ข้อสอบ PAT 1 - มีนาคม 2556

$$\begin{gathered} \mathrm{จากสูตร} \mathrm{อนุกรมอนันต์ของเรขาคณิต} {\mathrm{S}}_{\infty }=\frac{{\mathrm{a}}_1}{1-\mathrm{r}}\to \boxed{\mathrm{A}} \\ \mathrm{จากโจทย์} \\ \frac{{\mathrm{a}}_1+{\mathrm{a}}_3}{{\mathrm{a}}_2+{\mathrm{a}}_4}+\frac{{\mathrm{a}}_3+{\mathrm{a}}_5}{{\mathrm{a}}_4+{\mathrm{a}}_6}+\frac{{\mathrm{a}}_5+{\mathrm{a}}_7}{{\mathrm{a}}_6+{\mathrm{a}}_8}+...+\frac{{\mathrm{a}}_{2011}+{\mathrm{a}}_{2013}}{{\mathrm{a}}_{2012}+{\mathrm{a}}_{2014}}=2012\to \boxed{1} \end{gathered}$$
$$\begin{gathered} \mathbf{พิจารณา} \frac{{\mathrm{a}}_1+{\mathrm{a}}_3}{{\mathrm{a}}_2+{\mathrm{a}}_4}=\frac{{\mathrm{a}}_1+{\mathrm{a}}_1·{\mathrm{r}}^2}{{\mathrm{a}}_1·\mathrm{r}+{\mathrm{a}}_1·{\mathrm{r}}^3}=\frac{\bcancel{{\mathrm{a}}_1}\cancel{\left[1+{\mathrm{r}}^2\right]}}{\bcancel{{\mathrm{a}}_1}·\mathrm{r}\cancel{\left[1+{\mathrm{r}}^2\right]}}=\frac{1}{\mathrm{r}} \\ \hspace{0.17em}\hspace{0.17em} \frac{{\mathrm{a}}_3+{\mathrm{a}}_5}{{\mathrm{a}}_4+{\mathrm{a}}_6}=\frac{{\mathrm{a}}_1·{\mathrm{r}}^2+{\mathrm{a}}_1·{\mathrm{r}}^4}{{\mathrm{a}}_1·{\mathrm{r}}^3+{\mathrm{a}}_1·{\mathrm{r}}^5}=\frac{\bcancel{{\mathrm{a}}_1}·{\mathrm{r}}^2\cancel{\left[1+{\mathrm{r}}^2\right]}}{\bcancel{{\mathrm{a}}_1}·{\mathrm{r}}^3\cancel{\left[1+{\mathrm{r}}^2\right]}}=\frac{1}{\mathrm{r}} \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{1} \mathrm{จะได้} \frac{1}{\mathrm{r}}+\frac{1}{\mathrm{r}}+\frac{1}{\mathrm{r}}+...+\frac{1}{\mathrm{r}}=2012\to \boxed{2} \\ \mathrm{จำนวนตัวของ} \frac{1}{\mathrm{r}}=\frac{2012}{2}=1006 \mathrm{ตัว} \\ \mathrm{จาก} \boxed{2} \mathrm{จะได้} 1006\left(\frac{1}{\mathrm{r}}\right)=2012 \\ \mathrm{r}=\frac{1}{2} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \mathrm{ค่าของ} 1+5\mathrm{r}+12{\mathrm{r}}^2+22{\mathrm{r}}^3+... \\ \mathrm{กำหนดให้} {\mathrm{S}}_{\infty }=\mathrm{X} \\ \mathrm{จะได้} \mathrm{X}=1+5\mathrm{r}+12{\mathrm{r}}^2+22{\mathrm{r}}^3+...\to \boxed{3} \end{gathered}$$

$$\begin{gathered} \boxed{3} \mathrm{คูณ} \mathrm{r} \mathrm{ตลอดจะได้} \\ \mathrm{rX}=\mathrm{r}+5{\mathrm{r}}^2+12{\mathrm{r}}^3+22{\mathrm{r}}^4+...\to \boxed{4} \\ \boxed{3}-\boxed{4} \mathrm{จะได้} \\ \mathrm{X}-\mathrm{rX}=1+4\mathrm{r}+7{\mathrm{r}}^2+10{\mathrm{r}}^3+...\to \boxed{5} \end{gathered}$$

$$\begin{gathered} \boxed{5} \mathrm{คูณ} \mathrm{r} \mathrm{ตลอดจะได้} \\ \mathrm{r}\mathrm{X}-{\mathrm{r}}^2\mathrm{X}=\mathrm{r}+4{\mathrm{r}}^2+7{\mathrm{r}}^3+10{\mathrm{r}}^4+...\to \boxed{6} \\ \boxed{5}-\boxed{6} \mathrm{จะได้} \mathrm{X}-2\mathrm{rX}+{\mathrm{r}}^2\mathrm{X}=1+3\mathrm{r}+3{\mathrm{r}}^2+3{\mathrm{r}}^3+... \\ \mathrm{X}-2\mathrm{rX}+{\mathrm{r}}^2\mathrm{X}=1+3[\mathrm{r}+{\mathrm{r}}^2+{\mathrm{r}}^3+...] \\ \mathrm{แทนค่า} \mathrm{r} \mathrm{ในสมการ} \end{gathered}$$

$\mathrm{จะได้} \mathrm{X}-\cancel{2}\left(\frac{1}{\cancel{2}}\right)\mathrm{X}+{\left(\frac{1}{2}\right)}^2\mathrm{X}=1+3\left[\frac{1}{2}+{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^3+...\right]$
$$\begin{gathered} \mathrm{จาก} \boxed{\mathrm{A}} \mathrm{จะได้} \frac{1}{4}\mathrm{X}=1+3\left[\frac{\left({\displaystyle \frac{1}{2}}\right)}{1-{\displaystyle \frac{1}{2}}}\right] \\ \frac{1}{4}\mathrm{X}=1+3\left(1\right) \\ \mathrm{X}=16 \\ {\mathrm{S}}_{\infty }=16 \\ \mathrm{ดังนั้น} \mathrm{ค่าของ} 1+5\mathrm{r}+12{\mathrm{r}}^2+22{\mathrm{r}}^3+...\mathrm{เท่ากับ} 16 \end{gathered}$$