Chứng tỏ rằng :

(1+\(\dfrac{1}{3}\)+\(\dfrac{1}{5}\)+...+\(\dfrac{1}{99}\)) - (\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{6}\)+...+\(\dfrac{1}{100}\)) = \(\dfrac{1}{51}\)+ \(\dfrac{1}{52}\)+ \(\dfrac{1}{53}\)+ ...+ \(\dfrac{1}{100}\)

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