Ta đặt
\(A=\dfrac{1}{50}-\dfrac{1}{50\times49}-....-\dfrac{1}{2\times1}\)
\(A=\dfrac{1}{50}-\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{49\times50}\right)\)
\(A=\dfrac{1}{50}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(A=\dfrac{1}{50}-\left(1-\dfrac{1}{50}\right)\)
\(A=\dfrac{1}{50}-\dfrac{49}{50}\)
\(A=\dfrac{-48}{50}=\dfrac{-24}{25}\)
\(=\dfrac{1}{50}-\left(\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{50-49}{49.50}\right)=\)
\(=\dfrac{1}{50}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=\)
\(=\dfrac{1}{50}-\left(1-\dfrac{1}{50}\right)=\dfrac{2}{50}-1=\dfrac{1}{25}-1=-\dfrac{24}{25}\)
\(...=\dfrac{1}{50}-\left(\dfrac{1}{49}-\dfrac{1}{50}\right)-\left(\dfrac{1}{48}-\dfrac{1}{49}\right)-...\left(1-\dfrac{1}{2}\right)=\dfrac{1}{50}-\dfrac{1}{49}+\dfrac{1}{50}-\dfrac{1}{48}+\dfrac{1}{49}-...-1+\dfrac{1}{2}=0\)
\(\dfrac{1}{50}-\dfrac{1}{50.49}-\dfrac{1}{49.48}-....-\dfrac{1}{2.1}\)
=\(\dfrac{1}{50}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
=\(\dfrac{1}{50}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
=\(\dfrac{1}{50}-\left(1-\dfrac{1}{50}\right)\)
=\(\dfrac{1}{50}-\dfrac{49}{50}\)
=\(\dfrac{-24}{25}\)
Đính chính kết quả \(=-1\) không phải \(=0\)
