대칭함수의 합의 성질을 이용한 시그마

함수 $\displaystyle f \left( x \right) = \frac{2 ^{x}}{2 ^{x} + \sqrt{2}}$에서 $f \left( 1 - x \right)$$\displaystyle = \frac{2 ^{1 - x}}{2 ^{1 - x} + \sqrt{2}}$$\displaystyle = \frac{\sqrt{2}}{2 ^{x} + \sqrt{2}}$ 이므로 $f \left( x \right) + f \left( 1 - x \right)$$= 1$ $\cdots$$\cdots$ ㉠ $\displaystyle f \left( 0 \right) = \frac{1}{1 + \sqrt{2}} = \sqrt{2} - 1$, $\displaystyle f \left( \frac{1}{2} \right) = \frac{1}{2}$ $\cdots$$\cdots$ ㉡ $\displaystyle a _{n} = f \left( \frac{1}{n} \right) + f \left( \frac{2}{n} \right) + f \left( \frac{3}{n} \right) + \cdots + f \left( \frac{n - 1}{n} \right) + f \left( \frac{n}{n} \right)$에서 자연수 $k$에 대하여 $a _{2 k}$$\displaystyle = f \left( \frac{1}{2 k} \right) + f \left( \frac{2}{2 k} \right) + \cdots + f \left( \frac{k}{2 k} \right) + \cdots + f \left( \frac{2 k}{2 k} \right)$ $\displaystyle = f \left( \frac{0}{2 k} \right) + f \left( \frac{1}{2 k} \right) + \cdots + f \left( \frac{2 k}{2 k} \right) - f \left( 0 \right)$ $\displaystyle = k + \frac{1}{2} - \sqrt{2} + 1$ ($\because$ ㉠, ㉡) $a _{2 k - 1}$$\displaystyle = f \left( \frac{1}{2 k - 1} \right) + f \left( \frac{2}{2 k - 1} \right) + \cdots + f \left( \frac{2 k - 1}{2 k - 1} \right)$ $\displaystyle = f \left( \frac{0}{2 k - 1} \right) + f \left( \frac{1}{2 k - 1} \right) + \cdots + f \left( \frac{2 k - 1}{2 k - 1} \right) - f \left( 0 \right)$ $\displaystyle = k - \sqrt{2} + 1$ ($\because$ ㉠, ㉡) $\therefore$ $\displaystyle \sum\limits _{n = 1} ^{20} a _{n}$$\displaystyle = \sum\limits _{k = 1} ^{10} \left( a _{2 k - 1} + a _{2 k} \right)$ $\displaystyle = \sum\limits _{k = 1} ^{10} \left( 2 k + \frac{5}{2} - 2 \sqrt{2} \right)$ $\displaystyle = 2 \times \frac{10 \times 11}{2} + 10 \times \frac{5}{2} - 20 \sqrt{2}$ $\displaystyle = 135 - 20 \sqrt{2}$ 이므로 $p = 135$, $q = - 20$ $\therefore$ $p + q = 115$