Question

\(\dfrac{4}{3.5}+\dfrac{4}{5.7}+\dfrac{4}{7.9}+......+\dfrac{4}{97.95}\)

Answer 1

\(\dfrac{4}{3.5}+\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{97.95}\)

\(=2\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{95.97}\right)\)

\(=2\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)

\(=2\left(\dfrac{1}{3}-\dfrac{1}{97}\right)\)

\(=2.\dfrac{94}{291}=\dfrac{188}{291}\)

Answer 2

\(=2\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{95\cdot97}\right)\\ =2\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\\ =2\left(\dfrac{1}{3}-\dfrac{1}{97}\right)=2\cdot\dfrac{94}{291}=\dfrac{188}{291}\)