F=10/50+10/140+10/260+..+10/140
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\(\)\(\dfrac{10}{56}+\dfrac{10}{140}+...+\dfrac{10}{1400}\)
\(=\dfrac{5}{28}+\dfrac{5}{70}+...+\dfrac{5}{700}\)
\(=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)
\(=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{28}\right)\)
\(=\dfrac{5}{3}\cdot\dfrac{6}{28}=2\cdot\dfrac{5}{28}=\dfrac{10}{28}=\dfrac{5}{14}\)
\(=\dfrac{5}{28}+\dfrac{5}{70}+\dfrac{5}{130}+...+\dfrac{5}{700}\\ =\dfrac{5}{3}\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+...+\dfrac{3}{25\cdot28}\right)\\ =\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\\ =\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{28}\right)=\dfrac{5}{3}\cdot\dfrac{3}{14}=\dfrac{5}{14}\)
\(F=\frac{15}{90.94}+\frac{15}{94.98}+\frac{15}{98.102}+....+\frac{15}{146.150}\)
\(\Rightarrow F=\frac{15}{4}\left(\frac{15}{90}-\frac{15}{94}+\frac{15}{94}-\frac{15}{98}+....+\frac{15}{146}-\frac{15}{150}\right)\)
\(\Rightarrow F=\frac{15}{4}\left(\frac{1}{6}-\frac{1}{10}\right)=\frac{15}{4}\left(\frac{5}{30}-\frac{3}{30}\right)=\frac{15}{4}.\frac{1}{15}=\frac{15}{60}=\frac{1}{4}\)
\(A=\dfrac{5}{28}+\dfrac{5}{70}+\dfrac{5}{130}+...+\dfrac{5}{700}\)
\(\dfrac{3A}{5}=\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{25.28}\)
\(\dfrac{3A}{5}=\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}\)
\(\dfrac{3A}{5}=\dfrac{1}{4}-\dfrac{1}{28}=\dfrac{3}{14}\)
⇒ \(A=\dfrac{5}{14}\)
\(=\frac{20}{112}+\frac{20}{280}+\frac{20}{520}+...+\frac{20}{2800}=20\left(\frac{1}{8.14}+\frac{1}{14.20}+\frac{1}{20.26}+...+\frac{1}{50.56}\right)\)
\(=20\left(\frac{1}{8}-\frac{1}{56}\right)=20.\frac{3}{28}=\frac{15}{7}\)
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