$\frac{1}{2.3}+\frac{1}{3.4}+\mathrm{.......}+\frac{1}{\left(n+1\right)\left(n+2\right)}=$

$Given : \frac{1}{2.3}+\frac{1}{3.4}+\mathrm{.......}+\frac{1}{\left(n+1\right)\left(n+2\right)}$

Put $n=1, \frac{1}{\left(1+1\right)\left(1+2\right)}=\frac{1}{2.3}=\frac{1}{6}$

$$\begin{gathered} Put n=1 , in option \left(A\right) \\ \frac{1}{1+1}=\frac{1}{2} \end{gathered}$$

$$\begin{gathered} Put n=1 , in option \left(B\right) \\ \frac{1}{4\left(1+4\right)}=\frac{1}{20} \end{gathered}$$

$$\begin{gathered} Put n=1 , in option \left(C\right) \\ \frac{1(1+1)}{1+4}=\frac{2}{5} \end{gathered}$$

$$\begin{gathered} Put n=1 , in option \left(D\right) \\ \frac{n}{2(n+2)}=\frac{1}{2.3}=\frac{1}{6} \end{gathered}$$

Hence , option (D) is correct