Question

Answer 1

\(\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{17}{104}\)

\(\Rightarrow\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{17}{104}\)

\(\Rightarrow\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{17}{104}\)

\(\Rightarrow\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{x+3}\right)=\dfrac{17}{104}\)

\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+3}=\dfrac{51}{104}\Rightarrow x=101\)