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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













5












$begingroup$


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 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 $$sum_{n=0}^infty left(frac{1}{4}right)^n$$










share|cite|improve this question











$endgroup$

















    5












    $begingroup$


    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 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 $$sum_{n=0}^infty left(frac{1}{4}right)^n$$










    share|cite|improve this question











    $endgroup$















      5












      5








      5





      $begingroup$


      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 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 $$sum_{n=0}^infty left(frac{1}{4}right)^n$$










      share|cite|improve this question











      $endgroup$




      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 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 $$sum_{n=0}^infty left(frac{1}{4}right)^n$$







      sequences-and-series algebra-precalculus geometric-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 12 hours ago









      Blue

      49.5k870157




      49.5k870157










      asked 14 hours ago









      user477465user477465

      130113




      130113






















          5 Answers
          5






          active

          oldest

          votes


















          2












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            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 $s = frac{1}{4}(frac12 + s + 1 + frac12)$.






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              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 $S=frac{a_1}{1-r}$.






              share|cite|improve this answer









              $endgroup$





















                0












                $begingroup$

                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 $frac{2}{3}.$






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  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



















                0












                $begingroup$

                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:



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






                share|cite|improve this answer









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                  5 Answers
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                  5 Answers
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                  active

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                  2












                  $begingroup$

                  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.






                  share|cite|improve this answer









                  $endgroup$


















                    2












                    $begingroup$

                    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.






                    share|cite|improve this answer









                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$

                      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.






                      share|cite|improve this answer









                      $endgroup$



                      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.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 14 hours ago









                      Ross MillikanRoss Millikan

                      301k24200375




                      301k24200375























                          2












                          $begingroup$

                          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 $s = frac{1}{4}(frac12 + s + 1 + frac12)$.






                          share|cite|improve this answer









                          $endgroup$


















                            2












                            $begingroup$

                            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 $s = frac{1}{4}(frac12 + s + 1 + frac12)$.






                            share|cite|improve this answer









                            $endgroup$
















                              2












                              2








                              2





                              $begingroup$

                              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 $s = frac{1}{4}(frac12 + s + 1 + frac12)$.






                              share|cite|improve this answer









                              $endgroup$



                              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 $s = frac{1}{4}(frac12 + s + 1 + frac12)$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 7 hours ago









                              orlporlp

                              7,6491433




                              7,6491433























                                  1












                                  $begingroup$

                                  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 $S=frac{a_1}{1-r}$.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    1












                                    $begingroup$

                                    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 $S=frac{a_1}{1-r}$.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$

                                      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 $S=frac{a_1}{1-r}$.






                                      share|cite|improve this answer









                                      $endgroup$



                                      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 $S=frac{a_1}{1-r}$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 13 hours ago









                                      farruhotafarruhota

                                      21.8k2842




                                      21.8k2842























                                          0












                                          $begingroup$

                                          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 $frac{2}{3}.$






                                          share|cite|improve this answer









                                          $endgroup$













                                          • $begingroup$
                                            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
















                                          0












                                          $begingroup$

                                          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 $frac{2}{3}.$






                                          share|cite|improve this answer









                                          $endgroup$













                                          • $begingroup$
                                            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














                                          0












                                          0








                                          0





                                          $begingroup$

                                          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 $frac{2}{3}.$






                                          share|cite|improve this answer









                                          $endgroup$



                                          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 $frac{2}{3}.$







                                          share|cite|improve this answer












                                          share|cite|improve this answer



                                          share|cite|improve this answer










                                          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


















                                          • $begingroup$
                                            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











                                          0












                                          $begingroup$

                                          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:



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






                                          share|cite|improve this answer









                                          $endgroup$


















                                            0












                                            $begingroup$

                                            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:



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






                                            share|cite|improve this answer









                                            $endgroup$
















                                              0












                                              0








                                              0





                                              $begingroup$

                                              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:



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






                                              share|cite|improve this answer









                                              $endgroup$



                                              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:



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







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered 13 hours ago









                                              Michael RybkinMichael Rybkin

                                              4,164422




                                              4,164422






























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