ข้อสอบคณิต 9 สามัญ - ปี 2563

$$\begin{gathered} \mathrm{จากโจทย์} \\ \sum _{n=0}^{\infty }{\left(\sqrt{2}\sin \frac{\mathrm{\pi }}{12}\right)}^{2n} = 1+{\left(\sqrt{2}\sin \frac{\mathrm{\pi }}{12}\right)}^2+{\left(\sqrt{2}\sin \frac{\mathrm{\pi }}{12}\right)}^4+{\left(\sqrt{2}\sin \frac{\mathrm{\pi }}{12}\right)}^6+\dots \end{gathered}$$

$$\begin{gathered} \mathrm{จะเห็นว่า} \mathrm{เป็นอนุกรมเรขาคณิตอนันต์ที่มี} \\ a_1 = 1 \mathrm{และ} r = {\left(\sqrt{2}\sin \frac{\mathrm{\pi }}{12}\right)}^2 \\ \mathrm{ต้องตรวจสอบว่า} \left|r\right| < 1 \mathrm{หรือไม่ก่อนใช้สูตร} \\ S_{\infty } = \frac{a_1}{1-r} \mathrm{เมื่อ} \left|r\right| < 1 \end{gathered}$$

$$\begin{gathered} \mathrm{จะเห็นว่า} 0 \le {\left(\sqrt{2}\sin \frac{\mathrm{\pi }}{12}\right)}^2 < {\left(\sqrt{2}\sin \frac{\mathrm{\pi }}{4}\right)}^2 = {\left(\sqrt{2}·\frac{\sqrt{2}}{2}\right)}^2 = 1 \\ 0 \le r < 1 \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} S_{\infty } = \frac{1}{1-{\left(\sqrt{2}\sin {\displaystyle \frac{\mathrm{\pi }}{12}}\right)}^2} \\ = \frac{1}{1-2{\sin }^2{\displaystyle \frac{\mathrm{\pi }}{12}}} \\ = \frac{1}{\cos \left(2·{\displaystyle \frac{\mathrm{\pi }}{12}}\right)} \end{gathered}$$

$$\begin{gathered} = \frac{1}{\cos {\displaystyle \frac{\mathrm{\pi }}{6}}} = \frac{1}{{\displaystyle \frac{\sqrt{3}}{2}}} = \frac{2\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}} = \frac{2\sqrt{3}}{3} \\ \mathrm{ดังนั้น} \mathrm{ตอบ} 2 \end{gathered}$$