Prove the statement by the Principle of Mathematical Induction: $n(n^{2}+5)$ is divisible by $6$,for each natural number $n$.

(A) Let $P(n): n(n^{2}+5)$ is divisible by $6$ for all $n \in N$.
Step $1$: For $n=1$,$P(1) = 1(1^{2}+5) = 6$,which is divisible by $6$. Thus,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N$,i.e.,$k(k^{2}+5) = 6m$ for some integer $m$. $(i)$
Step $3$: We need to prove $P(k+1)$ is true,i.e.,$(k+1)((k+1)^{2}+5)$ is divisible by $6$.
Consider $(k+1)((k+1)^{2}+5) = (k+1)(k^{2}+2k+1+5) = (k+1)(k^{2}+2k+6)$
$= k(k^{2}+5) + k(2k+1) + 1(k^{2}+2k+6)$
$= k(k^{2}+5) + 2k^{2} + k + k^{2} + 2k + 6$
$= k(k^{2}+5) + 3k^{2} + 3k + 6$
$= 6m + 3k(k+1) + 6$
Since $k(k+1)$ is the product of two consecutive integers,it is always even,i.e.,$k(k+1) = 2p$ for some integer $p$.
$= 6m + 3(2p) + 6 = 6m + 6p + 6 = 6(m+p+1)$.
This is clearly divisible by $6$. Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the Principle of Mathematical Induction,$P(n)$ is true for all $n \in N$.