Or search by topic
Extension:
Can you prove that when you have $n$ consecutive even numbers, the difference between the product of the first and last numbers, and the product of the second and penultimate numbers, will be $4n-8$?
Sunhari from British School Muscat sent in this proof:
Let the n numbers be: $2(a), 2(a+1),…2(a+n-2), 2(a+n-1)$
First product: $2a(2a+2n-2) = 4a^2 + 4an - 4a$
Second product:
$\begin{split}(2a+2)(2a+2n-4)& = 4a^2 + 4an - 8a + 4a + 4n - 8\\&= 4a^2 + 4an - 4a + 4n - 8 \end{split}$
Difference: $4a^2 + 4an - 4a + 4n - 8 - 4a^2 - 4an + 4a= 4n-8$