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If \(n\) is an integer (a whole number), then the expression \(2n\) represents an even number, because even numbers are the multiples of 2. The expressions \(2n - 1\) and \(2n + 1\) can represent odd numbers, as an odd number is one less, or one more than an even number.

Try some examples: \(2 + 2 = 4\), \(4 + 12 = 16\), \(1002 + 3024 = 4026\).

Write the first of the two even numbers as \(2n\), where \(n\) is an integer. The second even number can be written as \(2m\), where \(m\) is also an integer. We cannot use the same letter for both expressions, otherwise they would represent the same even number.

Adding the two even numbers gives \(2n + 2m\).

This can be factorised to give \(2n + 2m = 2(n + m)\)

Since \(n\) and \(m\) are both integers, then \(n + m\) will also be an integer, so the expression \(2(n + m)\) represents an even number.

This shows that whenever two even numbers are added, the total is also an even number because \(2n + 2m = 2(n + m)\).

Product is the value obtained by multiplying. Try some examples: \(3 \times 5 = 15\), \(7 \times 9 = 63, 11 \times 13 = 143\).

For these examples, odd x odd = odd. To prove that it is true for all odd numbers, we can write two odd numbers as \(2n + 1\) and \(2m + 1\), where \(n\) and \(m\) are integers.

\((2n + 1)(2m + 1) = 4nm + 2n + 2m + 1\)

\(2(2nm + n + m) + 1\)

Since \(n\) and \(m\) are integers, the expression inside the bracket, \(2nm + n + m\), will also be an integer. This means that the expression \(2(2nm + n + m) + 1 \) represents an odd number, as it is 2 multiplied by an integer plus 1.

So the product of two odd numbers is always odd because \((2n + 1)(2m + 1) = 2(2nm + n + m) + 1\).

Consecutive integers are whole numbers that follow each other without gaps. For example, 15, 16, 17 are consecutive integers. If \(n\) is an integer, then the consecutive integers starting at \(n\) are \(n\), \(n + 1\), \(n + 2\), \(n + 3\) and so on.

Try some examples: \(1 + 2 + 3 = 6\), \(5 + 6 + 7 = 18\), \(102 + 103 + 104 = 309\). This shows the sum of three consecutive integers is a multiple of 3 in these cases, but to prove it is true in all cases, we can use algebra.

We can write three consecutive integers as \(n\), \(n + 1\) and \(n + 2\), so the sum of three consecutive integers can be written as: \(n + (n + 1) + (n + 2)\)

\(n + (n + 1) + (n + 2) = 3n + 3\)

This can be factorised to give \(3n + 3 = 3(n + 1)\) which will be a multiple of 3 for all integer values of \(n\).