Question

Tính tổng S(n) = \(\dfrac{1}{2,5}+\dfrac{1}{5,8}+...+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\).

Answer 1

Bn viết để sai rồi, mk sửa lại :)

\(S\left(n\right)=\dfrac{1}{2.5}+\dfrac{1}{5.8}+.........+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)

\(\Leftrightarrow3S\left(n\right)=\dfrac{3}{2.5}+\dfrac{3}{5.8}+.........+\dfrac{3}{\left(3n-1\right)\left(3n+2\right)}\)

\(\Leftrightarrow3S\left(n\right)=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+......+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\)

\(\Leftrightarrow3S\left(n\right)=\dfrac{1}{2}-\dfrac{1}{3n+2}\)

\(\Leftrightarrow S\left(n\right)=\dfrac{\dfrac{1}{2}-\dfrac{1}{3n+2}}{3}\)