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

Explanation

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.