Equation - help needed

**mat1990** · December 17th 2010, 09:59 AM

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

**snowtea** · December 17th 2010, 10:09 AM

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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