amaths - mathematical induction

First of all, the solution given by laiyatwah is wrong, a mistake is made in the Induction step:

>> Consider p(k+1),
>> L.H.S. = (k+1)(k+1+1)(k+1+2)(k+1+3)
>> = (k+1)(k+2)(k+3)(k+4)
>> = 24Q(k+4) [by assumption p(k) is true) <- this step is wrong, check it!
>> = 24(P) where P = Q(k+4) is a integer.
>> = R.H.S.


Then, I shall give the full solution:

The case n = 1:
L. H. S = 1(2)(3)(4) = 24 which is divisible by 24.

The induction step:
Suppose that k(k+1)(k+2)(k+3) is divisible by 24, that is k(k+1)(k+2)(k+3) = 24M, where M is some positive integer. Then we consider the case when n = k+1:
(k+1)(k+2)(k+3)(k+4)
= k(k+1)(k+2)(k+3) + 4(k+1)(k+2)(k+3) << expanding the last bracket
= 24M + 4(k+1)(k+2)(k+3) << by induction hypothesis

Since (k+1)(k+2)(k+3) is the product of 3 consecutive positive integers. Within k+1, k+2, k+3, they must contain a multiple of 2 and a multiple of 3. Therefore, (k+1)(k+2)(k+3) must be divisible by 6. That is (k+1)(k+2)(k+3) = 6N, where N is some other positive integer. So continuing:

= 24M + 4(6N)
= 24M + 24N
= 24 (M+N)
which is divisible by 24 since M+N is also a positive integer.

By the principle of Mathematical Induction, we know that n(n+1)(n+2)(n+3) is divisible by 24 for all positive integers n.

The question is a typical M.I. question.
The solution is as follows:

Let p(n): n(n+1)(n+2)(n+3) = 24M where M is an integer.

when n=1.
L.H.S. = 1(1+1)(1+2)(1+3) = 24 = 24(1) = R.H.S.
so p(1) is true.

Assume p(k) is true, i.e. k(k+1)(k+2)(k+3) = 24Q where Q is an integer.

Consider p(k+1),
L.H.S. = (k+1)(k+1+1)(k+1+2)(k+1+3)
= (k+1)(k+2)(k+3)(k+4)
= 24Q(k+4) [by assumption p(k) is true)
= 24(P) where P = Q(k+4) is a integer.
= R.H.S.

So p(k+1) is true, if p(k) is true.

By M.I., p(n) is true for all positive integer n.