ข้อสอบ PAT 1 - ตุลาคม 2558

$$\begin{gathered} \mathrm{สูตรอนุกรม} {\mathrm{S}}_{\infty }=\frac{{\mathrm{a}}_1}{1-\mathrm{r}} \mathrm{โดย} \left|\mathrm{r}\right|<1\to \boxed{\mathrm{a}} \\ \mathrm{จากโจทย์} {\mathrm{a}}_{\mathrm{n}} = \frac{\mathrm{n}·2^{3\mathrm{n}}}{3^{2\mathrm{n} + 1}} \\ \mathrm{n}=1 \mathrm{จะได้} \hspace{0.17em}{\mathrm{a}}_1=\frac{1·2^{3\left(1\right)}}{3^{2\left(1\right)+1}}=\frac{1·2^3}{3^3} \\ \mathrm{n}=2 \mathrm{จะได้} \hspace{0.17em}{\mathrm{a}}_2=\frac{2·2^{3\left(2\right)}}{3^{2\left(2\right)+1}}=\frac{2·2^6}{3^5} \\ \mathrm{n}=3 \mathrm{จะได้} \hspace{0.17em}{\mathrm{a}}_3=\frac{3·2^{3\left(3\right)}}{3^{2\left(3\right)+1}}=\frac{3·2^9}{3^7} \\ \mathrm{n}=4 \mathrm{จะได้} \hspace{0.17em}{\mathrm{a}}_4=\frac{4·2^{3\left(4\right)}}{3^{2\left(4\right)+1}}=\frac{4·2^{12}}{3^9} \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \sum _{\mathrm{n}=1}^{\infty }{\mathrm{a}}_{\mathrm{n}}={\mathrm{a}}_1+{\mathrm{a}}_2+{\mathrm{a}}_3+{\mathrm{a}}_4+... ; \mathrm{แทน} {\mathrm{a}}_1,{\mathrm{a}}_2,{\mathrm{a}}_3,{\mathrm{a}}_4 \\ =\frac{1·2^3}{3^3}+\frac{2·2^6}{3^5}+\frac{3·2^9}{3^7}+\frac{4·2^{12}}{3^9}+... \\ \mathbf{โดย} \sum _{\mathrm{n}=1}^{\infty }{\mathrm{a}}_{\mathrm{n}}={\mathrm{S}}_{\infty } ; \\ \mathrm{กำหนดให้} {\mathrm{S}}_{\infty }=\frac{1·2^3}{3^3}+\frac{2·2^6}{3^5}+\frac{3·2^9}{3^7}+\frac{4·2^{12}}{3^9}+...\to \boxed{1} \end{gathered}$$

$$\begin{gathered} \mathrm{นำ} \boxed{1}\times \frac{2^3}{3^2}; \hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \frac{2^3}{3^2}·{\mathrm{S}}_{\infty }=\frac{1·2^6}{3^5}+\frac{2·2^9}{3^7}+\frac{3·2^{12}}{3^9}+\frac{4·2^{15}}{3^{11}}+... \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \frac{8}{9}·{\mathrm{S}}_{\infty }=\frac{1·2^6}{3^5}+\frac{2·2^9}{3^7}+\frac{3·2^{12}}{3^9}+\frac{4·2^{15}}{3^{11}}+...\to \boxed{2} \\ \mathrm{นำ} \boxed{1}-\boxed{2} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \frac{1}{9}·{\mathrm{S}}_{\infty }=\frac{1·2^3}{3^3}+\frac{1·2^6}{3^5}+\frac{1·2^9}{3^7}+\frac{1·2^{12}}{3^9}+... \\ \mathrm{โดย} a_1=\frac{1·2^3}{3^3} , \mathrm{r}=\frac{2^3}{3^2} ; \\ \mathrm{จาก} \boxed{\mathrm{a}} \mathrm{จะได้} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \frac{1}{9}·{\mathrm{S}}_{\infty }=\frac{{\displaystyle \frac{1·2^3}{3^3}}}{1-{\displaystyle \frac{2^3}{3^2}}} \end{gathered}$$
$$\begin{gathered} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \frac{1}{9}·{\mathrm{S}}_{\infty }=\frac{{\displaystyle \frac{8}{27}}}{1-{\displaystyle \frac{8}{9}}} \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \frac{1}{9}·{\mathrm{S}}_{\infty }=\frac{{\displaystyle \frac{8}{27}}}{{\displaystyle \frac{1}{9}}} \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \frac{1}{9}·{\mathrm{S}}_{\infty }=\frac{8}{3} \end{gathered}$$

$$\begin{gathered} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em} {\mathrm{S}}_{\infty }=24 \\ \mathrm{จาก} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \sum _{\mathrm{n}=1}^{\infty }{\mathrm{a}}_{\mathrm{n}}={\mathrm{S}}_{\infty } \\ \mathrm{ดังนั้น}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \sum _{\mathrm{n}=1}^{\infty }{\mathrm{a}}_{\mathrm{n}}=24 \end{gathered}$$