\(C=\dfrac{4}{1.3}+\dfrac{4}{3.5}+...+\dfrac{4}{97.99}-\dfrac{1}{49.99}\\ =2\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{97.99}\right)-\dfrac{1}{49.99}\\ =2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)-\dfrac{1}{49.99}\\ =2.\left(1-\dfrac{1}{99}\right)-\dfrac{1}{49.99}\\ =\dfrac{2.98}{99}-\dfrac{1}{49.99}\\ =\dfrac{196}{99}-\dfrac{1}{4851}\\ =\dfrac{97}{49}\)
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