Problem 6.

Let $S_{n} = \dfrac {1}{1\dot{}2\dot{}3\dot{}4} + \dfrac {1}{2\dot{}3\dot{}4\dot{}5} + \dfrac {1}{3\dot{}4\dot{}5\dot{}6} + ... + \dfrac {1}{n(n+1)(n+2)(n+3)} \\ \quad = \dfrac {0!}{4!} + \dfrac {1!}{5!} + \dfrac {2!}{6!} + ... + \dfrac {(n-1)!}{(n+3)!}$

We note that:

$S_1 = \dfrac{1}{4!} = \dfrac {1}{24} = \dfrac {24-6}{18\dot{}24} = \dfrac {2\dot{}3\dot{}4-6}{18\dot{}2\dot{}3\dot{}4}$

$S_2 = \dfrac {1}{4!} + \dfrac {1}{5!} = \dfrac{6}{5!} = \dfrac {9\dot{}6}{9\dot{}2\dot{}3\dot{}4\dot{}5} = \dfrac {60-6}{18\dot{}60} = \dfrac {3\dot{}4\dot{}5-6}{18\dot{}3\dot{}4\dot{}5}$

$S_3 = \dfrac{6}{5!} + \dfrac{2!}{6!} = \dfrac{38}{6!} = \dfrac {4\dot{}5\dot{}6-6}{18\dot{}4\dot{}5\dot{}6} = \dfrac {114}{2160} = \dfrac {38}{6!}$

$S_4 = \dfrac{38}{6!} + \dfrac{3!}{7!} = \dfrac {272}{7!} = \dfrac {5\dot{}6\dot{}7-6}{18\dot{}5\dot{}6\dot{}7} = \dfrac {204}{3780} = \dfrac {272}{7!}$

Therefore,

$S_n = \dfrac {(n+1)(n+2)(n+3)-6}{18(n+1)(n+2)(n+3)} = \dfrac {1}{18} - \dfrac {1}{3(n+1)(n+2)(n+3)}$

$S_{50} = \dfrac {1}{18} - \dfrac {1}{3\dot{}51\dot{}52\dot{}53} = \boxed{0.055553184}$

The expression can be written in this form: $\sum_{n=1}^{50}\left(\dfrac{1}{n(n+1)(n+2)(n+3)}\right)\\ =\dfrac{1}{3}\sum_{n=1}^{50}\left(\dfrac{(n+3)-(n)}{n(n+1)(n+2)(n+3)}\right)\\ =\dfrac{1}{3}\sum_{n=1}^{50}\left(\dfrac{1}{n(n+1)(n+2)}-\dfrac{1}{(n+1)(n+2)(n+3)}\right)$ .Now we have converted the expression into a telescoping series,and thus,can easily find the answer.