ข้อสอบคณิต 9 สามัญ - ปี 2561

$\mathrm{จากโจทย์} \sum _{n=0}^{\infty }{\cos }^n(\frac{\mathrm{\pi }}{3}+n\mathrm{\pi })$    
$={\cos }^0(\frac{\mathrm{\pi }}{3}+0\mathrm{\pi })+{\cos }^1\left(\frac{\mathrm{\pi }}{3}+\mathrm{\pi }\right)+{\cos }^2\left(\frac{\mathrm{\pi }}{3}+2\mathrm{\pi }\right)+{\cos }^3\left(\frac{\mathrm{\pi }}{3}+3\mathrm{\pi }\right)+...$
$$\begin{gathered} =1+(-\cos \frac{\mathrm{\pi }}{3})+(\cos \frac{\mathrm{\pi }}{3}{)}^2+(-\cos \frac{\mathrm{\pi }}{3}{)}^3+... \\ =1+\left(\frac{-1}{2}\right)+{\left(\frac{1}{2}\right)}^2+{\left(\frac{-1}{2}\right)}^3+...\to \overline{)1} \end{gathered}$$

$$\begin{gathered} \mathrm{จากสูตร} S_{\infty }=\frac{a_1}{1-r} \mathrm{โดย} \left|r\right|<1 \\ \mathrm{โดย} a_1=1 , r=\frac{-1}{2} \end{gathered}$$

 $$\begin{gathered} \mathrm{จาก}\overline{)1} \mathrm{จะได้} \sum _{n=0}^{\infty }{\cos }^n(\frac{\mathrm{\pi }}{3}+n\mathrm{\pi }) =\frac{1}{1-\left({\displaystyle \frac{-1}{2}}\right)} \\ =\frac{1}{{\displaystyle \frac{3}{2}}} \\ =\frac{2}{3} \end{gathered}$$