$\begin{align}  & when\text{ }n=1 \\ & n(n+1)(n+2)=1(1+1)(1+2)=6 \\ & \text{which is divisble by 6} \\ & \text{when }n=2 \\ & n(n+1)(n+2)=2(2+1)(2+2)=24 \\ & \text{which is divisible by 6} \\ & \text{Let assume that }n(n+1)(n+2)\text{ is divisible by 6, then} \\ & \text{we need to show that when }n=n+1,\text{ the relation still holds} \\ & \text{when }n=k \\ & k(k+1)(k+2)=6p,\text{ where }p\text{ is any integer} \\ & {{k}^{3}}+3{{k}^{2}}+2k=6p \\ & {{k}^{3}}=6p-3{{k}^{2}}-2k \\ & \text{when }n=k+1 \\ & (k+1)(k+2)(k+3)=(k+1)({{k}^{2}}+5k+6) \\ & \text{                             }={{k}^{3}}+6{{k}^{2}}+11k+6 \\ & \text{But }{{k}^{3}}=6p-3{{k}^{2}}-2k \\ & (k+1)(k+2)(k+3)=6p-3{{k}^{2}}-2k+6{{k}^{2}}+11k+6 \\ & (k+1)(k+2)(k+3)=6p+3{{k}^{2}}+9k+6 \\ & (k+1)(k+2)(k+3)=\frac{1}{6}\left( p+\frac{{{k}^{2}}}{2}+\frac{3k}{2}+1 \right) \\ & \text{which when divisible by 6 yields integer} \\ & \left( p+\frac{{{k}^{2}}}{2}+\frac{3k}{2}+1 \right)\text{ }note:\frac{{{k}^{2}}}{2}+\frac{3k}{2}\text{ is an integer value } \\\end{align}$