`#3107.101107`
\(B=\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-...-\dfrac{1}{5\cdot3}-\dfrac{1}{3\cdot1}\\ =\dfrac{1}{99\cdot97}-\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{95\cdot97}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{97\cdot99}\right)-\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{95\cdot97}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{97}-\dfrac{1}{99}\right)-\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{97}-\dfrac{1}{99}\right)-\dfrac{1}{2}\cdot\left(1-\dfrac{1}{97}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{9603}-\dfrac{1}{2}\cdot\dfrac{96}{97}\\ =\dfrac{1}{2}\cdot\left(\dfrac{2}{9603}-\dfrac{96}{97}\right)\\ =\dfrac{1}{2}\cdot\left(-\dfrac{9502}{9603}\right)\\ =-\dfrac{4751}{9603}\)
Vậy, `B = -4751/9603.`
\(B=\dfrac{1}{99.97}-\dfrac{1}{97.95}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)
\(B=\dfrac{1}{97.99}-\left(\dfrac{1}{95.97}+...+\dfrac{1}{3.5}+\dfrac{1}{1.3}\right)\)
Đặt \(C=\dfrac{1}{95.97}+...+\dfrac{1}{3.5}+\dfrac{1}{1.3}\)
\(C=\dfrac{1}{95.97}+...+\dfrac{1}{3.5}+\dfrac{1}{1.3}\)
\(C=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{95.97}\)
\(C=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{95.97}\right):2\)
\(2C=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{95.97}\)
\(2C=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5} +...+\dfrac{1}{95}-\dfrac{1}{97}\)
\(2C=\dfrac{1}{1}-\dfrac{1}{97}\)
\(2C=\dfrac{96}{97}\)
\(C=\dfrac{96}{97}:2=\dfrac{48}{97}\)
Thay C vào ta được:
\(B=\dfrac{1}{97.99}-\dfrac{48}{97}\)
\(99B=\dfrac{99}{97.99}-\dfrac{48.99}{97}\)
\(99B=\dfrac{1}{97}-\dfrac{4752}{97}\)
\(99B=-\dfrac{4751}{97}\)
\(B=-\dfrac{4751}{97}:99=-\dfrac{4751}{9603}\)
