Asked by Harry
The question is: Prove algebraically that the difference between the squares of any two
consecutive even numbers is always a multiple of 4.
My working was: Let the numbers be n and n + 2
(n+2)^2 - (n)^2
=n^2 + 4n + 4 - n^2
=4n + 4
=4(n + 1)
But the answers say the numbers should be 2n + 2 and 2n, why is this or is mine right too?
Thanks
consecutive even numbers is always a multiple of 4.
My working was: Let the numbers be n and n + 2
(n+2)^2 - (n)^2
=n^2 + 4n + 4 - n^2
=4n + 4
=4(n + 1)
But the answers say the numbers should be 2n + 2 and 2n, why is this or is mine right too?
Thanks
Answers
Answered by
Steve
any even number is a multiple of 2.
When you pick any old n, it might not be even.
You have proven that any two numbers which differ by 2 have their squares differing by a multiple of 4.
So, technically you have not done what was asked, but have proved an even stronger result.
Using 2n, the proof is even easier:
(2n+2)^2 - (2n)^2
= 4(n+1)^2 - 4n^2
= 4((n+1)^2 - n^2)
QED
When you pick any old n, it might not be even.
You have proven that any two numbers which differ by 2 have their squares differing by a multiple of 4.
So, technically you have not done what was asked, but have proved an even stronger result.
Using 2n, the proof is even easier:
(2n+2)^2 - (2n)^2
= 4(n+1)^2 - 4n^2
= 4((n+1)^2 - n^2)
QED
Answered by
Harry
I understand thanks!
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