Prove algebraically that the difference between the squares of any two consecutive odd numbers is always a multiple of 8. - Edexcel - GCSE Maths - Question 16 - 2018 - Paper 3

Next, we will calculate the squares of both the first and the second odd numbers:

1. The square of the first odd number is:

   $(2k + 1)^2 = 4k^2 + 4k + 1$

2. The square of the second odd number is:

   $(2k + 3)^2 = 4k^2 + 12k + 9$

We now compute the difference between the squares of the two consecutive odd numbers:

$\text{Difference} = (2k + 3)^2 - (2k + 1)^2$

$= (4k^2 + 12k + 9) - (4k^2 + 4k + 1)$

Simplifying this gives us:

$= 12k + 9 - 4k - 1 = 8k + 8$

Which can be factored as:

$= 8(k + 1)$

Since $k$ is an integer, $k + 1$ is also an integer. Thus, the difference between the squares of any two consecutive odd numbers can be expressed as a multiple of 8.

Hence, we have proven algebraically that the difference between the squares of any two consecutive odd numbers is always a multiple of 8.