\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}=\frac{50}{51}\)
=> \(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}=\frac{50}{51}\)
=> \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n-1}-\frac{1}{2n+1}=\frac{50}{51}\)
=> \(1-\frac{1}{2n+1}=\frac{50}{51}\)
=> \(\frac{1}{2n+1}=1-\frac{50}{51}=\frac{1}{51}\)
=> 2n + 1 = 51
=> 2n = 50
=> n = 25
Vậy n = 25