Prove that the difference between squares of consecutive even numbers is always a multiple of 4.

Note: Let n stand for any integer in your working.

Proved: the difference between squares of consecutive even numbers is always a multiple of 4.

Step-by-step explanation:

Let's consider two consecutive even numbers.

If n = a positive integer

Then the 1st even number would be = 2n

Let the next consecutive even number = 2 (n+1)

The consecutive even numbers would be:

2n, 2 (n+1)

The square of each consecutive even number would be:

(2n) ² = 2n * 2n = 4n²

[2 (n+1) ]² = (2n+2) ² = (2n+2) (2n+2)

Expanding the brackets

= 4n² + 4n + 4n+4

[2 (n+1) ]² = 4n²+8n+4

To determine if the difference between squares of consecutive even numbers is always a multiple of 4, we would assign values for n and find the difference of the squares of each consecutive even numbers.

Let n = 1, 2, 3

For n = 1

4n² = 4 (1) ² = 4*1 = 4

4n²+8n+4 = 4 (1) ² + 8 (1) + 4

= 4+8+4 = 16

Difference between the squares of consecutive even numbers = 16-4 = 12

For n = 2

4n² = 4 (2) ² = 4*4 = 16

4n²+8n+4 = 4 (2) ² + 8 (2) + 4

= 16+16+4 = 36

Difference between the squares of consecutive even numbers = 36-16 = 20

For n = 3

4n² = 4 (3) ² = 4*9 = 36

4n²+8n+4 = 4 (3) ² + 8 (3) + 4

= 36+24+4 = 64

Difference between the squares of consecutive even numbers = 64-36 = 28

12 = 4*3

20 = 4*5

28 = 4*7

From the results above, we can see that 12, 20, and 28 are multiples of 4.

Therefore, the difference between squares of consecutive even numbers is always a multiple of 4.