Tổng \(S_n=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+......+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\) là
\(\dfrac{n}{2\left(3n+2\right)}\).\(\dfrac{3n}{2\left(3n+2\right)}\).\(\dfrac{2n}{4\left(3n+2\right)}\).\(\dfrac{n+1}{2\left(3n+2\right)}\).Hướng dẫn giải:\(S_n=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+......+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}\right)+\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}\right)+....+\dfrac{1}{3}\left(\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+......+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)\)
\(=\dfrac{1}{3}.\dfrac{3n+2-2}{2\left(3n+2\right)}=\dfrac{n}{2\left(3n+2\right)}\).