Ta có: \(\dfrac{5}{3\cdot7}+\dfrac{5}{7\cdot11}+\dfrac{5}{11\cdot15}+...+\dfrac{5}{\left(4n-1\right)\left(4n+3\right)}\)
\(=\dfrac{5}{4}\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\)
\(=\dfrac{5}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{4n-1}-\dfrac{1}{4n+3}\right)\)
\(=\dfrac{5}{4}\left(\dfrac{4n+3-3}{3\left(4n+3\right)}\right)\)
\(=\dfrac{5}{4}\cdot\dfrac{4n}{3\left(4n+3\right)}\)
\(=\dfrac{5n}{3\left(4n+3\right)}\)
\(\dfrac{5}{3.7}+\dfrac{5}{7.11}+\dfrac{5}{11.15}+...+\dfrac{5}{\left(4n-1\right)\left(4n+3\right)}=\dfrac{5}{4}\left(\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{\left(4n-1\right)\left(4n+3\right)}\right)=\dfrac{5}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{4n-1}-\dfrac{1}{4n+3}\right)=\dfrac{5}{4}.\left(\dfrac{1}{3}-\dfrac{1}{4n+3}\right)=\dfrac{5}{4}.\dfrac{4n}{12n+9}=\dfrac{5n}{12n+9}\)
Mình thấy đề sai sai hay sao ấy
