$\begin{align}  & (1\times 2\times 3)+(2\times 3\times 4)+\cdot \cdot \cdot +n(n+1)(n+2)\text{=}\frac{n(n+1)(n+2)(n+3)}{4} \\ & when\text{ }n=1 \\ & L.H.S=1\times 2\times 3=6,\text{ } \\ & R.H.S=\frac{1(1+1)(1+2)(1+3)}{4}=\frac{1(2)(3)(4)}{4}=6 \\ & when\text{ }n=2 \\ & L.H.S=(1\,\times 2\times 3)+(2\times 3\times 4)=6+24=30 \\ & R.H.S=\frac{2(2+1)(2+2)(2+3)}{4}=\frac{2(3)(4)(5)}{4}=30 \\ & \text{So the formula is valid for }n=1,2 \\ & \text{Let assume that the first }n=k\text{ term is} \\ & \frac{k(k+1)(k+2)(k+3)}{4}\text{ } \\ & \text{we need toshow that sum of }n=k+1\text{ is } \\ & \frac{(k+1)(k+1+)(k+1+2)(k+1+3)}{4}----(i) \\ & kth\text{ term}=k(k+1)(k+2) \\ & (k+1)th\text{ term }(k+1)(k+2)(k+3) \\ & {{S}_{k}}=(1\times 2\times 3)+(2\times 3\times 4)+\cdot \cdot \cdot +k(k+1)(k+2)=\frac{k(k+1)(k+2)(k+3)}{4} \\ & \text{Sum of }n=k+1\text{ terms is the sum }n=k\text{ term and} \\ & n=k+1\text{ terms} \\ & {{S}_{k+1}}=\underbrace{(1\times 2\times 3)+(2\times 3\times 4)+\cdot \cdot \cdot +k(k+1)(k+2)}_{{{S}_{k}}}+(k+1)(k+2)(k+3) \\ & {{S}_{k+1}}=\frac{k(k+1)(k+2)(k+3)}{4}+(k+1)(k+2)(k+3) \\ & {{S}_{k+1}}=(k+1)(k+2)(k+3)\left[ \frac{k}{4}+1 \right] \\ & {{S}_{k+1}}=\frac{(k+1)(k+2)(k+3)(k+4)}{4} \\ & {{S}_{k+1}}=\frac{(k+1)(k+1+1)(k+1+2)(k+1+3)}{4}---(ii) \\ & \text{The relation holds }n=k\text{ }n=k+1,\text{ it holds for all  values of }n \\\end{align}$