Sigma? Sum! 3

$\displaystyle \sum_{r=1}^n {T_r} = \frac{n(n+1)(n+2)(n+3)}{8}$

$\begin{aligned} \Rightarrow T_k & = \displaystyle \sum_{r=1}^k {T_r} - \sum_{r=1}^{k-1} {T_r} \\ & = \frac{k(k+1)(k+2)(k+3)}{8} - \frac{(k-1)k(k+1)(k+2)}{8} \\ & = \frac{k(k+1)(k+2)(k+3 - k +1)}{8} \\ & = \frac{k(k+1)(k+2)}{2} \end{aligned}$

$\begin{aligned} \Rightarrow \displaystyle \sum_{r=1}^{50} {\frac{1}{T_r}} & = \sum_{r=1}^{50} {\frac{2}{r(r+1)(r+2)}} \\ & = \sum_{r=1}^{50} {\left( \frac{1}{r} - \frac{2}{r+1} + \frac{1}{r+2}\right)} \\ & = \sum_{r=1}^{50} {\frac{1}{r}} - \sum_{r=2}^{51} {\frac{1}{r}} - \sum_{r=2}^{51} {\frac{1}{r}} + \sum_{r=3}^{52} {\frac{1}{r}} \\ & = \frac{1}{1} - \frac{1}{51} - \frac{1}{2} + \frac{1}{52} \\ & = \boxed{0.4996} \end{aligned}$