How do you show algebraically that the sum of any three consecutive even numbers is always a multiple of 6?

The exact question states:


';The nth even number is 2n. The next even number after 2n is 2n + 2.





Show algebraically that the sum of any three consecutive even numbers is always a multiple of 6.';





Show all working and please explain each step!How do you show algebraically that the sum of any three consecutive even numbers is always a multiple of 6?
The three consecutive even numbers are:


2n


2n+2 (Adding 2 to previous number, i.e. 2n)


2n+4 (Adding 2 to previous number, i.e. 2n + 2)





Adding all three,


Sum = 2n + 2n + 2 + 2n + 4


Rearranging equation,


Sum = (2n + 2n + 2n) + (2+4)


Sum = 6n + 6


Taking 6 common from both terms,


Sum = 6 (n +1)


This shows that the sum of 3 consecutive even numbers will always be a multiple of 6.How do you show algebraically that the sum of any three consecutive even numbers is always a multiple of 6?
Could it be something like;





n + (n+1) + (n+2) = 6





?
n + (n+2) + (n+4) = 6x