B=\(\dfrac{3}{3.5}.\dfrac{3}{5.7}.....\dfrac{3}{47.49}\)
B=\(\dfrac{3}{2}.\left(\dfrac{2}{3.5}.\dfrac{2}{5.7}.....\dfrac{2}{47.49}\right)\)
B=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{47}-\dfrac{1}{49}\right)\)
B=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)
B=\(\dfrac{3}{2}.\dfrac{46}{147}\)
B=\(\dfrac{23}{49}\)
a) Ta có: \(A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{2499}{2500}\)
\(=\dfrac{1\cdot3}{2^2}\cdot\dfrac{2\cdot4}{3^2}\cdot\dfrac{3\cdot5}{4^2}\cdot...\cdot\dfrac{49\cdot51}{50^2}\)
\(=\dfrac{1}{50}\cdot\dfrac{51}{2}=\dfrac{51}{100}\)
b) Ta có: \(B=\dfrac{3}{3\cdot5}+\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+...+\dfrac{3}{47\cdot49}\)
\(=\dfrac{3}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{47\cdot49}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{46}{147}=\dfrac{138}{294}=\dfrac{23}{49}\)
