ข้อสอบ PAT 1 - กุมภาพันธ์ 2563

$$\begin{gathered} \mathrm{จากโจทย์} \\ \mathrm{สังเกตว่าแต่ละพจน์} \mathrm{จะมีเศษส่วนน้อยกว่า} 1 \mathrm{เสมอ} \\ \mathrm{จะได้} \frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\dots \\ = \frac{2-1}{2}+\frac{4-1}{4}+\frac{8-1}{8}+\frac{16-1}{16}+\dots \\ = 1-\frac{1}{2}+1-\frac{1}{4}+1-\frac{1}{8}+1-\frac{1}{16}+\dots \\ = \left(1+1+1+1+\dots \right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\dots \right) \end{gathered}$$

$$\begin{gathered} \mathrm{บวกกัน} n \mathrm{ตัว} \mathrm{จะได้} S_n = n-\frac{{\displaystyle \frac{1}{2}\left(1-\frac{1}{2^n}\right)}}{1-{\displaystyle \frac{1}{2}}} \\ = n-1+\frac{1}{2^n} \end{gathered}$$

$$\begin{gathered} \mathrm{ดังนั้น} \underset{n\to \infty }{\lim }\frac{S_n}{S_{2n}} = \underset{n\to \infty }{\lim }\frac{n-1+{\displaystyle \frac{1}{2^n}}}{2n-1+{\displaystyle \frac{1}{2^n}}} \\ = \frac{\mathrm{สปส}. n}{\mathrm{สปส}. 2n} = \frac{1}{2} \\ \mathrm{ตอบ} 4 \end{gathered}$$