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

(N/A) Let the given statement be $P(n)$,i.e.,
$P(n): 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \ldots + n(n+1) = \frac{n(n+1)(n+2)}{3}$
For $n=1$,we have:
$P(1): 1 \cdot 2 = 2 = \frac{1(1+1)(1+2)}{3} = \frac{1 \cdot 2 \cdot 3}{3} = 2$,which is true.
Assume $P(k)$ is true for some positive integer $k$,i.e.,
$1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \ldots + k(k+1) = \frac{k(k+1)(k+2)}{3}$ $(i)$
We shall now prove that $P(k+1)$ is true.
Consider the sum up to $(k+1)$ terms:
$1 \cdot 2 + 2 \cdot 3 + \ldots + k(k+1) + (k+1)(k+2)$
$= \frac{k(k+1)(k+2)}{3} + (k+1)(k+2)$ [Using $(i)$]
$= (k+1)(k+2) \left( \frac{k}{3} + 1 \right)$
$= (k+1)(k+2) \left( \frac{k+3}{3} \right)$
$= \frac{(k+1)(k+2)(k+3)}{3}$
This is the form of $P(k+1)$.
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 $n \in N$.