Using principle of mathematical induction, prove that for all `n in N, n(n+1)(n+5)` is a multiple of 3.

Using principle of mathematical induction, prove that for all `n in N,

1.	Using principle of mathematical induction, prove that for all `n in N, n(n+1)(n+5)` is a multiple of 3.
Answer» We have: ` P(n) : n(n+1)(n+5) = 3x, x in N` For `n = 1, 1 xx (1+1)(1+5) = 12`, which is a multiple of 3. So, `P(1)` is true. Let `P(n)` be true for some `n = k`. i.e., `k(k+1) (k+5) = 3m`, where `m in N"….."(1)` We have to prove that `P(k+1)` is true. i.e., `(k+1)(k+2)(k+6)=3p, p in N` Now, `(k+1)(k+2)(k+6)-k(k+1)(k+5)` `= (k+1)[(k+2)(k+6)-k(k+5)]` `= (k+1)[k^(2)+8k+12-k^(2)-5k]` ltbr gt `= (k+1)(3k+12)` `= 3q`, where `q in N` Thus, `(k+1)(k+2)(k+6)` `= k (k+1)(k+5)+3q` `= 3m + 3q`, which is multiple of 3. Thus, `P(k+1)` is true whenever `P(k)` is true. Hence, by the principle of mathematical induction, statement `P(n)` is true for all natural numbers.