\(\dfrac{12}{1.5}+\dfrac{12}{5.9}+...+\dfrac{12}{93.97}+\dfrac{12}{97.101}\)
= \(3.\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{93.97}+\dfrac{4}{97.101}\right)\)
= \(3.\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{93}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{101}\right)\)
= \(3.\left(1-\dfrac{1}{101}\right)\)
= \(3.\dfrac{100}{101}\)
= \(\dfrac{300}{101}=2\dfrac{98}{101}\)