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Answers ( 3 )
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    \(\dfrac{1}{1.5}+\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{1991.1995}\)

    \(=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{1991}-\dfrac{1}{1995}\)

    \(=1-\dfrac{1}{1995}\)

    \(=\dfrac{1994}{1995}\)

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    \(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{1999.2001}\)

    \(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{1999}-\dfrac{1}{2001}\)

    \(=1-\dfrac{1}{2001}\)

    \(=\dfrac{2000}{2001}\)

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    \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}\)

    \(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\)

    \(=1-\dfrac{1}{103}\)

    \(=\dfrac{102}{103}\)

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