\(A=3.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\right)\)
\(A=3.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(A=3.\left(1-\dfrac{1}{100}\right)\)
\(A=3.\dfrac{99}{100}=\dfrac{297}{100}\)
\(A=\dfrac{3^2}{1\times4}+\dfrac{3^2}{4\times7}+\dfrac{3^2}{7\times10}+\dfrac{3^2}{10\times13}+\dfrac{3^2}{13\times16}...+\dfrac{3^2}{97\times100}\)
\(=3\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\dfrac{3}{10\times13}+\dfrac{3}{13\times16} +...+\dfrac{3}{97\times100}\right)\)
\(=3\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)\(=3\times\left(1-\dfrac{1}{100}\right)\)
\(=3\times\dfrac{99}{100}\)
\(=\dfrac{297}{100}\)
\(=2\dfrac{97}{100}\)
Vậy \(A=2\dfrac{97}{100}\)
