Question

\(\dfrac{1}{10.12}\)+\(\dfrac{1}{12.14}\)+\(\dfrac{1}{14.16}+...+\dfrac{1}{98.100}\)

Answer 1

\(\dfrac{1}{10\cdot12}+\dfrac{1}{12\cdot14}+...+\dfrac{1}{98\cdot100}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{10\cdot12}+\dfrac{2}{12\cdot14}+...+\dfrac{2}{98\cdot100}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{14}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{10}-\dfrac{1}{100}\right)=\dfrac{1}{2}\cdot\dfrac{9}{100}=\dfrac{9}{200}\)