Answer:
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.
Answer:
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
