Here are the first four squares in a sequence. Andrew says, “If I take any two squares in the sequence, then the difference between
their areas is always a multiple of 40.” Prove that Andrew is correct.
In order to prove this statement, we need to use algebra. It isn’t enough to just demonstrate that the statement is correct for particular
squares. We can call any two squares in the sequence the 𝑛th square and the 𝑛 plus 𝑥th
square, where 𝑛 and 𝑥 are both positive integers. By assuming that 𝑥 is a positive integer, this just means that the 𝑛th square will
always be the smaller of these two squares.
To find the areas of these two squares, we need expressions for their side
lengths. Let’s look at the sequence of side lengths more closely. The sequence of side lengths is 13, 23, 33, 43. The first difference between terms in this sequence is 10. As this difference is constant, this means that the sequence is linear. So its 𝑛th term is of the form 𝑎𝑛 plus 𝑏.
The value of 𝑎 is the difference between the terms in the sequence. So it’s 10. To find the value of 𝑏, we need to see what we have to add or subtract from the
sequence 10𝑛 to get our sequence. The sequence 10𝑛 is the 10 times table. So its first few terms are 10, 20, 30, and 40. To get from each term in the 10 times table to our sequence, we have to add three
which means that the 𝑛th term for the sequence of side lengths is 10𝑛 plus
Now, we can consider the areas of our two squares. To find the area of a square, we square its side length. So for the 𝑛th square, the area will be 10𝑛 plus three all squared. Remember this is equal to 10𝑛 plus three multiplied by 10𝑛 plus three.
And we can use FOIL to expand these brackets. Multiplying the first terms in each bracket together, we have 10𝑛 multiplied by
10𝑛, which is 100𝑛 squared. Multiplying the outer terms, we have 10𝑛 multiplied by three, which is 30𝑛. For the inner terms, we have three multiplied by 10𝑛, which is 30𝑛, and for the
last terms, three multiplied by three, which is nine. Don’t forget to simplify by grouping like terms together. So the area of the 𝑛th square is 100𝑛 squared plus 60𝑛 plus nine.
Now, you need to show all of the stages of your working out as they’re an important
part of the algebraic proof. However, I’m going to need some more space on screen for the next part of this
calculation. So I’m going to delete the working I’ve just done, but remember you must keep it.
Now, I need to find the area of the 𝑛 plus 𝑥th square. And this one is a little bit more complicated. We substitute 𝑛 plus 𝑥 for 𝑛 in the 𝑛th term of our side lengths. So the area is 10 multiplied by 𝑛 plus 𝑥 plus three all squared. Remember that we’re multiplying this bracket by itself. So we write it out twice. And once again, we’ll use the FOIL method.
Multiplying the first terms together, we have 10 multiplied by 10 which is 100 and 𝑛
plus 𝑥 multiplied by 𝑛 plus 𝑥 which is 𝑛 plus 𝑥 squared. We’ll expand this bracket later. Multiplying the outer terms together, we have 30 𝑛 plus 𝑥. Multiplying the inner terms together, we have another lot of 30 𝑛 plus 𝑥. And finally, multiplying the last terms together, we have plus nine.
Now, let’s just think about how to simplify this first term, which is 100 multiplied
by 𝑛 plus 𝑥 all squared. 𝑛 plus 𝑥 all squared is equal to 𝑛 plus 𝑥 multiplied by 𝑛 plus 𝑥. And again, we can use FOIL to expand this. It gives 𝑛 squared plus 𝑛𝑥 plus 𝑛𝑥 plus 𝑥 squared, which simplifies to 𝑛
squared plus two 𝑛𝑥 plus 𝑥 squared.
Multiplying all of these terms by 100 gives 100𝑛 squared plus 200𝑛𝑥 plus 100𝑥
squared. We can also group together the two terms of 30 𝑛 plus 𝑥 to give 60 lots of 𝑛 plus
𝑥. And we can expand this bracket to give 60𝑛 plus 60𝑥. The simplified version of the area of the 𝑛 plus 𝑥th square is 100𝑛 squared plus
200𝑛𝑥 plus 100𝑥 squared plus 60𝑛 plus 60𝑥 plus nine.
Now, remember the statement that Andrew makes is about the difference between the
areas of these two squares. So we need to subtract the area of the 𝑛th square from our expression for the area
of the 𝑛 plus 𝑥th square. So the difference is given by 100𝑛 squared plus 200𝑛𝑥 plus 100𝑥 squared plus 60𝑛
plus 60𝑥 plus nine minus 100𝑛 squared plus 60𝑛 plus nine.
This looks pretty complicated. But some of the terms cancel straight away. 100𝑛 squared minus 100𝑛 squared gives zero, 60𝑛 minus 60𝑛 gives zero, and nine
minus nine gives zero. The difference between the areas, therefore, simplifies to 200𝑛𝑥 plus 100𝑥 squared
plus 60𝑥. These three terms have a common factor of 20𝑥. So we can factorize by it, giving 20𝑥 multiplied by 10𝑛 plus five 𝑥 plus
three. You could expand this bracket to check the factorization if you want.
Now, remember we’ve been asked to prove that this difference is always a multiple of
40. We can see that it is a multiple of 20 as we have this factor of 20 here. As 40 is equal to 20 multiplied by two, this means we need to show that the remaining
factor which is 𝑥 multiplied by 10𝑛 plus five 𝑥 plus three is a multiple of two
or an even number.
There are two cases to consider depending on whether 𝑥 is even or odd. If 𝑥 is even first of all, then we can write it as two 𝑦 for some number 𝑦. This means that the expression 20𝑥 multiplied by 10𝑛 plus five 𝑥 plus three can be
written as 20 multiplied by two 𝑦 multiplied by 10𝑛 plus five 𝑥 plus three, which
gives 40𝑦 10𝑛 plus five 𝑥 plus three. This whole expression is then a multiple of 40 due to this factor of 40 here.
If 𝑥 is odd, let’s consider the terms inside the bracket. 10𝑛 will be even regardless of whether 𝑛 is even or odd as 10 is even. Five 𝑥 will be odd as an odd multiplied by an odd gives an odd. And three is odd. Inside the bracket, we, therefore, have an even number plus an odd number plus an odd
An odd number plus another odd number gives an even number. And two even numbers added together also gives an even number. So if 𝑥 is odd, the whole bracket will be even. We can, therefore, call this bracket two 𝑧 for some number 𝑧. The full expression for the difference will, therefore, be 20𝑥 multiplied by two 𝑧,
which is 40𝑥𝑧. This is always a multiple of 40 due to the presence of this factor of 40 here.
Therefore, the difference between any two squares in the sequence is always a
multiple of 40. And we’ve proven that Andrew is correct.
Remember that all of the stages of algebraic working out are essential in your proof
here. I’ve had to delete them in order to make space on the screen, but you must include
every stage in your answer.