일반항의 합 구하기

$\displaystyle \sqrt{3 n + 1} = x$, $\displaystyle \sqrt{3 n - 2} = y$라 하면 $\displaystyle \sqrt{9 n ^{2} - 3 n - 2} = \sqrt{\left( 3 n - 1 \right) \left( 3 n + 2 \right)} = xy$, $6 n - 1 = \left( 3 n - 1 \right) + \left( 3 n + 2 \right) = x ^{2} + y ^{2}$, $x ^{2} - y ^{2} = 3$이므로 $a _{n}$$\displaystyle = \frac{\sqrt{9 n ^{2} - 3 n - 2} + 6 n - 1}{\sqrt{3 n + 1} + \sqrt{3 n - 2}}$ $\displaystyle = \frac{xy + x ^{2} + y ^{2}}{x + y}$ $\displaystyle = \frac{( x - y ) ( x ^{2} + xy + y ^{2} )}{( x - y ) ( x + y )}$ $\displaystyle = \frac{x ^{3} - y ^{3}}{x ^{2} - y ^{2}}$ $\displaystyle = \frac{x ^{3} - y ^{3}}{3}$ $\displaystyle = \frac{( \sqrt{3 n + 1} ) ^{3} - ( \sqrt{3 n - 2} ) ^{3}}{3}$ $\therefore$$\displaystyle \sum\limits _{n = 1} ^{16} \frac{( \sqrt{3 n + 1} ) ^{3} - ( \sqrt{3 n - 2} ) ^{3}}{3}$ $\displaystyle = \frac{1}{3} \{ \left\{ ( \sqrt{4} ) ^{3} - ( \sqrt{1} ) ^{3} \right\} + \left\{ ( \sqrt{7} ) ^{3} - ( \sqrt{4} ) ^{3} \right\}$ $\displaystyle + \cdots + \left\{ ( \sqrt{49} ) ^{3} - ( \sqrt{46} ) ^{3} \right\}$ $\displaystyle = \frac{1}{3} ( 7 ^{3} - 1 ^{3} )$ $= 114$