Using the principle of mathematical induction prove that for all n belongs to N: n(n+1)(n+2)(n+3) is a multiple of 24

The given statement is:

P(n): n(n+1)(n+2)(n+3) is a multiple of 24

 

P(1) = 1(1+1)(1+2)(1+3) = 1.2.3.4 = 24, a multiple of 24

 

Let P(m) be true.

That is, m(m+1)(m+2)(m+3) is a multiple of 24

m(m+1)(m+2)(m+3) = 24k

 

Now, P(m+1) = (m+1)(m+2)(m+3)(m+4) = m (m+1)(m +2)(m +3) + 4 (m+1)(m +2)(m +3)

                   = 24 k + 4 (m+1)(m +2)(m +3)       ... (1)

 

Now, if (m+1)(m +2)(m +3) is a multiple of 6, then P(m+1) will be a multiple of 24.

 

So, by principle of mathematical induction, we will prove that (m+1)(m +2)(m +3) is a multiple of 6.

 

Let Q(m): (m+1)(m +2)(m +3) is a multiple of 6

Q(1) = 2.3.4 = 24, a multiple of 6

 

Let Q(t) be true.

(t+1)(t +2)(t +3) = 6p

Now, we will prove that Q(t+1) is true.

(t+2)(t +3)(t +4) = (t+2)(t +3)(t +3+1) = (t+2)(t +3)(t +1) + (t+2)(t +3)(3)

                                                              = 6p + 3 (a multiple of 2)

                                                              = 6p + a multiple of 6

                                                               = 6q

Thus, Q(t+1) is true.

Therefore, by principle of mathematical induction, (m+1)(m +2)(m +3) is a multiple of 6.

From (1), P(m+1) = 24 k + 4 (m+1)(m +2)(m +3)

                          = 24 + 4 (a multiple of 6)

                          = a multiple of 24

 

Thus, P(m+1) is true.

 

Hence, by principle of mathematical induction, n(n+1)(n+2)(n+3) is a multiple of 24.