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

Verify the base case, then show that $P(k+1)-P(k)$ (as a difference of the two products) is always a multiple of 3, so if $P(k)$ is a multiple of 3, so is $P(k+1)$.

Let $P(n): n(n+1)(n+5)$ is a multiple of $3$.

Base case ($n=1$):

$P(1): 1\times2\times6 = 12 = 3\times4$

This is a multiple of $3$, so $P(1)$ is true.

Inductive step: Assume $P(k)$ is true, i.e. $k(k+1)(k+5) = 3m$ for some integer $m$.

We must show $P(k+1)$ is true, i.e. $(k+1)(k+2)(k+6)$ is a multiple of $3$.

Consider the difference:

$(k+1)(k+2)(k+6) - k(k+1)(k+5) = (k+1)\Big[(k+2)(k+6) - k(k+5)\Big]$

Expand the bracket:

$(k+2)(k+6) = k^2+8k+12,\qquad k(k+5) = k^2+5k$

$(k+2)(k+6)-k(k+5) = 3k+12$

So:

$(k+1)(k+2)(k+6) - k(k+1)(k+5) = (k+1)(3k+12) = 3(k+1)(k+4)$

Therefore: …