Answer
$\therefore \sum_{k=1}^{50}\left(-\frac{1}{1+k}+\frac{1}{k}\right)=\frac{50}{51}$
Work Step by Step
We find:
$\begin{aligned} \sum_{k=1}^{50}\left(-\frac{1}{1+k}+\frac{1}{k}\right) &=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{50}-\frac{1}{51}\right) \\ &=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{50}-\frac{1}{51}\right) \\ &=-\frac{1}{51}+1 \\ &=\left[\frac{50}{51}\right] \end{aligned}$
$\therefore \sum_{k=1}^{50}\left(-\frac{1}{1+k}+\frac{1}{k}\right)=\frac{50}{51}$