cách hay cho những người kh biết làm là tìm A=?, B=? rồi chia vs nhau
Ta có: A = \(\dfrac{1}{1\times101}+\dfrac{1}{2\times102}+....+\dfrac{1}{25\times125}\)
=> 100.A = \(\dfrac{100}{1\times101}+\dfrac{100}{2\times102}+....+\dfrac{100}{25\times125}\)
=> 100.A = \(\dfrac{1}{1}-\dfrac{1}{101}+\dfrac{1}{2}-\dfrac{1}{102}+....+\dfrac{1}{25}-\dfrac{1}{125}\)
=> A =\(\dfrac{1}{100}\). \(\left[\left(\dfrac{1}{1}+\dfrac{1}{2}+....+\dfrac{1}{25}\right)-\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{125}\right)\right]\)
Ta lại có: B = \(\dfrac{1}{1\times26}+\dfrac{1}{2\times27}+....+\dfrac{1}{100\times125}\)
=> 25.B = \(\dfrac{25}{1\times26}+\dfrac{25}{2\times27}+....+\dfrac{25}{100\times125}\)
=> 25.B = \(\dfrac{1}{1}-\dfrac{1}{26}+\dfrac{1}{2}-\dfrac{1}{27}+....+\dfrac{1}{100}-\dfrac{1}{125}\)
=> 25B = \(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+....\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{100}+...+\dfrac{1}{125}-\dfrac{1}{26}-\dfrac{1}{27}-...-\dfrac{1}{100}-\dfrac{1}{101}-..-\dfrac{1}{125}\)
=> B = \(\dfrac{1}{25}\).\(\left[\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{125}\right)\right]\)
=> \(\dfrac{A}{B}\) =\(\dfrac{\dfrac{1}{100}\left[\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{125}\right)\right]}{\dfrac{1}{25}\left[\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{125}\right)\right]}\)
=> \(\dfrac{A}{B}\) = \(\dfrac{1}{100}:\dfrac{1}{25}=\dfrac{25}{100}=\dfrac{1}{4}\)
