Question

Answer 1

\(A=\dfrac{2}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{\left(3n+1\right)\left(3n+4\right)}\right)\)

\(=\dfrac{2}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{3n+1}-\dfrac{1}{3n+4}\right)\)

\(=\dfrac{2}{3}\cdot\dfrac{3n+4-1}{3n+4}=\dfrac{2}{3}\cdot\dfrac{3n+3}{3n+4}=\dfrac{2n+2}{3n+4}\)