Stupid factorial question.

Question

If i have $(3(n+1))!$ can I say:

$(3(n+1))! = 3(n+1) \times (3n)!$

but if I expand by first multiplying the expression in the parenthesis;

$(3(n+1))! = (3n+3) \times (3n+2) \times (3n+1) \times (3n)!$

Which one is it? I think the second one.

Answer 1

$$(3(n+1))! = 3(n+1) \times (3(n+1)-1)! = 3(n+1) \times (3n +2)!$$ Repeating the above, you get the second answer.

Answer 2

I suppose that your confusion become from the recursive definition of factorial as: $$ 0!=1 \, \quad (s(n))!=s(n) \cdot n! $$ where $s(n)=n+1$ is the successor of $n$.

Note that $3(n+1) $ is not the successor of $3n$ but of $3n+2$.

So you have the correct result: $$ (3(n+1))! =3(n+1)\cdot (3n+2)! $$

The successor of $3n$ is $3n+1$.