\(\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{97.101}\\ =1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{13}+...+\dfrac{1}{97}-\dfrac{1}{101}\\ =1-\dfrac{1}{101}\\ =\dfrac{100}{101}\)
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