원 밖 삼각형 넓이 극한

[출제의도] 무한수열의 극한을 활용하여 문제해결하기 직각삼각형 $\mathrm{P} _{\mathit{n}} \mathrm{OQ} _{\mathit{n}}$에서 $\displaystyle \mathrm{\angle} P _{\mathit{n}} OQ _{\mathit{n}} = \frac{\pi}{3}$이고 $\displaystyle \mathrm{\overline{OP _{\mathit{n}}}} = \mathit{n} + 2$이므로 $\displaystyle \mathrm{\overline{OQ _{\mathit{n}}}} = \frac{\mathit{1}}{2} \left( n + 2 \right)$, $\displaystyle \mathrm{\overline{P _{\mathit{n}} Q _{\mathit{n}}}} = \frac{\mathit{\sqrt{3}}}{2} \left( n + 2 \right)$ $\displaystyle \triangle \mathrm{P} _{\mathit{n}} \mathrm{OQ} _{\mathit{n}} = \frac{1}{2} \times {\overline{\mathrm{O} _{\mathit{}} \mathrm{Q} _{\mathit{n}} \mathit{}}} \times {\overline{\mathrm{P} _{\mathit{n}} \mathrm{Q} _{\mathit{n}} \mathit{}}} = \frac{\mathit{\sqrt{3} \left( n + 2 \right) ^{2}}}{8}$ $S _{n}$은 $\triangle \mathrm{P} _{\mathit{n}} \mathrm{OQ} _{\mathit{n}}$의 넓이에서 원 $C _{n}$의 넓이의 $\displaystyle \frac{1}{6}$을 빼면 된다. $\therefore$ $\displaystyle S _{n} = \frac{\sqrt{3} \left( n + 2 \right) ^{2}}{8} - \frac{n \pi}{6}$ $\displaystyle \lim\limits _{n \rightarrow \infty} {\frac{n ^{2}}{S _{n}}} = \lim\limits _{n \rightarrow \infty} {\frac{n ^{2}}{\frac{\sqrt{3} \left( n + 2 \right) ^{2}}{8} - \frac{n \pi}{6}} = \frac{8}{\sqrt{3}}} = a$ 따라서 $3 a ^{2} = 64$