로그를 이용한 수열의 합

$\displaystyle \frac{9}{8} + 9 + 9 ^{2} + \cdots + 9 ^{n} = b _{n + 1}$라고 하면, $\displaystyle b _{n + 1} = \frac{9}{8} + 9 + 9 ^{2} + \cdots + 9 ^{n} = \frac{9}{8} + \frac{9}{8} ( 9 ^{n} - 1 ) = \frac{9}{8} 9 ^{n}$ 따라서 $\displaystyle a _{n} = \left( \frac{9}{8} \right) ^{n} \cdot 9 ^{\frac{n ( n - 1 )}{2}} = \left( \frac{9}{8} \right) ^{n} 3 ^{n ( n - 1 )}$$\displaystyle = \left( \frac{9}{8} \times 3 ^{n - 1} \right) ^{n}$ $\displaystyle \frac{\log a _{n}}{n} = \log \frac{9}{8} \times 3 ^{n - 1} = \log \frac{3 ^{n + 1}}{2 ^{3}}$ $\displaystyle \sum\limits _{k = 1} ^{10} \frac{\log a _{k}}{k} = \sum\limits _{k = 1} ^{10} \log \frac{3 ^{k + 1}}{2 ^{3}} = \log \frac{3 ^{65}}{2 ^{30}}$이다. 그러므로 $\displaystyle A = \frac{3 ^{65}}{2 ^{30}}$이다.