ago
$begingroup$
The first sum is $frac{frac{1}{4}}{1 - frac{1}{4}}=frac{1}{3}$.
$endgroup$
– mjw
13 hours ago
|
show 1 more comment
$begingroup$
In your first example, with the squares, you color in $(frac{1}{4})^n$ with each new square. These squares add to $sum_{n=1}^infty (frac{1}{4})^n = frac{1}{3}.$
In your second example, with the rectangles, the first rectangle is $frac{1}{2}$ of the square, but your second rectangle is $frac{1}{4}cdotfrac{1}{2}$, so this sum is $frac{1}{2} sum_{n=0}^infty (frac{1}{4})^n = frac{2}{3}.$
Your triangles also sum in the same way to $frac{2}{3}.$
$endgroup$
$begingroup$
im having trouble understanding the difference between the first image and the second and third
$endgroup$
– user477465
14 hours ago
$begingroup$
i dont understand how you get the formula for the second image as different from the first one
$endgroup$
– user477465
14 hours ago
$begingroup$
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
14 hours ago
$begingroup$
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
14 hours ago
$begingroup$
The first sum is $frac{frac{1}{4}}{1 - frac{1}{4}}=frac{1}{3}$.
$endgroup$
– mjw
13 hours ago
|
show 1 more comment
$begingroup$
In your first example, with the squares, you color in $(frac{1}{4})^n$ with each new square. These squares add to $sum_{n=1}^infty (frac{1}{4})^n = frac{1}{3}.$
In your second example, with the rectangles, the first rectangle is $frac{1}{2}$ of the square, but your second rectangle is $frac{1}{4}cdotfrac{1}{2}$, so this sum is $frac{1}{2} sum_{n=0}^infty (frac{1}{4})^n = frac{2}{3}.$
Your triangles also sum in the same way to $frac{2}{3}.$
$endgroup$
In your first example, with the squares, you color in $(frac{1}{4})^n$ with each new square. These squares add to $sum_{n=1}^infty (frac{1}{4})^n = frac{1}{3}.$
In your second example, with the rectangles, the first rectangle is $frac{1}{2}$ of the square, but your second rectangle is $frac{1}{4}cdotfrac{1}{2}$, so this sum is $frac{1}{2} sum_{n=0}^infty (frac{1}{4})^n = frac{2}{3}.$
Your triangles also sum in the same way to $frac{2}{3}.$
answered 14 hours ago
mjwmjw
2437
2437
$begingroup$
im having trouble understanding the difference between the first image and the second and third
$endgroup$
– user477465
14 hours ago
$begingroup$
i dont understand how you get the formula for the second image as different from the first one
$endgroup$
– user477465
14 hours ago
$begingroup$
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
14 hours ago
$begingroup$
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
14 hours ago
$begingroup$
The first sum is $frac{frac{1}{4}}{1 - frac{1}{4}}=frac{1}{3}$.
$endgroup$
– mjw
13 hours ago
|
show 1 more comment
$begingroup$
im having trouble understanding the difference between the first image and the second and third
$endgroup$
– user477465
14 hours ago
$begingroup$
i dont understand how you get the formula for the second image as different from the first one
$endgroup$
– user477465
14 hours ago
$begingroup$
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
14 hours ago
$begingroup$
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
14 hours ago
$begingroup$
The first sum is $frac{frac{1}{4}}{1 - frac{1}{4}}=frac{1}{3}$.
$endgroup$
– mjw
13 hours ago
$begingroup$
im having trouble understanding the difference between the first image and the second and third
$endgroup$
– user477465
14 hours ago
$begingroup$
im having trouble understanding the difference between the first image and the second and third
$endgroup$
– user477465
14 hours ago
$begingroup$
i dont understand how you get the formula for the second image as different from the first one
$endgroup$
– user477465
14 hours ago
$begingroup$
i dont understand how you get the formula for the second image as different from the first one
$endgroup$
– user477465
14 hours ago
$begingroup$
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
14 hours ago
$begingroup$
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
14 hours ago
$begingroup$
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
14 hours ago
$begingroup$
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
14 hours ago
$begingroup$
The first sum is $frac{frac{1}{4}}{1 - frac{1}{4}}=frac{1}{3}$.
$endgroup$
– mjw
13 hours ago
$begingroup$
The first sum is $frac{frac{1}{4}}{1 - frac{1}{4}}=frac{1}{3}$.
$endgroup$
– mjw
13 hours ago
|
show 1 more comment
$begingroup$
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: $frac{1}{4}cdotfrac{1}{4}$. The area of the third square would be a fourth of that: $frac{1}{4}cdotfrac{1}{4}cdotfrac{1}{4}$. Do you see the pattern?
$$
frac{1}{4}+left(frac{1}{4}cdotfrac{1}{4}right)+left(frac{1}{4}cdotfrac{1}{4}cdotfrac{1}{4}right)+...=\
left(frac{1}{4}right)^1+left(frac{1}{4}right)^2+left(frac{1}{4}right)^3+...=\
sum_{n=1}^{infty}left(frac{1}{4}right)^n=
sum_{n=0}^{infty}left(frac{1}{4}right)^n-1=
frac{1}{1-frac{1}{4}}-1=frac{4}{3}-1=frac{1}{3}.
$$
2) The first rectangle is area $frac{1}{2}$. The second is a half of the original area divided by four $frac{1}{2}cdotfrac{1}{4}$. The third part is one fourt of that $frac{1}{2}cdotfrac{1}{4}cdotfrac{1}{4}$:
$$
frac{1}{2}+left(frac{1}{2}cdotfrac{1}{4}right)+left(frac{1}{2}cdotfrac{1}{4}cdotfrac{1}{4}right)+...=\
frac{1}{2}left(frac{1}{4}right)^0+frac{1}{2}left(frac{1}{4}right)^1+
frac{1}{2}left(frac{1}{4}right)^2+...=\
sum_{n=0}^{infty}frac{1}{2}left(frac{1}{4}right)^n=frac{1}{2}cdotfrac{1}{1-frac{1}{4}}=frac{2}{3}.
$$
3) The first triangle is area $frac{1}{2}$. The second triangle is area $frac{1}{2}cdotfrac{1}{4}$ (a fourth part of a half of the original triangle). The third triangle is going to have area $frac{1}{2}cdotfrac{1}{4}cdotfrac{1}{4}$. I think you see that the pattern is the same as in the previous case:
$$
frac{1}{2}+left(frac{1}{2}cdotfrac{1}{4}right)+left(frac{1}{2}cdotfrac{1}{4}cdotfrac{1}{4}right)+...=\
frac{1}{2}left(frac{1}{4}right)^0+frac{1}{2}left(frac{1}{4}right)^1+
frac{1}{2}left(frac{1}{4}right)^2+...=\
sum_{n=0}^{infty}frac{1}{2}left(frac{1}{4}right)^n=frac{1}{2}cdotfrac{1}{1-frac{1}{4}}=frac{2}{3}.
$$
$endgroup$
add a comment |
$begingroup$
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: $frac{1}{4}cdotfrac{1}{4}$. The area of the third square would be a fourth of that: $frac{1}{4}cdotfrac{1}{4}cdotfrac{1}{4}$. Do you see the pattern?
$$
frac{1}{4}+left(frac{1}{4}cdotfrac{1}{4}right)+left(frac{1}{4}cdotfrac{1}{4}cdotfrac{1}{4}right)+...=\
left(frac{1}{4}right)^1+left(frac{1}{4}right)^2+left(frac{1}{4}right)^3+...=\
sum_{n=1}^{infty}left(frac{1}{4}right)^n=
sum_{n=0}^{infty}left(frac{1}{4}right)^n-1=
frac{1}{1-frac{1}{4}}-1=frac{4}{3}-1=frac{1}{3}.
$$
2) The first rectangle is area $frac{1}{2}$. The second is a half of the original area divided by four $frac{1}{2}cdotfrac{1}{4}$. The third part is one fourt of that $frac{1}{2}cdotfrac{1}{4}cdotfrac{1}{4}$:
$$
frac{1}{2}+left(frac{1}{2}cdotfrac{1}{4}right)+left(frac{1}{2}cdotfrac{1}{4}cdotfrac{1}{4}right)+...=\
frac{1}{2}left(frac{1}{4}right)^0+frac{1}{2}left(frac{1}{4}right)^1+
frac{1}{2}left(frac{1}{4}right)^2+...=\
sum_{n=0}^{infty}frac{1}{2}left(frac{1}{4}right)^n=frac{1}{2}cdotfrac{1}{1-frac{1}{4}}=frac{2}{3}.
$$
3) The first triangle is area $frac{1}{2}$. The second triangle is area $frac{1}{2}cdotfrac{1}{4}$ (a fourth part of a half of the original triangle). The third triangle is going to have area $frac{1}{2}cdotfrac{1}{4}cdotfrac{1}{4}$. I think you see that the pattern is the same as in the previous case:
$$
frac{1}{2}+left(frac{1}{2}cdotfrac{1}{4}right)+left(frac{1}{2}cdotfrac{1}{4}cdotfrac{1}{4}right)+...=\
frac{1}{2}left(frac{1}{4}right)^0+frac{1}{2}left(frac{1}{4}right)^1+
frac{1}{2}left(frac{1}{4}right)^2+...=\
sum_{n=0}^{infty}frac{1}{2}left(frac{1}{4}right)^n=frac{1}{2}cdotfrac{1}{1-frac{1}{4}}=frac{2}{3}.
$$
$endgroup$
add a comment |
$begingroup$
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: $frac{1}{4}cdotfrac{1}{4}$. The area of the third square would be a fourth of that: $frac{1}{4}cdotfrac{1}{4}cdotfrac{1}{4}$. Do you see the pattern?
$$
frac{1}{4}+left(frac{1}{4}cdotfrac{1}{4}right)+left(frac{1}{4}cdotfrac{1}{4}cdotfrac{1}{4}right)+...=\
left(frac{1}{4}right)^1+left(frac{1}{4}right)^2+left(frac{1}{4}right)^3+...=\
sum_{n=1}^{infty}left(frac{1}{4}right)^n=
sum_{n=0}^{infty}left(frac{1}{4}right)^n-1=
frac{1}{1-frac{1}{4}}-1=frac{4}{3}-1=frac{1}{3}.
$$
2) The first rectangle is area $frac{1}{2}$. The second is a half of the original area divided by four $frac{1}{2}cdotfrac{1}{4}$. The third part is one fourt of that $frac{1}{2}cdotfrac{1}{4}cdotfrac{1}{4}$:
$$
frac{1}{2}+left(frac{1}{2}cdotfrac{1}{4}right)+left(frac{1}{2}cdotfrac{1}{4}cdotfrac{1}{4}right)+...=\
frac{1}{2}left(frac{1}{4}right)^0+frac{1}{2}left(frac{1}{4}right)^1+
frac{1}{2}left(frac{1}{4}right)^2+...=\
sum_{n=0}^{infty}frac{1}{2}left(frac{1}{4}right)^n=frac{1}{2}cdotfrac{1}{1-frac{1}{4}}=frac{2}{3}.
$$
3) The first triangle is area $frac{1}{2}$. The second triangle is area $frac{1}{2}cdotfrac{1}{4}$ (a fourth part of a half of the original triangle). The third triangle is going to have area $frac{1}{2}cdotfrac{1}{4}cdotfrac{1}{4}$. I think you see that the pattern is the same as in the previous case:
$$
frac{1}{2}+left(frac{1}{2}cdotfrac{1}{4}right)+left(frac{1}{2}cdotfrac{1}{4}cdotfrac{1}{4}right)+...=\
frac{1}{2}left(frac{1}{4}right)^0+frac{1}{2}left(frac{1}{4}right)^1+
frac{1}{2}left(frac{1}{4}right)^2+...=\
sum_{n=0}^{infty}frac{1}{2}left(frac{1}{4}right)^n=frac{1}{2}cdotfrac{1}{1-frac{1}{4}}=frac{2}{3}.
$$
$endgroup$
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: $frac{1}{4}cdotfrac{1}{4}$. The area of the third square would be a fourth of that: $frac{1}{4}cdotfrac{1}{4}cdotfrac{1}{4}$. Do you see the pattern?
$$
frac{1}{4}+left(frac{1}{4}cdotfrac{1}{4}right)+left(frac{1}{4}cdotfrac{1}{4}cdotfrac{1}{4}right)+...=\
left(frac{1}{4}right)^1+left(frac{1}{4}right)^2+left(frac{1}{4}right)^3+...=\
sum_{n=1}^{infty}left(frac{1}{4}right)^n=
sum_{n=0}^{infty}left(frac{1}{4}right)^n-1=
frac{1}{1-frac{1}{4}}-1=frac{4}{3}-1=frac{1}{3}.
$$
2) The first rectangle is area $frac{1}{2}$. The second is a half of the original area divided by four $frac{1}{2}cdotfrac{1}{4}$. The third part is one fourt of that $frac{1}{2}cdotfrac{1}{4}cdotfrac{1}{4}$:
$$
frac{1}{2}+left(frac{1}{2}cdotfrac{1}{4}right)+left(frac{1}{2}cdotfrac{1}{4}cdotfrac{1}{4}right)+...=\
frac{1}{2}left(frac{1}{4}right)^0+frac{1}{2}left(frac{1}{4}right)^1+
frac{1}{2}left(frac{1}{4}right)^2+...=\
sum_{n=0}^{infty}frac{1}{2}left(frac{1}{4}right)^n=frac{1}{2}cdotfrac{1}{1-frac{1}{4}}=frac{2}{3}.
$$
3) The first triangle is area $frac{1}{2}$. The second triangle is area $frac{1}{2}cdotfrac{1}{4}$ (a fourth part of a half of the original triangle). The third triangle is going to have area $frac{1}{2}cdotfrac{1}{4}cdotfrac{1}{4}$. I think you see that the pattern is the same as in the previous case:
$$
frac{1}{2}+left(frac{1}{2}cdotfrac{1}{4}right)+left(frac{1}{2}cdotfrac{1}{4}cdotfrac{1}{4}right)+...=\
frac{1}{2}left(frac{1}{4}right)^0+frac{1}{2}left(frac{1}{4}right)^1+
frac{1}{2}left(frac{1}{4}right)^2+...=\
sum_{n=0}^{infty}frac{1}{2}left(frac{1}{4}right)^n=frac{1}{2}cdotfrac{1}{1-frac{1}{4}}=frac{2}{3}.
$$
answered 13 hours ago
Michael RybkinMichael Rybkin
4,164422
4,164422
add a comment |
add a comment |
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