# Lesson video

In progress...

Hi, I'm Miss Davies.

In this lesson, we're going to be proving that an expression will be a multiple of a given number.

How can we represent different types of numbers algebraically? If n is an integer that is more than or equal to zero, we would represent an integer using n.

An even number is a multiple of two, so we would represent this by using two n.

An odd number is one more than an even number, so an odd number is two n add one, and a multiple of three is three n.

We're going to prove that the sum of three consecutive numbers is always a multiple of three.

The consecutive numbers appear one after the other.

For example, three, four, and five are three consecutive numbers.

The number or the integer that appears after n is n add one.

The integer that follows this is n add two.

We are working with the sum of the three consecutive numbers.

This means we are going to add the three numbers together.

This simplifies to give three n add three, or three bracket n add one.

One multiplied by three is three.

Two multiplied by three is six.

Three multiplied by three is nine.

This means that three consecutive numbers, when added together, gives a multiple of three.

We have been asked to prove that this expression is always a multiple of seven for all positive integer values of n.

Let's start by expanding the brackets.

If we're subtracting that second bracket, we're going to rewrite it as subtract n squared add four n subtract four.

This then simplifies to 14 n add 21.

Both the n coefficient and the constant are multiples of seven.

This means that seven can be taken out as a factor of our expression, meaning that our original expression is always a multiple of seven.

Can you spot the mistake with this question? Pause the video to complete your task and resume once you're finished.

The mistakes appear both here and here.

Let's have a look at what it should look like.

We're subtracting n squared add two n add one from that first expansion.

This can be rewritten as subtract n squared subtract two n subtract one.

This then simplifies to give negative 12 n add 24.

We can take 12 out of the factor to show that this is always a multiple of 12 for all positive integer values of n.

Here are some questions for you to try.

Pause the video to complete your task and resume once you're finished.

Remember that you are subtracting all terms that's been formed by the second expansion.

Here are some questions for you to try.

Pause the video to complete your task and resume once you're finished.

Remember, three n add two squared means three n add two multiplied by three n add two, this gives the result of nine n squared add 12 n add four.

Our next example asks us to prove that the difference between the squares of any two terms in the sequence is always a multiple of eight.

The first thing we're going to do with this question is to find the nth term of this sequence.

The nth term is four n add five.

As we need two terms, we're going to use n and p as integers that are greater than zero.

Our first term in the sequence is four n add five.

Our second term is four p add five.

If we square both of these terms, it gives us 16 n squared add 40 n add 25, and 16 p squared add 40 p add 25.

We're subtracting the bracket.

This can be rewritten as subtract 16 p squared subtract 40 p subtract 20.

This expression simplifies to give 16 n squared add 40 n subtract 16 p squared subtract 40 p.

All of the coefficients in this expression are multiples of eight, therefore eight can be taken out as a factor, showing that the difference between the squares of any two terms in the sequence is always a multiple of eight.

Here is a question for you to try.

Pause the video to complete your task and resume once you're finished.

The nth term of the sequence is eight n add 12.

The two expressions that you are squaring should use different variables, as we could be using any two terms from the sequence.

Here is a question for you to try.

Pause the video to complete your task and resume once you're finished.