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.

for mathswatch

Step-by-step explanation:

(2n+2)²=4n²+8n+4

4n²+8n+4-4n²=8n+4

4(2n+1)

4n²+16-4n²+4

(2n+6)²=4n²+16-4n²-8

4

step-by-step explanation:

answer: howdy

step-by-step explanation: