\displaystyle\sum_{k=1}^{n}(\frac{1}{2k}-\frac{1}{1+k}+\frac{1}{2(2+k)})=\\
\begin{array}{ccc}
\frac{1}{2} & -\frac{1}{2} & +\frac{1}{6}\\
+\frac{1}{4}&- \frac{1}{3} & +\frac{1}{8}\\
+\frac{1}{6}&- \frac{1}{4} & +\frac{1}{10}\\
\vdots & \vdots & \vdots \\
+\frac{1}{2(n-2)}&- \frac{1}{-1+n} & +\frac{1}{2n}\\
+\frac{1}{2(n-1)}&- \frac{1}{n} & +\frac{1}{2(1+n)}\\
+\frac{1}{2n}&- \frac{1}{1+n} & +\frac{1}{2(2+n)}\\
\end{array}