Ta có :
\(A=\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{25}+\dfrac{1}{35}+...+\dfrac{1}{1985}\)
\(A=\dfrac{1}{5}+\dfrac{1}{3.5}+\dfrac{1}{5.5}+\dfrac{1}{7.5}+...+\dfrac{1}{397.5}\)
\(\Rightarrow5A=1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{397}\)
\(5A-1=\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{397}\)
\(5A-1=\dfrac{1}{3}+\left(\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{9}\right)+\left(\dfrac{1}{11}+\dfrac{1}{13}+...+\dfrac{1}{27}\right)+\)
\(\left(\dfrac{1}{29}+\dfrac{1}{31}+...+\dfrac{1}{81}\right)+\left(\dfrac{1}{83}+\dfrac{1}{85}+...+\dfrac{1}{243}\right)+...+\dfrac{1}{397}\)
\(\Rightarrow5A-1>\dfrac{1}{3}+\dfrac{1}{9}.3+\dfrac{1}{27}.9+\dfrac{1}{81}.27+\dfrac{1}{243}.81=\dfrac{1}{3}.5=\dfrac{5}{3}\)
\(\Rightarrow5A-1>\dfrac{5}{4}\Rightarrow5A>\dfrac{9}{4}\)
\(\Rightarrow A>\dfrac{9}{4}:5=\dfrac{9}{20}\Rightarrow\left(dpcm\right)\)
