\(A=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)
=> \(A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+-+\dfrac{1}{99}-\dfrac{1}{100}\)
=> \(A=\dfrac{1}{2}-\dfrac{1}{100}\)
=> \(A=\dfrac{49}{100}\)
======================--
\(B=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{92.95}\)
=> \(B=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\)
=> \(B=\dfrac{1}{2}-\dfrac{1}{95}\)
=> \(B=\dfrac{93}{190}\)
======================-
\(C=\dfrac{7}{1.3}+\dfrac{7}{3.5}+\dfrac{7}{5.7}+\dfrac{7}{7.9}+...+\dfrac{7}{49.51}\)
=> \(2C=\dfrac{7.2}{1.3}+\dfrac{7.2}{3.5}+\dfrac{7.2}{3.7}+\dfrac{7.2}{7.9}+...+\dfrac{7.2}{49.51}\)
=> \(2C=7.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{49.51}\right)\)
=> \(2C=7.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
=> \(2C=7.\left(1-\dfrac{1}{51}\right)\)
=> \(2C=7.\dfrac{50}{51}\)
=> \(2C=\dfrac{350}{51}\)
=> \(C=\dfrac{350}{51}:2\)
=> \(C=\dfrac{175}{51}\)
