부분분수 합

$\displaystyle a _{n} = \frac{1}{\left( 3 n - 2 \right) \left( 3 n + 1 \right)} = \frac{1}{3} \left( \frac{1}{3 n - 2} - \frac{1}{3 n + 1} \right)$ $\displaystyle \sum\limits _{k = 1} ^{8} a _{k}$$\displaystyle = \frac{1}{3} \times \sum\limits _{k = 1} ^{8} \left( \frac{1}{3 k - 2} - \frac{1}{3 k + 1} \right)$ $\displaystyle = \frac{1}{3} \times \left\{ \left( \frac{1}{1} - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{7} \right) + \cdots \right.$$\displaystyle \left. + \left( \frac{1}{19} - \frac{1}{22} \right) + \left( \frac{1}{22} - \frac{1}{25} \right) \right\}$ $\displaystyle = \frac{1}{3} \times \left( 1 - \frac{1}{25} \right) = \frac{8}{25}$