Prove the following by using the principle of mathematical induction for all $n \in N$:
$1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \ldots + n(n+1)(n+2) = \frac{n(n+1)(n+2)(n+3)}{4}$

Let the given statement be $P(n)$,i.e.,
$P(n): 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \ldots + n(n+1)(n+2) = \frac{n(n+1)(n+2)(n+3)}{4}$
For $n=1$,we have
$P(1): 1 \cdot 2 \cdot 3 = 6 = \frac{1(1+1)(1+2)(1+3)}{4} = \frac{1 \cdot 2 \cdot 3 \cdot 4}{4} = 6$,which is true.
Let $P(k)$ be true for some positive integer $k$,i.e.,
$1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \ldots + k(k+1)(k+2) = \frac{k(k+1)(k+2)(k+3)}{4}$ ........$(i)$
We shall now prove that $P(k+1)$ is true.
Consider
$1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \ldots + k(k+1)(k+2) + (k+1)(k+2)(k+3)$
$= \{1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \ldots + k(k+1)(k+2)\} + (k+1)(k+2)(k+3)$
$= \frac{k(k+1)(k+2)(k+3)}{4} + (k+1)(k+2)(k+3)$ [Using $(i)$]
$= (k+1)(k+2)(k+3) \left( \frac{k}{4} + 1 \right)$
$= \frac{(k+1)(k+2)(k+3)(k+4)}{4}$
$= \frac{(k+1)((k+1)+1)((k+1)+2)((k+1)+3)}{4}$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all natural numbers $n \in N$.