Question

A = \(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+...+10}\)

Answer 1

\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{55}\)

\(\dfrac{A}{2}=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{110}=\)

\(=\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+...+\dfrac{1}{10x11}=\)

\(=\dfrac{3-2}{2x3}+\dfrac{4-3}{3x4}+\dfrac{5-4}{4x5}+...+\dfrac{11-10}{10x11}=\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{10}-\dfrac{1}{11}=\)

\(=\dfrac{1}{2}-\dfrac{1}{11}=\dfrac{9}{22}\Rightarrow A=\dfrac{9}{11}\)