Suppose f is a function satisfying $\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{y}\right)$ for  all $\mathrm{x},\mathrm{y}\in \mathrm{\mathbb{N}}$ and $\mathrm{f}\left(1\right)=\frac{1}{5}$. If $\sum _{n=1}^m\frac{f\left(n\right)}{n(n+1)(n+2)}=\frac{1}{12}$ then $\mathrm{m}$ is equal to ______.

Given,

$f\left(x+y\right)=f\left(x\right)+f\left(y\right) \& f\left(1\right)=\frac{1}{5}$

Now solving the given functional equation by taking $x=1 \& y=1$ we get,

$f\left(1+1\right)=f\left(1\right)+f\left(1\right)$

$\Rightarrow f\left(2\right)=\frac{2}{5} \left\{\text{as }f\left(1\right)=\frac{1}{5}\right\}$

Now taking $x=2 \& y=1$ we get,

$f\left(3\right)=\frac{2}{5}+\frac{1}{5}\Rightarrow f\left(3\right)=\frac{3}{5}$

Hence, $f\left(x\right)=\frac{x}{5}$

Now solving  $\sum _{n=1}^m\frac{f\left(n\right)}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{12}$

$\Rightarrow \sum _{n=1}^m\frac{{\displaystyle \frac{n}{5}}}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{12}$

$\Rightarrow \frac{1}{5}\sum _{n=1}^m\frac{{\displaystyle 1}}{\left(n+1\right)\left(n+2\right)}=\frac{1}{12}$

$\Rightarrow \frac{1}{5}\sum _{n=1}^m\frac{{\displaystyle 1}}{\left(n+1\right)}-\frac{{\displaystyle 1}}{\left(n+2\right)}=\frac{1}{12}$

$\Rightarrow \frac{1}{5}\left[\frac{{\displaystyle 1}}{2}-\frac{{\displaystyle 1}}{3}+\frac{{\displaystyle 1}}{3}-\frac{{\displaystyle 1}}{4}+\frac{{\displaystyle 1}}{4}.....-\frac{{\displaystyle 1}}{m+2}\right]=\frac{1}{12}$

$\Rightarrow \frac{1}{5}\left[\frac{{\displaystyle 1}}{2}-\frac{{\displaystyle 1}}{m+2}\right]=\frac{1}{12}$

$\Rightarrow \left[\frac{{\displaystyle m}}{2\left(m+2\right)}\right]=\frac{5}{12}$

$\Rightarrow 12m=10\left(m+2\right)$

$\Rightarrow m=10$