Ta có \(P=\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+...+50}\)
\(=\dfrac{2}{2.\left(1+2\right)}+\dfrac{2}{2.\left(1+2+3\right)}+...+\dfrac{2}{2.\left(1+2+...+50\right)}\)\(=2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{50.51}\right)\)
\(=2.\dfrac{49}{102}\)
=\(\dfrac{49}{51}\)
xong rầu đó khỏi cảm ơn