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Ta có:
\(A=\dfrac{1}{1.1981}+\dfrac{1}{2.1982}+...+\dfrac{1}{n\left(1980+n\right)}+...+\dfrac{1}{25.2005}\)
\(=\dfrac{1}{1980}\left(\dfrac{1981-1}{1.1981}+\dfrac{1982-2}{2.1982}+...+\dfrac{1980+n-n}{n\left(1980+n\right)}+...+\dfrac{2005-25}{25.2005}\right)\)
\(=\dfrac{1}{1980}\left(1-\dfrac{1}{1981}+\dfrac{1}{2}-\dfrac{1}{1982}+...+\dfrac{1}{n}-\dfrac{1}{1980+n}+...+\dfrac{1}{25}-\dfrac{1}{2005}\right)\)
\(=\dfrac{1}{1980}\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{1981}+\dfrac{1}{1982}+...+\dfrac{1}{2005}\right)\right]\)
Lại có:
\(B=\dfrac{1}{1.26}+\dfrac{1}{2.27}+...+\dfrac{1}{m\left(m+25\right)}+...+\dfrac{1}{1980.2005}\)
\(=\dfrac{1}{25}\left(\dfrac{26-1}{1.26}+\dfrac{27-2}{2.27}+...+\dfrac{25+m-m}{m\left(25+m\right)}+...+\dfrac{2005-1980}{1980.2005}\right)\)
\(=\dfrac{1}{25}\left(\dfrac{1}{1}-\dfrac{1}{26}+\dfrac{1}{2}-\dfrac{1}{27}+...+\dfrac{1}{m}-\dfrac{1}{25+m}+...+\dfrac{1}{1980}-\dfrac{1}{2005}\right)\)
\(=\dfrac{1}{25}\left[\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{1980}\right)-\left(\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{2005}\right)\right]\)
\(=\dfrac{1}{25}\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{1981}+\dfrac{1}{1982}+...+\dfrac{1}{2005}\right)\right]\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{\dfrac{1}{1980}}{\dfrac{1}{25}}=\dfrac{5}{396}\)
Vậy tỉ số của \(A\) và \(B\) là \(\dfrac{5}{396}\)
25B=\(\frac{25}{1.26}+\frac{25}{2.27}+.......+\frac{25}{1980.2005}\)
\(=1-\frac{1}{26}+\frac{1}{2}-\frac{1}{27}+..........+\frac{1}{1980}-\frac{1}{2005}\)
\(=1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{25}-\frac{1}{1981}-\frac{1}{1982}-....-\frac{1}{2005}\)
1980A=\(\frac{1980}{1.1981}+\frac{1980}{2.1982}+.........+\frac{1980}{25.2005}\)
\(=1-\frac{1}{1981}+\frac{1}{2}-\frac{1}{1982}+......+\frac{1}{25}-\frac{1}{2005}\)
\(=1+\frac{1}{2}+...+\frac{1}{25}-\frac{1}{1981}-\frac{1}{1982}-...-\frac{1}{2005}\)
\(\Rightarrow25B=1980A\)\(\Rightarrow\frac{A}{B}=\frac{25}{1980}\)
