자연계열 1번

(1)① $\displaystyle N(x)=\left(\frac12\right)^x\times\sin(\theta x)\times\sin\left(\frac\pi2-\theta x\right)=\left(\frac12\right)^x\times\sin(\theta x)\times\cos(\theta x)=\left(\frac12\right)^{x+1}\times\sin(2\theta x)$② $\displaystyle \lim_{x\to0}\frac{N(x)}x=\frac12\times\lim_{x\to0}\left(\frac12\right)^x\times\lim_{x\to0}\frac{\sin2\theta x}x=\frac12\times\lim_{x\to0}\frac{\sin2\theta x}x$③ $\displaystyle t=2\theta x$로 놓으면 $\displaystyle x\to0$일 때 $\displaystyle t\to0$이므로④ $\displaystyle \lim_{x\to0}\frac{\sin2\theta x}x=\lim_{x\to0}2\theta\times\frac{\sin2\theta x}{2\theta x}=2\theta\times\lim_{t\to0}\frac{\sin t}t=2\theta$⑤ $\displaystyle \lim_{x\to0}\frac{N(x)}x=\frac12\times2\theta=\theta$

(2)① $\displaystyle \theta=\frac\pi4$일 때, $\displaystyle N(x)=\left(\frac12\right)^{x+1}\times\sin\left(\frac\pi2x\right)$② $\displaystyle \sin\left(\frac\pi2x\right)$는 $\displaystyle x$가 1부터 1씩 증가함에 따라 $\displaystyle 1,0,-1,0,1,0,\cdots$로 나타난다③ 따라서 $\displaystyle \sum_{n=1}^{\infty}|N(n)|=\left(\frac12\right)^2+0+\left(\frac12\right)^4+0+\left(\frac12\right)^6+\cdots$④ 주어진 급수는 첫째항이 $\displaystyle \frac14$이고 공비가 $\displaystyle \frac14$인 무한등비급수이다.⑤ $\displaystyle \sum_{n=1}^{\infty}|N(n)|=\sum_{n=1}^{\infty}\frac14\left(\frac14\right)^{n-1}=\frac{\frac14}{1-\frac14}=\frac13$

(3)① $\displaystyle N(x)=\left(\frac12\right)^{x+1}\times\sin(2\theta x)$에서$\displaystyle \left(\frac12\right)^{x+1}+\cos(2\theta x)\times N(x)=\left(\frac12\right)^{x+1}+\cos(2\theta x)\times\left(\frac12\right)^{x+1}\times\sin(2\theta x)=\left(\frac12\right)^{x+1}\left(1+\frac12\sin(4\theta x)\right)=\left(\frac12\right)^{x+2}(2+\sin(4\theta x))$② $\displaystyle \log_2\left(\left(\frac12\right)^{x+1}+\cos(2\theta x)\times N(x)\right)=\log_2\left(\frac12\right)^{x+2}(2+\sin(4\theta x))=-(x+2)+\log_2(2+\sin(4\theta x))$③ $\displaystyle \theta=\frac\pi8$일 때, $\displaystyle 2+\sin(4\theta x)=2+\sin\left(\frac\pi2x\right)$이므로$\displaystyle \sum_{n=1}^{20}\left(-(n+2)+\log_2\left(2+\sin\left(\frac\pi2n\right)\right)\right)=\sum_{n=1}^{20}(-(n+2))+\sum_{n=1}^{20}\log_2\left(2+\sin\left(\frac\pi2n\right)\right)=-\frac{20\times21}2-2\times20+\sum_{n=1}^{20}\log_2\left(2+\sin\left(\frac\pi2n\right)\right)$④ $\displaystyle 2+\sin\left(\frac\pi2n\right)$은 $\displaystyle n$이 1부터 1씩 증가함에 따라 $\displaystyle 3,2,1,2,3,2,1,2\ldots$로 나타나므로$\displaystyle \sum_{n=1}^{20}\log_2\left(2+\sin\left(\frac\pi2n\right)\right)=(\log_23+1+0+1)\times5=5\log_23+10$이다.⑤ 따라서, 구하는 식의 값은 $\displaystyle 5\log_23-240$이다.