Question

$( \frac{3}{1·4}+ \frac{3}{4·7y}+ \frac{3}{7·10}+...+ \frac{3}{52·55})+( \frac{1}{10·11}+ \frac{1}{11·12}+ \frac{1}{12·13}+...+ \frac{1}{54·55})$ AJUTATIMA VA ROG!!!!!

Answer 1

       
$\frac{3}{1*4} = \frac{1}{1} - \frac{1}{4} \\ \\ \frac{3}{4*7} = \frac{1}{4} - \frac{1}{7} \\ \\ \frac{3}{52*55} = \frac{1}{52} - \frac{1}{55} \\ \\ =>\frac{3}{1*4}+\frac{3}{4*7}+\frac{3}{7*10}+...+\frac{3}{52*55}= \\ \\ = \frac{1}{1} - \frac{1}{4}+ \frac{1}{4} - \frac{1}{7}+ \frac{1}{7} - \frac{1}{10}+...+\frac{1}{52} - \frac{1}{55}=\frac{1}{1}-\frac{1}{55} = \frac{54}{55}$


$\frac{1}{10*11} = \frac{1}{10} - \frac{1}{11} \\ \\ \frac{1}{11*12} = \frac{1}{11} - \frac{1}{12} \\ \\ \frac{1}{12*13} = \frac{1}{112} - \frac{1}{13} \\ \\ \frac{1}{54*55} = \frac{1}{54} - \frac{1}{55} \\ \\ => \frac{1}{10*11}+\frac{1}{11*12}+\frac{1}{12*13}+...+\frac{1}{54*55} = \\ \\ =\frac{1}{10} - \frac{1}{11}+ \frac{1}{11} - \frac{1}{12} +\frac{1}{112} - \frac{1}{13}+...+ \frac{1}{54} - \frac{1}{55}=\frac{1}{10}-\frac{1}{55} = \frac{11-2}{110} =\frac{9}{110}$



$\frac{54}{55}+ \frac{9}{110} = \frac{2*54+9}{110} = \frac{108+9}{110}= \boxed{ \frac{117}{110}}$