Need help finding the series in order to prove by induction I need to prove that for all natural numbers $n$, $\frac{n(n+1)(n+2)}{6}$ is a natural number. The problem is that I can't seem to figure out what the series is, the LHS. Any help would be appreciated.
Hint: The numerator is a product of $3$ consecutive numbers, so we can say something about the prime factors of at least two of the numbers (think of the prime factors for $6$).
You seem to be completely misunderstanding the question, thinking, apparently, that the fraction you are given is the sum of some series.  Whether it is or not is irrelevant to the question.  You have n(n+1)(n+ 2).

Now, either n is even or n is odd.  If n is even, it is divisible by 2- it can be written a n= 2k and so is n(n+ 1)(n+ 2)= 2k(2k+1)(2k+ 2) which is divisible by 2.  If n is odd, it can be written n= 2k+ 1. n(n+1)(n+2)= (2k+1)(2k+1+1)(2k+1+2)= (2k+1)(2k+2)(2k+3)= 2(2k+1)(k+1)(2k+3).  So what ever n is, n(n+ 1)(n+2) is divisible by 2.

Now, "3" is just a little harder.  Either n is divisible by 3 or it has remainder 1 when divided by 3 or it has remainder 2 when divided by 3.
If n is divisible by 3, n= 3k so n(n+1)(n+2)= 3k(3k+1)(3k+2) and is divisible by 3.
If n has remainder 1 when divided by 3, it is of the form n= 3k+1 so n(n+1)(n+2)= (3k+1)(3k+1+1)(3k+1+2)= (3k+1)(3k+2)(3k+3)= 3(3k+1)(3k+2)(k+1) and is divisible by 3.
If n has remainder 2 when divided by 3, it is of the form n= 3k+2 so n(n+1)(n+2)= (3k+2)(3k+2+1)(3k+2+2)= (3k+2)(3k+3)(3k+4)= 3(3k+2)(k+1)(3k+4) and is divisible by 3.

So for any integer n, n(n+1)(n+3) is divisible by both 2 and 3 and so divisible by 6.