Question

1/99-1/99.97-1/97.95-1/95.93-...+1/5.3-1/3.1

Answer 1

\(=\dfrac{1}{99}-\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{95\cdot97}+\dfrac{1}{97\cdot99}\right)\\ =\dfrac{1}{99}-\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{95\cdot97}+\dfrac{2}{97\cdot99}\right)\\ =\dfrac{1}{99}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\\ =\dfrac{1}{99}-\dfrac{1}{2}\left(1-\dfrac{1}{99}\right)=\dfrac{1}{99}-\dfrac{1}{2}\cdot\dfrac{98}{99}\\ =\dfrac{1}{99}-\dfrac{49}{99}=-\dfrac{48}{99}=-\dfrac{16}{33}\)