Prove the following by using the principle of mathematical induction for all $n \in N$ :
$\frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{3 \cdot 4 \cdot 5} + \ldots + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}$

(N/A) Let the given statement be $P(n),$ i.e.,
$P(n): \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{3 \cdot 4 \cdot 5} + \ldots + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}$
For $n=1,$ we have
$P(1): \frac{1}{1 \cdot 2 \cdot 3} = \frac{1(1+3)}{4(1+1)(1+2)} = \frac{4}{4 \cdot 2 \cdot 3} = \frac{1}{1 \cdot 2 \cdot 3},$ which is true.
Assume $P(k)$ is true for some positive integer $k$,i.e.,
$\frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \ldots + \frac{1}{k(k+1)(k+2)} = \frac{k(k+3)}{4(k+1)(k+2)}$ ........$(i)$
We shall now prove that $P(k+1)$ is true.
Consider the sum up to $(k+1)$ terms:
$\left[\frac{1}{1 \cdot 2 \cdot 3} + \ldots + \frac{1}{k(k+1)(k+2)}\right] + \frac{1}{(k+1)(k+2)(k+3)}$
$= \frac{k(k+3)}{4(k+1)(k+2)} + \frac{1}{(k+1)(k+2)(k+3)}$ [Using $(i)$]
$= \frac{1}{(k+1)(k+2)} \left\{ \frac{k(k+3)}{4} + \frac{1}{k+3} \right\}$
$= \frac{1}{(k+1)(k+2)} \left\{ \frac{k(k+3)^2 + 4}{4(k+3)} \right\}$
$= \frac{k^3 + 6k^2 + 9k + 4}{4(k+1)(k+2)(k+3)}$
Factoring the numerator $k^3 + 6k^2 + 9k + 4 = (k+1)^2(k+4)$:
$= \frac{(k+1)^2(k+4)}{4(k+1)(k+2)(k+3)} = \frac{(k+1)(k+4)}{4(k+2)(k+3)}$
$= \frac{(k+1)((k+1)+3)}{4((k+1)+1)((k+1)+2)}$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,$P(n)$ is true for all $n \in N.$