Question

Ex se afla in imagine....... .

Answer 1

$S_5=\frac{1}{4}(\frac{1}{1\cdot2}+\frac{1}{2\cdot 3}+...\frac{1}{1007\cdot 1008})=\\ \frac{1}{4}(\frac{2-1}{1\cdot2}+\frac{3-2}{2\cdot3}+\frac{4-3}{3\cdot4}+...\frac{1008-1007}{1007\cdot1008})=\\ \frac{1}{4}(\frac{2}{1\cdot2}-\frac{1}{2}+\frac{3}{2\cdot3}-\frac{2}{2\cdot3}+\frac{4}{3\cdot4}-\frac{3}{3\cdot4}+...\frac{1008}{1007\cdot1008}-\frac{1007}{1007\cdot1008})=$ $\frac{1}{4}(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{1007}-\\\frac{1}{1008})=\frac{1}{4}(1-\frac{1}{1008})=$
$\frac{1}{4}\cdot\frac{1007}{1008}=\frac{1007}{4032}$