ข้อสอบ PAT 1 - พฤศจิกายน 2557

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em} \left\{{\mathrm{a}}_{\mathrm{n}}\right\} \mathrm{เป็นลำดับเลขคณิต} \mathrm{โดยที่} {\mathrm{a}}_1=2 \\ \mathrm{จากสูตร} {\mathrm{a}}_{\mathrm{n}}={\mathrm{a}}_1+\left(\mathrm{n}-1\right)\mathrm{d} ; \mathrm{แทน} {\mathrm{a}}_1=2 \\ {\mathrm{a}}_{\mathrm{n}}=2+\left(\mathrm{n}-1\right)\mathrm{d}\to \boxed{1} \\ \mathrm{n}=2 {\mathrm{a}}_2=2+1\mathrm{d} \\ \mathrm{n}=4 {\mathrm{a}}_4=2+3\mathrm{d} \\ \mathrm{n}=8 {\mathrm{a}}_8=2+7\mathrm{d} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em} {\mathrm{a}}_2,{\mathrm{a}}_4,{\mathrm{a}}_8 \mathrm{เรียงกันเป็นลำดับเรขาคณิต} \\ \mathrm{จะได้} \frac{{\mathrm{a}}_4}{{\mathrm{a}}_2}=\frac{{\mathrm{a}}_8}{{\mathrm{a}}_4} \\ {\left({\mathrm{a}}_4\right)}^2={\mathrm{a}}_2{\mathrm{a}}_8 ; \mathrm{แทนค่า} {\mathrm{a}}_2,{\mathrm{a}}_4,{\mathrm{a}}_8 \\ {\left(2+3\mathrm{d}\right)}^2=\left(2+\mathrm{d}\right)\left(2+7\mathrm{d}\right) \\ 4+12\mathrm{d}+9{\mathrm{d}}^2=4+16\mathrm{d}+7{\mathrm{d}}^2 \\ 2{\mathrm{d}}^2-4\mathrm{d}=0 \\ 2\mathrm{d}\left(\mathrm{d}-2\right)=0 \\ \mathrm{d}=0,2 \end{gathered}$$​

$$\begin{gathered} - \mathrm{d}=0 \mathrm{ไม่ได้} \mathrm{เพราะ} {\mathrm{a}}_1<{\mathrm{a}}_2<{\mathrm{a}}_3<... \\ \mathrm{แสดงว่า} \mathrm{d}=2 ; \mathrm{แทนใน} \boxed{1} \\ \mathrm{จะได้} {\mathrm{a}}_{\mathrm{n}}=2+\left(\mathrm{n}-1\right)2 \\ {\mathrm{a}}_{\mathrm{n}}=2\mathrm{n} \\ \mathrm{แทน} \mathrm{n}=2 {\mathrm{a}}_2=2\left(2\right)=4 \\ \mathrm{แทน} \mathrm{n}=4 {\mathrm{a}}_4=2\left(4\right)=8 \\ \mathrm{แทน} \mathrm{n}=8 {\mathrm{a}}_8=2\left(8\right)=16 \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \frac{({\mathrm{a}}_1-1{)}^3+({\mathrm{a}}_2-1{)}^3+...+({\mathrm{a}}_{\mathrm{n}}-1{)}^3}{{{\mathrm{a}}_1}^3+{{\mathrm{a}}_2}^3+...{{\mathrm{a}}_{\mathrm{n}}}^3}=\frac{391}{450} \\ -\mathrm{แทนค่า} {\mathrm{a}}_1,{\mathrm{a}}_2,{\mathrm{a}}_{\mathrm{n}} \\ \mathrm{จะได้} \frac{(2-1{)}^3+(4-1{)}^3+...+(2\mathrm{n}-1{)}^3}{{\left(2\right)}^3+{\left(4\right)}^3+...+{\left(2\mathrm{n}\right)}^3}=\frac{391}{450} \\ \frac{1^3+3^3+5^3+...+{\left(2\mathrm{n}-1\right)}^3}{2^3+4^3+6^3+...+{\left(2\mathrm{n}\right)}^3}=\frac{391}{450} \end{gathered}$$

$$\begin{gathered} \frac{\left[1^3+2^3+3^3+...+{\left(2\mathrm{n}\right)}^3\right]-\left[2^3+4^3+6^3+...+{\left(2\mathrm{n}\right)}^3\right]}{2^3+4^3+6^3+...+{\left(2\mathrm{n}\right)}^3}=\frac{391}{450} \\ \frac{\left[1^3+2^3+3^3+...+{\left(2\mathrm{n}\right)}^3\right]}{2^3+4^3+6^3+...+{\left(2\mathrm{n}\right)}^3}-1=\frac{391}{450} \\ \frac{\left[1^3+2^3+3^3+...+{\left(2\mathrm{n}\right)}^3\right]}{{\left(2·1\right)}^3+{\left(2·2\right)}^3+{\left(2·3\right)}^3+...+{\left(2·\mathrm{n}\right)}^3}-1=\frac{391}{450} \end{gathered}$$

$$\begin{gathered} \frac{\left[1^3+2^3+3^3+...+{\left(2\mathrm{n}\right)}^3\right]}{2^3\left(1^3+2^3+3^3+...+{\mathrm{n}}^3\right)}-1=\frac{391}{450} \\ \frac{{\left[{\displaystyle \frac{\left(2\mathrm{n}\right)\left(2\mathrm{n}+1\right)}{2}}\right]}^2}{2^3{\left[{\displaystyle \frac{\left(\mathrm{n}\right)\left(\mathrm{n}+1\right)}{2}}\right]}^2}-1=\frac{391}{450} \\ \frac{{\left(2\mathrm{n}+1\right)}^2}{2{\left(\mathrm{n}+1\right)}^2}-1=\frac{391}{450} \\ {\left(\frac{2\mathrm{n}+1}{\mathrm{n}+1}\right)}^2=\frac{841}{225} \\ \frac{2\mathrm{n}+1}{\mathrm{n}+1}=\pm \sqrt{\frac{841}{225}} \end{gathered}$$

$$\begin{gathered} \mathrm{n} \mathrm{เป็นจำนวนเต็มบวกเสมอ} ; \frac{2\mathrm{n}+1}{\mathrm{n}+1}=\frac{29}{15} \\ 30\mathrm{n}+15=29\mathrm{n}+29 \\ \mathrm{n}=14 \\ \mathrm{ดังนั้น} \mathrm{ค่าของ} \mathrm{n}=14 \end{gathered}$$