Question

Answer 1

a: \(=\dfrac{1}{3}+\dfrac{2}{3}+\dfrac{3}{3}+...+\dfrac{10}{3}+\dfrac{11}{3}=22\)

b: \(A=9+\dfrac{9}{2}+\dfrac{9}{4}+...+\dfrac{9}{256}\)

=>\(2A=18+9+...+\dfrac{9}{128}\)

=>\(A=18-\dfrac{9}{256}\)

c: \(=\dfrac{5}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{101\cdot103}\right)\)

\(=\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{101}-\dfrac{1}{103}\right)\)

\(=\dfrac{5}{2}\cdot\dfrac{102}{103}=\dfrac{255}{103}\)