Infinite Series - 2

The figure consists of an infinite number of similar "L" shapes; one of them is marked above. In each "L", there are three congruent squares, one which is shaded green. Therefore the area that is green is $\frac{1}{3}$ of the entire figure.

Let the area of the original be 1. Then the area of first green square is $\frac 14$ , that of second green square is $\frac 14 \times \frac 14$ , the third, $\frac 14 \times \frac 14 \times \frac 14$ and so on. Then the total green area is given by:

$\begin{aligned} A_{\color{#20A900}\text{green}} & = \frac 14 + \frac 1{4^2} + \frac 1{4^3} + \cdots \\ & = \frac 14 \color{#3D99F6} \left(1+\frac 14 + \frac 1{4^2} + \frac 1{4^3} + \cdots \right) & \small \color{#3D99F6} \text{Sum of geometric progression (see note).} \\ & = \frac 14 \color{#3D99F6} \left(\frac 1{1-\frac 14} \right) \\ & = \boxed{\dfrac 13} \end{aligned}$

Note: Sum of geometric progression .

$\begin{aligned} S_n & = 1 + r + r^2 + r^3 + \cdots + r^n \\ rS_n & = r + r^2 + r^3 + r^4 \cdots + r^{n+1} \\ S_n - rS_n & = 1- r^{n+1} \\ \implies S_n & = \frac {1-r^{n+1}}{1-r} \\ \implies S_\infty & = \lim_{n \to \infty} \frac {1-r^{n+1}}{1-r} = \frac 1{1-r} & \text{for }|r| < 1 \end{aligned}$

Simple,

Green Area: $0.25$ + $0.25*0.25$ + $0.25*0.25*0.25$ + $...$

(It's a geometric progression with first term ( $a$ ) $0.25$ and common ratio ( $r$ ) $0.25$ , the infinite sum is given by $\frac{a}{1-r}$ )

Answer : $\frac{0.25}{1-0.25}$ = $\frac{0.25}{0.75}$ = $\frac{1}{3}$

:)

If the fraction of the square that is green is $f$ , then by considering each quarter of the square, we write down an equation: $f = \frac{1}{4} + \frac{f}{4}$ . This is because the top-right quarter is green, and the same fraction of the bottom-right quarter is green as the fraction of the whole square that is green.

Therefore, by multiplying by 4, $4f = 1 + f$ , so $3f = 1$ , and we have that $f = \frac{1}{3}$ .

Since the first green rectangle is a quarter of the hole/1 we can write its area as $\frac{1}{4}$ , the term for the next square is a quarter of a quarter so $\frac{1}{4}$ x $\frac{1}{4}$ which is $\frac{1}{4}$ to the power of 2, the third will be $\frac{1}{4}$ to the power of 3 the fourth $\frac{1}{4}$ to the power of 4 and so on.. (my images are different but the problem works the same)

We can write the series of numbers as: If we move the green rectangles one step down/right in my example this happens: Now we can see that the green rectangles are lining up perfectly with one set of the unoccupied white squares. We can also perform a vertical translation of the green squares from their original position, to fill the other set of white squares: Since shifting doesn't change the area of the squares, we can see that by performing both shifts, we can fill the unit square completely with three copies of our series: Thus we can see that the area of the original green rectangles were exactly equal to $\frac{1}{3}$ I didn't come up with this solution but I think its a nice way of visually confirming the answer without using any complicated algebra, I wouldn't answer this on a test though :)

Each green square is placed between two white squares of equal area . So out of three squares two are white and one is green . This series goes on continuously till infinity .

Therefore , $\frac{1}{3}$ of the outer square will be green .

I just figured it had to be the fraction that, when written out in decimals, just kept repeating forever. Does that make sense? And that only works because this was a multiple choice question.

There’s a nice illustration of this sequence in the Number Theory course, chapter: Exploring Infinity that clearly illustrates that the series approaches 1/3 ...

Some of the solutions have used the general formula for finding the limit of this type of series. For the benefit of those who are not familiar with it here is a rough derivation:

Let $x = r + r^2 + r^3 + \cdots$ Notice that if you (a) multiply the series by $r$ or (b) subtract $r$ from the series you obtain the same result:

(a) $rx = r^2 + r^3 + r^4 + \cdots$ (b) $x - r = r^2 + r^3 + r^4 + \cdots$

Thus $rx$ and $x-r$ are equal, giving a simple equation that can be rearranged to give a general solution for $x$ : $rx = x-r$ $x - rx = r$ $x(1-r) = r$ Formula $x = \frac{r}{1-r}$

Let the area of the original be 1. The first green square's area is $\frac{1}{4}$ , the second green square's area is $\frac{1}{4^2}$ , the third green square's area is $\frac{1}{4^3}$ , and so on. So the total green area, represented by $x$ is:

$\begin{aligned} x & = \frac{1}{4} + \frac{1}{4^2} + \frac{1}{4^3} \dots \\ & = \frac{1}{4}\left( 1+\frac{1}{4} + \frac{1}{4^2} + \frac{1}{4^3} \dots \right) \\ & = \frac{1}{4} (1+x) \end{aligned}$

Since we have $x=\frac{1}{4}(1+x),$ we can solve for $x$ :

$x=\frac{1}{4}(1+x) \Rightarrow \\ 4x = 1+x \Rightarrow \\ 3x = 1 \Rightarrow \\ \boxed{x=\frac{1}{3}}$

Therefore, the answer is $\frac{1}{3}$ .

Each of the L shapes have 2 white squares and 1 green square, so the answer is 1/3. This problem is a proof without words of the geometric progression when r=1/4. For more interesting proofs without words, see here

Using intuitive thought, I remembered that between all the solutions given to me, only 1/3 continues to have a "repetitive" pattern (i.e. 0.33333...), while 1/4, 3/8, and 2/5 are non-repeating decimals. Using this thought I concluded that the solution was 1/3 due to its infinite nature.

Let the area of the biggest square be equal to $1$ . Then, we can write the green area $S$ as the sum :

$S = \frac{1}{4} + \frac{1}{16}\ + \frac{1}{64} + \dots$

By putting $\frac{1}{4}$ in evidence, we obtain :

$S = \frac{1}{4} \left( 1 + \frac{1}{4}\ + \frac{1}{16} + \dots \right)$

Recognizing the original sum $S$ in the parentheses, we can rewrite this last equation as :

$S = \frac{1}{4} \left( 1 + S \right)$

and solve it for $S$ :

$\begin{aligned} S = \frac{1}{4} \left( 1 + S \right) \Rightarrow 4S &= 1 + S \\ 3S &= 1 \\ S &= \frac{1}{3} \quad \scriptsize{\blacksquare} \end{aligned}$

$\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...=\dfrac{1}{4}(1+\dfrac{1}{4}+\dfrac{1}{4^2}+...)=\dfrac{1}{4}(\dfrac{1}{1-\frac{1}{4}})=\dfrac{1}{4-1}=\dfrac{1}{3}$

1/A + (1/A)^2 + (1/A)^3 ... (1/A)^n = [ 1-(1/A)^n ]÷(A-1)

You can check the above equation.

In this case n = infinite and A = 4 So the answer is 1÷(4-1) = 1/3

I just looked at the outermost square and it's 3/1

It only took less than a minute

either I did it to chance or did it with the right method

The Green area is in Geometric infinite series. Let the total green area equal 'S'. Then $\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\frac{1}{4^4}+\cdots=S$ Hence, S= $\frac{\frac{1}{4}}{1-\frac{1}{4}}$ = $\frac{\frac{1}{4}}{\frac{3}{4}}$ = $\frac{1}{3}$

απλώς κάνοντας πράξεις το άθροισμα τείνει στο 0.33 (1/4+1/16+1/64+...)

The shape is self replicating: The green area of the whole shape is equal to the green quarter square (top left) plus one fourth of itself (bottom right).

This gives us:

A = 1/4 + A/4

Which simplifies down to:

A = 1/3

Working out the first 3 squares is enough. 0.25 + 0.0625 + 0.015625 = 0.328125 ; progression will never reach 0.4 (2/5) so answer must be 1/3.

x = 1/4 + 1/4(1/4) + 1/4(1/4)(1/4) ... 4x = 1 + x 4=1/3

think of a fractal .. the fraction of area of the inner square is the same as the fraction of the whole

f = 1/4 + f•(1/4)

subtract f•(1/4) .. 3f/4 = 1/4

multiply through by 4 .. 3f = 1

divide by 3 .. f = 1/3

$\displaystyle \sum_{n=1}^\infty \frac{1}{4^n} = \frac{1}{3}$

But this works on TI-84 calcs

1
2
3

1000
  Σ (.25^X) ≈ 1/3
 X=1

Infinite geometric sequence

S= A/(1-r) S=sum A= initial r= ration between - 1 to 1 exclusive

Answer is 1/3

The other three options are rational numbers.

emm....I just drew a little precisely and by looking at it, I can tell the area is between 1/4 and 3/8, so it must be 1/3

Suppose the area of the largest square is a^2. then, first green square is 1/4 (a^2), second green square is 1/4 ((a/2)^2 and so on. total = (1/4+1/16+1/64+.....inf) (a^2) =(1/4)/(1-1/4) a^2 = 1/3

1/4 + 1/16 + 1/64 + 1/256 = 0.33... = 1/3.