\(1,=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{1}{2}-\dfrac{1}{6}=\dfrac{1.3}{2.3}-\dfrac{1}{6}=\dfrac{3-1}{6}=\dfrac{2}{6}=\dfrac{1}{3}\\ b,=\dfrac{2}{3}-\dfrac{2}{4}+\dfrac{2}{4}-\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{2}{6}+.....+\dfrac{2}{99}-\dfrac{2}{100}=\dfrac{2}{3}-\dfrac{2}{100}=\dfrac{2.100-2.3}{300}=\dfrac{200-6}{200}=\dfrac{97}{150}\)