We see that: \(\dfrac{1}{6.10}\)=\(\dfrac{1}{4}\). \(\dfrac{4}{6.10}\)=\(\dfrac{1}{4}\).  \(\dfrac{1}{6}\)- \(\dfrac{1}{10}\)

=>A=\(\dfrac{1}{4}\).  \(\dfrac{1}{6}\)- \(\dfrac{1}{10}\)+ \(\dfrac{1}{10}\)- \(\dfrac{1}{14}\)+...+\(\dfrac{1}{202}\)- \(\dfrac{1}{206}\)

       =\(\dfrac{1}{4}\). \(\dfrac{1}{6}\)- \(\dfrac{1}{206}\)

       =\(\dfrac{25}{618}\)

\(\dfrac{50}{309}\)is wrong

Lê Quốc Trần Anh is was wrong

Sorry, the last row is: \(\dfrac{50}{309}.\dfrac{1}{4}=\dfrac{25}{618}\)

A=\(\dfrac{1}{6}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+...+\dfrac{1}{202}-\dfrac{1}{206}=\dfrac{1}{6}-\dfrac{1}{206}=\dfrac{50}{309}\)

Help you solve math is answer question very good

Help you solve math was wrong

\(A=\dfrac{1}{6\cdot10}+\dfrac{1}{10\cdot14}+...+\dfrac{1}{202\cdot206}\)

\(A=\dfrac{1}{4}\left(\dfrac{1}{6}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+...+\dfrac{1}{202}-\dfrac{1}{206}\right)\)

\(A=\dfrac{1}{4}\left(\dfrac{1}{6}-\dfrac{1}{206}\right)\)

\(A=\dfrac{1}{4}\cdot\dfrac{50}{309}\)

\(A=\dfrac{25}{618}\)

So \(A=\dfrac{25}{618}\)