Question

Answer 1

a: \(=\dfrac{1}{1\cdot3}-\dfrac{1}{3\cdot7}+\dfrac{1}{3\cdot7}-\dfrac{1}{7\cdot9}+...+\dfrac{1}{13\cdot15}-\dfrac{1}{15\cdot19}\)

\(=\dfrac{1}{3}-\dfrac{1}{15\cdot19}=\dfrac{94}{285}\)

b: \(B=\dfrac{1}{6}\cdot A\)

=1/6*94/285

=47/855

c: \(C=\dfrac{1}{8}\left(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{48\cdot49\cdot50}\right)\)

\(=\dfrac{1}{16}\cdot\left(\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+...+\dfrac{2}{48\cdot49\cdot50}\right)\)

\(=\dfrac{1}{16}\cdot\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{48\cdot49}-\dfrac{1}{49\cdot50}\right)\)

\(=\dfrac{1}{16}\cdot\left(\dfrac{1}{2}-\dfrac{1}{2450}\right)=\dfrac{153}{4900}\)