자연계열 2번

$\displaystyle \angle\mathrm{OA}_2\mathrm{A}_1=180^\circ-30^\circ-15^\circ=135^\circ$이므로, 삼각형 $\displaystyle \mathrm{OA}_1\mathrm{A}_2$에 사인법칙을 적용하면$\displaystyle \frac{\overline{\mathrm{OA}_2}}{\sin15^\circ}=\frac{\overline{\mathrm{OA}_1}}{\sin135^\circ}.\qquad(1)$

$\displaystyle \sin135^\circ=\sin(180^\circ-45^\circ)=\sin45^\circ=\frac1{\sqrt2}$이므로, (1)에 의하여$\displaystyle \overline{\mathrm{OA}_2}=\frac{\frac14}{\frac1{\sqrt2}}\cdot1=\frac{\sqrt2}4.\qquad(2)$

삼각형 $\displaystyle \mathrm{OA}_2\mathrm{A}_3$에 사인법칙을 적용하면$\displaystyle \frac{\overline{\mathrm{OA}_3}}{\sin30^\circ}=\frac{\overline{\mathrm{OA}_2}}{\sin135^\circ}\qquad(3)$

이므로, (2)에 의하여$\displaystyle \overline{\mathrm{OA}_3}=\frac{\frac12}{\frac1{\sqrt2}}\cdot\frac{\sqrt2}4=\frac14.\qquad(4)$

제시문의 조건 (다), (라)에 의하여, 자연수 $\displaystyle n$에 대하여$\displaystyle \begin{aligned}\angle\mathrm{A}_{n+2}\mathrm{OA}_{n+3}&=\angle\mathrm{OA}_{n+1}\mathrm{A}_{n+2}=\angle\mathrm{A}_n\mathrm{OA}_{n+1},\\\angle\mathrm{OA}_{n+2}\mathrm{A}_{n+3}&=\angle\mathrm{A}_{n+1}\mathrm{OA}_{n+2}=\angle\mathrm{OA}_n\mathrm{A}_{n+1}\end{aligned}$이므로, 조건 (가), (나)에 의하여 <그림 5>과 같이$\displaystyle \angle\mathrm{A}_n\mathrm{OA}_{n+1}=30^\circ,\ \angle\mathrm{OA}_n\mathrm{A}_{n+1}=15^\circ,\quad n:$홀수,$\displaystyle \angle\mathrm{A}_n\mathrm{OA}_{n+1}=15^\circ,\ \angle\mathrm{OA}_n\mathrm{A}_{n+1}=30^\circ,\quad n:$짝수. $\displaystyle (15)$

따라서 삼각형 $\displaystyle \mathrm{OA}_n\mathrm{A}_{n+1}$에 사인법칙을 적용하면$\displaystyle \frac{a_{n+1}}{\sin15^\circ}=\frac{a_n}{\sin135^\circ},\quad n:$홀수,$\displaystyle \frac{a_{n+1}}{\sin30^\circ}=\frac{a_n}{\sin135^\circ},\quad n:$짝수 $\displaystyle (16)$

이므로,$\displaystyle a_{n+1}=\frac{\frac14}{\frac1{\sqrt2}}\cdot a_n=\frac{\sqrt2}4a_n,\quad n:$홀수,$\displaystyle a_{n+1}=\frac{\frac12}{\frac1{\sqrt2}}\cdot a_n=\frac{\sqrt2}2a_n,\quad n:$짝수. $\displaystyle (17)$

(17) 의하여$\displaystyle a_{n+2}=\frac{\sqrt2}4\cdot\frac{\sqrt2}2\cdot a_n=\frac14a_n,\quad n:$자연수 $\displaystyle (18)$

이고, $\displaystyle a_1=\overline{\mathrm{OA}_1}=1$이므로,수열 $\displaystyle a_1,a_3,a_5,\cdots$는 첫째항 $\displaystyle 1$, 공비 $\displaystyle \frac14$인 등비수열. $\displaystyle (19)$(2)에 의하여 $\displaystyle a_2=\overline{\mathrm{OA}_2}=\frac{\sqrt2}4$이므로, (17)에 의하여수열 $\displaystyle a_2,a_4,a_6,\cdots$는 첫째항 $\displaystyle \frac{\sqrt2}4$, 공비 $\displaystyle \frac14$인 등비수열. $\displaystyle (20)$

자연수 $\displaystyle n$에 대하여 삼각형 $\displaystyle \mathrm{OA}_n\mathrm{A}_{n+1}$의 넓이 $\displaystyle T_n$은$\displaystyle T_n=\frac12\cdot\overline{\mathrm{OA}_n}\cdot\overline{\mathrm{OA}_{n+1}}\cdot\sin(\angle\mathrm{A}_n\mathrm{OA}_{n+1})=\frac12a_n\cdot a_{n+1}\cdot\sin(\angle\mathrm{A}_n\mathrm{OA}_{n+1})$이므로, (15), (17)에 의하여$\displaystyle T_n=\frac12\cdot a_n\cdot\frac{\sqrt2}4a_n\cdot\sin30^\circ=\frac{\sqrt2}{16}a_n^2,\quad n:$홀수,$\displaystyle T_n=\frac12\cdot a_n\cdot\frac{\sqrt2}2a_n\cdot\sin15^\circ=\frac{\sqrt2}4\cdot\frac14a_n^2=\frac{\sqrt2}{16}a_n^2,\quad n:$짝수.[문제 2-2]의 결과인 (19), (20)에 의하여, 자연수 $\displaystyle n$에 대하여$\displaystyle a_{2n-1}=1\cdot\left(\frac14\right)^{n-1},\quad a_{2n}=\frac{\sqrt2}4\cdot\left(\frac14\right)^{n-1}$이므로, 자연수 $\displaystyle n$에 대하여$\displaystyle \begin{aligned}T_{2n-1}&=\frac{\sqrt2}{16}\cdot\left\{\left(\frac14\right)^{n-1}\right\}^2=\frac{\sqrt2}{16}\cdot\left(\frac1{16}\right)^{n-1},\\T_{2n}&=\frac{\sqrt2}{16}\cdot\left\{\frac{\sqrt2}4\left(\frac14\right)^{n-1}\right\}^2=\frac{\sqrt2}{128}\cdot\left(\frac1{16}\right)^{n-1}.\end{aligned}\qquad(39)$

(39)에 의하여 자연수 $\displaystyle n$에 대하여$\displaystyle \begin{aligned}\log_2 T_{2n-1}&=\log_2\left\{\frac{\sqrt2}{16}\left(\frac1{16}\right)^{n-1}\right\}=-4n+\frac12,\\\log_2 T_{2n}&=\log_2\left\{\frac{\sqrt2}{128}\left(\frac1{16}\right)^{n-1}\right\}=-4n-\frac52.\end{aligned}\qquad(40)$

따라서$\displaystyle \begin{aligned}\sum_{n=1}^{2N}\log_2 T_n&=\sum_{n=1}^N\log_2 T_{2n-1}+\sum_{n=1}^N\log_2 T_{2n}\\&=\sum_{n=1}^N\left(-4n+\frac12\right)+\sum_{n=1}^N\left(-4n-\frac52\right)=\sum_{n=1}^N(-8n-2)\\&=-8\cdot\frac12N(N+1)-2N=-4N^2-6N,\end{aligned}\quad N:$자연수. $\displaystyle (41)$