# Prove the following by using the principle of mathematical induction for all $n\in\mathbb{N}$ :Q : 11        $\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$

Let the given statement be p(n) i.e.
$p(n):\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$
For n = 1  we have
$p(1):\frac{1}{1.2.3}=\frac{1}{6}=\frac{1(1+3)}{4(1+1)(1+2)}=\frac{4}{4.2.3}=\frac{1}{6}$   ,   which is true

For  n = k  we have

$p(k):\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{k(k+1)(k+2)}=\frac{k(k+3)}{4(k+1)(k+2)} \ \ \ \ -(i)$   ,        Let's assume that this statement is true

Now,
For  n = k + 1  we have
$p(k+1):\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{(k+1)(k+2)(k+3)}$                                                                                                                                                                                                                  $=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{k(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)}$

$=\frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)} \ \ \ \ \ \ (using \ (i))$
$=\frac{1}{(k+1)(k+2)}\left ( \frac{k(k+3)}{4}+ \frac{1}{k+3} \right )$
$=\frac{1}{(k+1)(k+2)}\left ( \frac{k(k+3)^2+4}{4(k+3)} \right )$
$=\frac{1}{(k+1)(k+2)}\left ( \frac{k(k^2+9+6k)+4}{4(k+3)} \right )$
$=\frac{1}{(k+1)(k+2)}\left ( \frac{k^3+9k+6k^2+4}{4(k+3)} \right )$
$=\frac{1}{(k+1)(k+2)}\left ( \frac{k^3+2k^2+k+8k+4k^2+4}{4(k+3)} \right )$
$=\frac{1}{(k+1)(k+2)}\left ( \frac{k(k^2+2k+1)+4(k^2+2k+1)}{4(k+3)} \right )$
$=\frac{1}{(k+1)(k+2)}\left ( \frac{(k+1)^2(k+4)}{4(k+3)} \right )$
$= \frac{(k+1)((k+1)+3)}{4(k+1+1)(k+1+2)}$

Thus,  p(k+1)  is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n)  is true for all natural numbers n

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