ข้อสอบ PAT 1 - มีนาคม 2559

$$\begin{gathered} \mathrm{จากโจทย์} a_n = \frac{2}{4n^2-1}-{\left(-\frac{1}{3}\right)}^n \\ \mathrm{จะได้} \sum _{n=1}^{\infty }{a}_n = \sum _{n=1}^{\infty }\left(\frac{2}{4n^2-1}-{\left(-\frac{1}{3}\right)}^n\right) \\ = \sum _{n=1}^{\infty }\frac{2}{4n^2-1}-\sum _{n=1}^{\infty }{\left(-\frac{1}{3}\right)}^n \\ = \sum _{n=1}^{\infty }\frac{2}{\left(2n-1\right)\left(2n+1\right)}-\sum _{n=1}^{\infty }{\left(-\frac{1}{3}\right)}^n \end{gathered}$$
$$\begin{gathered} = \sum _{n=1}^{\infty }\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)-\sum _{n=1}^{\infty }{\left(-\frac{1}{3}\right)}^n \\ = \left(\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\dots \right) \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}-\left(\left(-\frac{1}{3}\right)+{\left(-\frac{1}{3}\right)}^2+{\left(-\frac{1}{3}\right)}^3+\dots \right) \\ = \frac{1}{1}-\left(\left(-\frac{1}{3}\right)+{\left(-\frac{1}{3}\right)}^2+{\left(-\frac{1}{3}\right)}^3+\dots \right) \end{gathered}$$

$$\begin{gathered} \mathrm{จากสูตร} S_{\infty } = \frac{a_1}{1-r} \\ \mathrm{จะได้} \sum _{n=1}^{\infty }{a}_n = 1-\left(\frac{-{\displaystyle \frac{1}{3}}}{1-\left(-{\displaystyle \frac{1}{3}}\right)}\right) \\ = \frac{5}{4} \\ \mathbf{ดังนั้น}\mathbf{ }\mathbf{ตอบ}\mathbf{ }\mathbf{1}\mathbf{)}\mathbf{ }\mathbf{อนุกรมลู่เข้าและมีผลบวกเท่ากับ}\mathbf{ }\frac{\mathbf{5}}{\mathbf{4}} \end{gathered}$$