An algebra problem by A Former Brilliant Member

Hint: apply telescopic series to get the result for sum of $n$ terms which is $S_n=\frac{1}{3}[\frac{1}{1×2×3}-\frac{1}{(n+1)(n+2)(n+3)}]$ and hence put $n=50$ to get sum equal to $\boxed{0.055}$

$\displaystyle \sum^{50}_{n=1}\frac{1}{n(n+1)(n+2)(n+3)}=\frac{1}{6} \displaystyle \sum^{50}_{n=1}\left(\frac{1}{n}-\frac{3}{n+1}+\frac{3}{n+2}-\frac{1}{n+3}\right) \\ \frac{1}{6}\left(\displaystyle \sum^{50}_{n=1}\frac{1}{n}-3 \displaystyle \sum^{50}_{n=1}\frac{1}{n+1}+3 \displaystyle \sum^{50}_{n=1}\frac{1}{n+2}- \displaystyle \sum^{50}_{n=1}\frac{1}{n+3}\right) \\ \frac{1}{6}\left(\displaystyle \sum^{50}_{n=1}\frac{1}{n}-3 \displaystyle \sum^{51}_{n=2}\frac{1}{n}+3 \displaystyle \sum^{52}_{n=3}\frac{1}{n}- \displaystyle \sum^{53}_{n=4}\frac{1}{n}\right) \\ \frac{1}{6}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{3}{52}-\frac{1}{51}-\frac{1}{52}-\frac{1}{53}\right) \\ \frac{1}{6}\left(\frac{883}{2562}\right) = \boxed{0.0555}$

The Cover Up Rule in Partial Fraction is the best method. However just as a different approach we have :-
$\displaystyle \sum^{50}_{n=1}f(n)= \sum^{50}_{n=1}\frac{1}{n(n+1)(n+2)(n+3)}.\\ Let ~ t=n^2+3n+1. ~\therefore ~f(n)=f(t)=\dfrac 1 {(t-1)(t+1)}=\frac 1 2 *\{\dfrac 1 {t - 1} - \dfrac 1 {t+1}\}\\ Substituting ~back ~n~ for~t, ~ f(n)=\frac 1 2 *\left (\dfrac{1}{n(n+3)} -\dfrac{1}{(n+1)(n+2)} \right ).\\ \dfrac{1}{n(n+3)} =\frac 1 3*\{\dfrac 1 n - \dfrac 1 {n+3} \} ~~~~\\ and ~~~\dfrac 1 {(n+1)(n+2)},\text{ we can calculate directly since (n+2)=(n+1) +1 . } \\ \therefore ~ \displaystyle \sum^{50}_{n=1}f(n) = \sum^{50}_{n=1}\color{#3D99F6}{ \frac 1 6*\left (\dfrac 1 n - \dfrac 1 {n+3} \right ) } - \frac 1 2*\left (\dfrac 1 {n+1} - \dfrac 1 {n+2} \right ) \\ \color{#3D99F6}{\text{In the first group,} } \text{ from n to n+3 , first three of n, }\\ \text{and last three of n+3 are only term not canceling}.\\ \implies ~\displaystyle \sum^{50}_{n=1}f(n)=\frac 1 6 *\left(\dfrac 1 1 +\dfrac 1 2 +\dfrac 1 3- \dfrac 1 { 51} -\dfrac 1 {52} - \dfrac 1 {53} \right ){\Large -} \frac 1 2 *\left(\dfrac 1 2 - \dfrac 1 { 51} \right )\\ = \Large ~~~\color{#D61F06}{\boxed {~~~~0.055741~~~} }$