Prove that the following is true for all $n \in N$ using the principle of mathematical induction:
$n(n+1)(n+5)$ is a multiple of $3.$

(N/A) Let the given statement be $P(n)$,i.e.,
$P(n): n(n+1)(n+5)$ is a multiple of $3$.
Step $1$: For $n=1$,$1(1+1)(1+5) = 1(2)(6) = 12$,which is a multiple of $3$. Thus,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some positive integer $k$,i.e.,$k(k+1)(k+5) = 3m$ for some $m \in N$.
Step $3$: We need to show $P(k+1)$ is true,i.e.,$(k+1)(k+2)(k+6)$ is a multiple of $3$.
Consider $(k+1)(k+2)(k+6) = (k+1)(k+2)(k+5+1)$
$= (k+1)(k+2)(k+5) + (k+1)(k+2)$
$= (k+1)(k+5)(k+2) + (k^2+3k+2)$
Since $k(k+1)(k+5) = 3m$,we have $(k+1)(k+5) = \frac{3m}{k}$. This approach is complex,so let's expand differently:
$(k+1)(k+2)(k+6) = (k^2+3k+2)(k+6) = k^3 + 6k^2 + 3k^2 + 18k + 2k + 12$
$= k^3 + 9k^2 + 20k + 12$
$= (k^3 + 6k^2 + 5k) + (3k^2 + 15k + 12)$
$= k(k^2 + 6k + 5) + 3(k^2 + 5k + 4)$
$= k(k+1)(k+5) + 3(k^2 + 5k + 4)$
Since $k(k+1)(k+5)$ is a multiple of $3$ (by assumption) and $3(k^2+5k+4)$ is clearly a multiple of $3$,their sum is also a multiple of $3$.
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,$P(n)$ is true for all $n \in N$.