자연계열 1-1번

[1.1] $\displaystyle n$을 $\displaystyle 4$로 나누었을 때의 나머지에 따라 분류해 보자.$\displaystyle n=4k-3$이면 $\displaystyle \sin^2\frac{(4k-3)\pi}4=\sin^2\frac\pi4=\frac12$$\displaystyle n=4k-2$이면 $\displaystyle \sin^2\frac{(4k-2)\pi}4=\sin^2\frac\pi2=1$$\displaystyle n=4k-1$이면 $\displaystyle \sin^2\frac{(4k-1)\pi}4=\sin^2\frac{3\pi}4=\frac12$$\displaystyle n=4k$이면 $\displaystyle \sin^2\frac{4k\pi}4=\sin^2\pi=0$이므로$\displaystyle \begin{aligned}\sum_{n=1}^{100}\left(-\frac23\right)^n\sin^2\frac{n\pi}4&=\sum_{k=1}^{25}\left\{\left(-\frac23\right)^{4k-3}\frac12+\left(-\frac23\right)^{4k-2}+\left(-\frac23\right)^{4k-1}\frac12\right\}\\&=\sum_{k=1}^{25}\left(-\frac{27}{16}+\frac94-\frac34\right)\left(-\frac23\right)^{4k}=-\frac3{16}\sum_{k=1}^{25}\left(\frac23\right)^{4k}\end{aligned}$이다.

따라서$\displaystyle \sum_{n=1}^{100}\left(-\frac23\right)^n\sin^2\frac{n\pi}4=-\frac3{16}\frac{\left(\frac23\right)^4\left\{1-\left(\frac23\right)^{100}\right\}}{1-\left(\frac23\right)^4}=\frac3{65}(A-1)$이다.