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Rút gọn biểu thức cho 3 ta được biểu thức mới là
\(\frac{4}{28}+\frac{4}{70}+\frac{4}{130}+...\frac{4}{700}\)
= \(\frac{4}{4.7}+\frac{4}{7.10}+\frac{4}{10.13}+...+\frac{4}{25.28}\)
=\(\frac{4}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{25.28}\right)\)
= \(\frac{4}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)
= \(\frac{4}{3}\left(\frac{1}{4}-\frac{1}{28}\right)\)
= \(\frac{4}{3}.\frac{3}{14}\)
= \(\frac{2}{7}\)
12/84 + 12/210 + 12/390 + ...+12/2100
= 4/28 + 4/70 + 4/130 + ...+ 4/700
= 4/4.7 + 4/7.10 + 4/10.13 + ... + 4/25.28
= (3/4.7 + 3/7.10 + 3/10.13 + ...+ 3/25.28).4/3
= (1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13 + ... + 1/25 - 1/28).4/3
= (1/4 - 1/28) .4/3
= (7/28 - 1/28).4/3
= 6/28 . 4/3
= 3.2.4/7.4.3
= 2/7
\(F=\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+...+\frac{1}{190}\)
\(\Rightarrow\)\(\frac{1}{2}F=\frac{1}{2}.\left(\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+...+\frac{1}{190}\right)\)
\(\Rightarrow\) \(\frac{1}{2}F=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{380}\)
\(\Rightarrow\) \(\frac{1}{2}F=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{19.20}\)
\(\Rightarrow\) \(\frac{1}{2}F=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{19}-\frac{1}{20}\)
\(\Rightarrow\) \(\frac{1}{2}F=\frac{1}{5}-\frac{1}{20}\)
\(\Rightarrow\) \(\frac{1}{2}F=\frac{4}{20}-\frac{1}{20}\)
\(\Rightarrow\) \(\frac{1}{2}F=\frac{3}{20}\)
\(\Rightarrow\)\(F=\frac{3}{20}\div\frac{1}{2}\)
\(\Rightarrow\) \(F=\frac{3}{20}.2\)
\(\Rightarrow\)\(F=\frac{3}{10}\)
\(F=\frac{1}{15}+\frac{ 1}{21}+...+\frac{1}{190}\)
\(F=\frac{2}{30}+\frac{2}{21}+...+\frac{2}{380}\)
\(F=\frac{2}{5.6}+...+\frac{2}{19.20}\)
\(F=2.\left(\frac{1}{5.6}+...+\frac{1}{19.20}\right)\)
\(F=2.\left(\frac{1}{5}-\frac{1}{6}+...+\frac{1}{19}-\frac{1}{20}\right)\)
\(F=2\left[\frac{1}{5}-\left(\frac{1}{6}-\frac{1}{6}\right)-...-\left(\frac{1}{19}-\frac{1}{19}\right)-\frac{1}{20}\right]\)
\(F=2.\left(\frac{1}{5}-\frac{1}{20}\right)\)
\(F=2.\frac{3}{20}\)
\(F=\frac{6}{20}=\frac{3}{10}\)
\(G=\frac{12}{84}+\frac{12}{210}+...+\frac{12}{2100}\)
\(G=\frac{4}{28}+\frac{4}{70}+...+\frac{4}{700}\)
\(G=\frac{4}{4.7}+\frac{4}{7.10}+...+\frac{4}{25.28}\)
\(G=\frac{4}{3}.\left(\frac{3}{4.7}+...+\frac{3}{25.28}\right)\)
\(G=\frac{4}{3}.\left(\frac{1}{4}-\frac{1}{7}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(G=\frac{4}{3}.\left(\frac{1}{4}-\frac{1}{28}\right)\)
\(G=\frac{4}{3}.\frac{6}{28}\)
\(G=\frac{2}{7}\)
Tổng của G và F là : \(\frac{3}{10}+\frac{2}{7}=\frac{21}{70}+\frac{20}{70}=\frac{41}{70}\)
