의예과 4번

치환적분법에 의해서 제시문 (ㄱ)의 함수 $\displaystyle f(x)$는 다음과 같다.$\displaystyle f(x)=2\int_0^x\tan\theta(\tan\theta)^{\prime}d\theta+1=\Biggl[\tan^2\theta\Biggr]_0^x+1=\tan^2x+1=\frac1{\cos^2x}$

따라서 제시문 (ㄴ)의 수열 $\displaystyle \{a_n\}$의 일반항은 다음과 같다.$\displaystyle a_n=\frac1{4^n\cos^2\left(\frac{\pi}{2^{n+2}}\right)}$

사인함수의 덧셈정리에 의해서$\displaystyle \begin{aligned}\frac1{\sin^2\left(\frac\pi{2^2}\right)}&=\frac1{4\sin^2\left(\frac\pi{2^{1+2}}\right)\times\cos^2\left(\frac\pi{2^{1+2}}\right)}\\&=\frac1{4\sin^2\left(\frac\pi{2^{1+2}}\right)}+\frac1{4\cos^2\left(\frac\pi{2^{1+2}}\right)}=\frac1{4\sin^2\left(\frac\pi{2^{1+2}}\right)}+a_1\end{aligned}$$\displaystyle \begin{aligned}\frac1{4\sin^2\left(\frac\pi{2^{1+2}}\right)}&=\frac1{4^2\sin^2\left(\frac\pi{2^{2+2}}\right)}+\frac1{4^2\cos^2\left(\frac\pi{2^{2+2}}\right)}\\&=\frac1{4^2\sin^2\left(\frac\pi{2^{2+2}}\right)}+a_2\end{aligned}$$\displaystyle \vdots$$\displaystyle \begin{aligned}\frac1{4^{n-2}\sin^2\left(\frac\pi{2^{n-2+2}}\right)}&=\frac1{4^{n-1}\sin^2\left(\frac\pi{2^{n-1+2}}\right)}+\frac1{4^{n-1}\cos^2\left(\frac\pi{2^{n-1+2}}\right)}\\&=\frac1{4^{n-1}\sin^2\left(\frac\pi{2^{n-1+2}}\right)}+a_{n-1}\end{aligned}$$\displaystyle \begin{aligned}\frac1{4^{n-1}\sin^2\left(\frac\pi{2^{n-1+2}}\right)}&=\frac1{4^n\sin^2\left(\frac\pi{2^{n+2}}\right)}+\frac1{4^n\cos^2\left(\frac\pi{2^{n+2}}\right)}\\&=\frac1{4^n\sin^2\left(\frac\pi{2^{n+2}}\right)}+a_n\end{aligned}$

모두 더하면$\displaystyle \sum_{k=1}^na_k=\frac1{\sin^2\left(\frac\pi{2^2}\right)}-\frac1{4^n\sin^2\left(\frac\pi{2^{n+2}}\right)}$

한편, $\displaystyle \lim_{x\to0}\frac{\sin x}x=1$이므로$\displaystyle \lim_{n\to\infty}\frac1{4^n\sin^2\left(\frac\pi{2^{n+2}}\right)}=\frac{16}{\pi^2}\lim_{n\to\infty}\left(\frac{\frac\pi{2^{n+2}}}{\sin\left(\frac\pi{2^{n+2}}\right)}\right)^2=\frac{16}{\pi^2}$

따라서 제시문 (ㄷ)의 $\displaystyle S$값은 다음과 같다.$\displaystyle S=\lim_{n\to\infty}\left(\sum_{k=1}^na_k\right)=\frac1{\sin^2\left(\frac\pi{2^2}\right)}-\lim_{n\to\infty}\frac1{4^n\sin^2\left(\frac\pi{2^{n+2}}\right)}=2-\frac{16}{\pi^2}$