Proszę​ o pomoc!!! potrzebuję to na jutro! zadanie w załączniku. (daję naj)

[latex]a)\\ cfrac{1}{12}+cfrac{1}{20}=cfrac{4-3}{3cdot 4}+cfrac{5-4}{4cdot 5}=cfrac{1}{3}-cfrac{1}{4}+cfrac{1}{4}-cfrac{1}{5}=cfrac{1}{3}-cfrac{1}{5}=cfrac{5-3}{5cdot 3}=cfrac{2}{15}\\\ b)\\ cfrac{1}{12}+cfrac{1}{20}+cfrac{1}{30}+cfrac{1}{42}= cfrac{1}{3}-cfrac{1}{4}+cfrac{1}{4}-cfrac{1}{5}+cfrac{1}{5}-cfrac{1}{6}+cfrac{1}{6}-cfrac{1}{7}=cfrac{1}{3}-cfrac{1}{7}=cfrac{4}{21}[/latex] [latex] c)\\ cfrac{1}{12}+cfrac{1}{20}+cfrac{1}{30}+cfrac{1}{42}+...+cfrac{1}{n(n+1)}=\\\ cfrac{1}{3}-cfrac{1}{4}+cfrac{1}{4}-cfrac{1}{5}+cfrac{1}{5}-cfrac{1}{6}+cfrac{1}{6}-cfrac{1}{7}+...+cfrac{1}{n}-cfrac{1}{n+1}=\\\ =cfrac{1}{3}-cfrac{1}{n+1}=cfrac{n+1-3}{3(n+1)}=cfrac{n-2}{3n+3}[/latex] np.: dla n=8 mamy [latex]cfrac{1}{12}+cfrac{1}{20}+cfrac{1}{30}+cfrac{1}{42}+...+cfrac{1}{8(8+1)}=\\\ cfrac{1}{3}-cfrac{1}{4}+cfrac{1}{4}-cfrac{1}{5}+cfrac{1}{5}-cfrac{1}{6}+cfrac{1}{6}-cfrac{1}{7}+...+cfrac{1}{8}-cfrac{1}{8+1}=\\\ =cfrac{1}{3}-cfrac{1}{8+1}=cfrac{8+1-3}{3(8+1)}=cfrac{8-2}{3cdot 8+3}=cfrac{6}{27}=cfrac{2}{9}[/latex] tada