(1) If n is divisible by 3, then n(n+ 1)(n+2)(n+3) can be divisible by 3;
(2) If n is divided by 3 and the remainder is 1, let n=3r+ 1(r is a natural number), then n+2=3(r+ 1), and then derive n (n+1) (n+2).
(3) If n is divisible by 3, it can be assumed that n=3r+2(r is a natural number), then n+ 1=3(r+ 1), and it is deduced that n(n+ 1)(n+2)(n+3) can be divisible by 3.
In a word, n(n+ 1)(n+2)(n+3) is divisible by 3.
Obviously, there must be two even numbers in four consecutive natural numbers, and their products can be divisible by 4, so n(n+ 1)(n+2)(n+3) can also be divisible by 4.
Because 3 and 4 are coprime, n(n+ 1)(n+2)(n+3) can be divisible by 12.
This question doesn't need mathematical induction ~
If you insist on mathematical induction?
(1) When n= 1, n (n+1) (n+2) (n+3) = 12, which is divisible by12;
(2) Suppose that when n=k, n (n+1) (n+2) (n+3) = k (k+1) (k+2) (k+3) is divisible by 12,
Then when n=k+ 1, n (n+1) (n+2) (n+3) = (k+1) (k+2) (k+3) (k+4) = k (k+)
As can be seen from the previous proof, one of the three consecutive natural numbers must be a multiple of 3, so 4(k+ 1)(k+2)(k+3) can be divisible by 12, assuming that k (k+1) (k+2) (k+.
Therefore, for any n, n(n+ 1)(n+2)(n+3) can be divisible by 12.