교대합과 부분분수 합

[출제의도] 부분분수전개를 사용하여 등차수열의 일반항을 구한다. 등차수열 $\left\{ a _{n} \right\}$의 첫째항을 $a$, 공차를 $d$라 하자. $\displaystyle S _{10} = \sum\limits _{k = 1} ^{10} ( - 1 ) ^{k} a _{k}$ $= - a _{1} + a _{2} - a _{3} + a _{4} - a _{5} + a _{6} - a _{7} + a _{8} - a _{9} + a _{10}$ $= ( a _{2} - a _{1} ) + ( a _{4} - a _{3} ) + ( a _{6} - a _{5} )$ $+ ( a _{8} - a _{7} ) + ( a _{10} - a _{9} )$ $= d + d + d + d + d$ $= 5 d$ $S _{10} = 25$이므로 $5 d = 25$ $\therefore$$d = 5$ $\displaystyle \frac{1}{a _{k} a _{k + 1}} = \frac{1}{a _{k + 1} - a _{k}} \left( \frac{1}{a _{k}} - \frac{1}{a _{k + 1}} \right)$ 그런데 수열 $\left\{ a _{n} \right\}$은 공차가 $d$인 등차수열이므로 $a _{k + 1} - a _{k} = d = 5$ 즉, $\displaystyle \frac{1}{a _{k + 1} - a _{k}} = \frac{1}{5}$ 그러므로 $\displaystyle \frac{1}{a _{k} a _{k + 1}} = \frac{1}{5} \left( \frac{1}{a _{k}} - \frac{1}{a _{k + 1}} \right)$ $\displaystyle T _{5} = \sum\limits _{k = 1} ^{5} \frac{1}{a _{k} a _{k + 1}}$ $\displaystyle = \frac{1}{5} \left\{ \left( \frac{1}{a _{1}} - \frac{1}{a _{2}} \right) + \left( \frac{1}{a _{2}} - \frac{1}{a _{3}} \right) + \left( \frac{1}{a _{3}} - \frac{1}{a _{4}} \right) \right.$ $\displaystyle \left. + \left( \frac{1}{a _{4}} - \frac{1}{a _{5}} \right) + \left( \frac{1}{a _{5}} - \frac{1}{a _{6}} \right) \right\}$ $\displaystyle = \frac{1}{5} \left( \frac{1}{a _{1}} - \frac{1}{a _{6}} \right)$ $\displaystyle = \frac{1}{5} \left( \frac{1}{a} - \frac{1}{a + 5 d} \right)$ $\displaystyle = \frac{1}{5} \left( \frac{1}{a} - \frac{1}{a + 25} \right)$ $\displaystyle = \frac{1}{5} \cdot \frac{25}{a ( a + 25 )}$ $\displaystyle = \frac{5}{a ( a + 25 )}$ $\displaystyle \frac{1}{T _{5}} = \frac{54}{5}$이므로 $\displaystyle \frac{a ( a + 25 )}{5} = \frac{54}{5}$ $a ( a + 25 ) = 54$ $a ^{2} + 25 a - 54 = 0$ $( a + 27 ) ( a - 2 ) = 0$ $\therefore$$a = 2 ( \because a > 0 )$ 따라서 $a _{6} = 2 + 5 \times 5 = 27$