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Math Help - Equation - help needed

  1. #1
    Newbie
    Joined
    Dec 2010
    Posts
    3

    Equation - help needed

    Hello guys, I have no idea how to do it;

    Prove that:

    \frac{1}{1\cdot 4}+\frac{1}{4\cdot 7}+\frac{1}{7\cdot 10}+...+\frac{1}{97\cdot 100}=\frac{33}{100}

    Thanks for your help
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  2. #2
    Senior Member
    Joined
    Dec 2010
    Posts
    470
    Using partial fractions

    \frac{1}{n(n+3)} = \frac{1}{3}(\frac{A}{n} + \frac{B}{n+3})

    so

    \frac{1}{1\cdot 4}+\frac{1}{4\cdot 7}+\frac{1}{7\cdot 10}+...+\frac{1}{97\cdot 100}<br />
= \frac{1}{3}(\frac{1}{1} - \frac{1}{4} + \frac{1}{4} - \frac{1}{7} + \frac{1}{7} - \frac{1}{10} + ... + \frac{1}{97} - \frac{1}{100})<br />
= \frac{1}{3}(\frac{1}{1} - \frac{1}{100}) = \frac{33}{100}
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