$\text{ If }\frac{1}{\mathrm{1.2.3.4}}+\frac{1}{\mathrm{2.3.4.5}}+\frac{1}{\mathrm{3.4.5.6}}+\dots \dots \text{ upto }\mathrm{n}\text{ terms }=\frac{1}{18}-\frac{1}{3\mathrm{f}\left(\mathrm{n}\right)}\text{ then }\mathrm{f}\left(1\right)-\mathrm{f}\left(3\right)\text{ is }$

$The Genereal term of the given series is t_n=\frac{1}{n(n+1)(n+2)(n+3)} Suppose that V_r=\frac{1}{r(r+1)(r+2)}$

$V_r-V_{r+1}=3t_r Hence, t_r=\frac{1}{3}\left(V_r-V_{r+1}\right)$

$$\begin{gathered} \sum _{r=1}^n{t}_n=\frac{1}{3}\left(V_1-V_{n+1}\right) \\ =\frac{1}{3}\left(\frac{1}{6}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\right) \\ =\frac{1}{18}-\frac{1}{3f\left(n\right)} \\ f\left(n\right)=\left(n+1\right)\left(n+2\right)\left(n+3\right) \end{gathered}$$

$f\left(1\right)-f\left(3\right)=24-120=-96$