Right, answers to yesterdays question!

So we started with a single square which had an area of 1. After the first iteration of the pattern, we had four new squares, each having an area of 1/3 X 1/3 = 1/9. On the next iteration we had 20 new boxes, each with an area of 1/9 X 1/9 = 1/81. Write down what you know so far:
Squares total
area sum
area total
1 4/9
1 56/81

As this shape is created using a pattern, there will be a pattern for the total area (that is a colloquial logic proof, but you use those principles if you want a ‘solid’ proof). All that we have to do is find the pattern that is forming from these areas, so I’m going to rewrite the area sums in a slightly different way for the 3rd iteration: 1 + (51-50)/91 + (52-51)/92. (I presume most people know 52=5X5 = 25).
The amount of boxes is also in a pattern, they are going up in powers of 5.
So, for the first iteration, the area = 1.
Second iteration = 1 + (51-50)/91.
Third iteration = second iteration + (52-51)/92
Fourth iteration = third iteration + (53-52)/93
So what will the sum be for any iteration?
Let’s try and find the sum for the nth iteration. As you will notice, the area of any iteration is equal to the area of all the previous ones + the area of the new one, meaning that the sum can be written as a pattern (which we’ve already shown) but this means that we can now use the summation symbol…
(∞,i=1) ∑ i
This symbol is an easier way of writing out the sum 1+2+3+4+….all the way to infinity (and beyond!). very simply, the lower limit (i=1) is the number you start with, ∞ is the upper limit (So you continue adding up to this number) and the pattern between each number is (in this case) 1.
If we had
(∞,i=1) ∑ (i ^2)
It would mean add up the numbers 1 + 4 +9 +16….all the way to infinity, whereas
(5,i=1) ∑( i^2)
Is equal to 1+22+32+42+52
So I hope you now understand summations.
Getting back to our problem: we are trying to find the sum of all the areas, we know that the area of the nth iteration is equal to the sum of all the previous iterations plus the new one, and we’re noticing that the 4th iteration is equal to the previous 3 plus (53-52)/93. Notice now that there are three powers, two of which are 3, the other is a 2. 3 is 4-1, 2 is 4-2. It is therefore possible to write this iteration as (54-1-54-2)94-1.
If you now replace the 4 with an n, you get the general form: (5n-1-5n-2)/9n-1.
Realising this, we can make the pattern for the area, the nth iteration =
So if we write it with a summation symbol:
1+(n,i=2) ∑((5^(n-1)-5^(n-2))/(9^(n-1))
Will give you the total area for whatever value of n you use.
But we can simplify it down. Looking at just the part in the summation symbol : (5n-1-5n-2)/9n-1 we can simplify it to 5n-1(1-5-1)/9n-1. 1-51 = 4/5, so =(4/5)*(5n-1)/(9n-1), so the summation can be written as
1+(4/5) (n,i=2)∑(5^(n-1)/9(n-1))
Note that this series is for the first iteration of the pattern (the first time the squares are added).
We could simplify it to
1+(4/5) (n,i=1) ∑(5^n/9^n)
(I appologise for the hideous format of the summations, i struggle with HTML….)
But the original question was what is the total area that this shape will make? To do this, we need to look at what the sum looks like at infinity, ie, we want to look at the area at the millionth iteration of the pattern (just an example, we actually want to use a number that is even bigger) so we want to find the answer to the equation
1+(4/5)(∞,i=1) ∑ (5^n/9^n)
So let’s have a closer look at the summation.
The first term is going to be 5/9, the second will be 25/81, 125/243…etc which is a geometric sequence.
A geometric sequence is a sequence where each term is a multiple of the term before, so 1,2,4,8,16,32,64….would be a geometric sequence. Adding up terms like these can be rather difficult without using a formula, so it’s rather…lucky?… that someone has found them ( I may at some stage do a blog on the history of the arithmetic and geometric sequences) and the formula for finding the sum is as follows: given that the ratio between the terms of a geometric series lies between -1 and 1, the sum of the terms to infinity can be expressed as a/(1-r), where a is the first term in the sequence and r is the ratio. If we had the series 1; 1/2 ; 1/4 ; 1/8 etc, a would be 1, and r would be 1/2 so the equation would be 1/(1/2)=2.
There is an interesting story about this particular series as well, but that will be another time….
Now, using the above formula for the equation given, we can solve just the summation as follows:
(5/9)/(1-(5/9)) = (5/9)/(4/9) =5/4
Substituting it in, we get 1 +(4/5)X(5/4)=2 …and this is the total area of the shape!


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