# Prove the following by using the principle of mathematical induction for all $n \in N$  $\large\frac{1}{1.2.3}$$+\large\frac{1}{2.3.4}$$+\large\frac{1}{3.4.5}$$+...+\large\frac{1}{n(n+1)(n+2)}$$=\large\frac{n(n+3)}{4(n+1)(n+2)}$

Let the given statement be $P(n)$, i.e.,
$P(n) : \large\frac{1}{1.2.3}$$+\large\frac{1}{2.3.4}$$+\large\frac{1}{3.4.5}$$+...+\large\frac{1}{n(n+1)(n+2)}$$=\large\frac{n(n+3)}{4(n+1)(n+2)}$
For $n = 1$, we have
$P(n) : \large\frac{1}{1.2.3}$$= \large\frac{1.(1+3)}{4(1+1)(1+2)}$$=\large\frac{1.4}{4.2.3}$$=\large\frac{1}{1.2.3}, which is true. Let P(k) be true for some positive integer k, i.e., \large\frac{1}{1.2.3}$$+\large\frac{1}{2.3.4}$$+\large\frac{1}{3.4.5}$$+...+\large\frac{1}{k(k+1)(k+2)}$$=\large\frac{k(k+3)}{4(k+1)(k+2)}----------(i) We shall now prove that P(k+1) is true. Consider \bigg[ \large\frac{1}{1.2.3}$$+\large\frac{1}{2.3.4}$$+\large\frac{1}{3.4.5}$$+...+\large\frac{1}{k(k+1)(k+2)} \bigg]$$+ \large\frac{1}{(k+1)(k+2)(k+3)} = \large\frac{k(k+3)}{4(k+1)(k+2)}$$+\large\frac{1}{(k+1)(k+2)(k+3)}$$\qquad [ Using (i) ] = \large\frac{1}{(k+1)(k+2)}$$ \bigg\{ \large\frac{k(k+3)}{4}$$+ \large\frac{1}{(k+3)} \bigg\} = \large\frac{1}{(k+1)(k+2)}$$ \bigg\{ \large\frac{k(k+3)^2+4}{4(k+3)}\bigg\}$
$= \large\frac{1}{(k+1)(k+2)}$$\bigg\{ \large\frac{k(k^2+6k+9)+4}{4(k+3)}\bigg\} = \large\frac{1}{(k+1)(k+2)}$$ \bigg\{ \large\frac{k^3+6k^2+9k+4}{4(k+3)}\bigg\}$
$= \large\frac{1}{(k+1)(k+2)}$$\bigg\{ \large\frac{k^3+2k^2+k+4k^2+8k+4}{4(k+3)}\bigg\} = \large\frac{1}{(k+1)(k+2)}$$ \bigg\{ \large\frac{k(k^2+2k+1)+4(k^2+2k+1)}{4(k+3)}\bigg\}$